ARAB ACADEMY FOR SCIENCE TECHNOLOGY AND MARITIME TRANSPORT

ARAB ACADEMY FOR SCIENCE TECHNOLOGY AND MARITIME TRANSPORT Course: Math For Engineering Winter 8 Lecture Notes By Dr. Mostafa Elogail

Page Lecture [ Functions / Graphs of Rational Functions] Functions Definition: A function f from a set A to a set B is a rule that assigns to each element x in A one an only one element y = f(x) in B. The set A is calle the omain of the function, an the set of assigne elements in B is calle the range. Graphically: The vertical Line Test A curve in the xy plane is the graph of a function of x if an only if no vertical line intersects the curve more than once. Domain an Range Domain(f) = the set of all possible input values x that we can substitute in the given function f(x). Domain(f) =the projection of the curve of f on the x-axis. Range(f) = the set of all output values y we obtaine by substituting the omain values x in the function f(x). Range(f) = the projection of the curve of f on the y-axis. x is calle the inepenent variable. y is calle the epenent variable.

Page Symmetry Even Function If a function f satisfies f( x) = f(x) for every number x In its omain, then f is calle an even function. ( graph of f is symmetric with respect to the y-axis ) O Function If a function f satisfies f( x) = f(x) for every number x In its omain, then f is calle an o function. ( graph of f is symmetric about the origin ) Example: Fin the omain an Range of; (a) f(x) = x +, (b) f(x) = x+ x 4, (c) f(x) = x Note that: The omain of a function is restricte to: ) You can not have zero in a enominator ) You can not take square root (or fourth root, sixth root, ) of a negative number. (a) Domain(f) is x ], [, Range(f) is y [, [. (b) We must have that x+ an x 4 x 4 Domain(f) is x ], ] ] 4, [, Range(f) is y [, [ ], [. (c) We must have that x x x x Domain(f) is x [, ] Range(f) is y [, ].

Polynomial Functions Page 3 A function of the form f(x) = a n x n + a n x n + + a x + a, where a, a,, a n are real numbers ( the coefficients of the polynomial with a n ) an n is an integer, is a polynomial function of egree n. Examples: Constant function f(x) = 4 (polynomial of egree zero), Linear function f(x) = 6x + (polynomial of egree one), Quaratic function(parabolas) f(x) = x + 7x (polynomial of egree two). Note: The omain of any polynomial is R = ], [. Graphs of some special polynomials Rational Functions The quotient of two polynomials f(x) = P(x) Q(x) ; such that P(x)an Q(x) are polynomials is calle a rational function. The omain of a rational function is R {Q(x) = }. Example: a) f(x) = Domain(f) is R {}. x b) f(x) = 7x6 + x 3 + 3x x Domain(f) is R {, }.

Asymptotes -Vertical Asymptotes Page 4 If at least one of the following statements is true, then the line x = c is a vertical asymptote of the graph of y = f(x) (a) lim x c f(x) =, (b) lim x c + f(x) =, (c) lim x c f(x) =, () lim x c + f(x) = For a rational function f(x) = P(x), where P(x) an Q(x) are polynomials. If c is a real number Q(x) such that Q(c) = an P(c), then the graph of f has a vertical asymptote at x = c. - Horizontal Asymptotes If either (a) lim x f(x) = L or (b) lim x f(x) = L Then the line y = L is a horizontal asymptote of the graph of y = f(x).

3- Slant Asymptote For a rational function Page 5 For a rational function f(x) = P(x), where the egree of P(x) is greater than the egree of Q(x), Q(x) then using long ivision to obtain f(x) = P(x) r(x) = (x) + Q(x) Q(x) It is clear that r(x) as x ±, so that f(x) is asymptotic to the graph of the Q(x) polynomial (x) as x ±. Asymptotes of Graphs of Rational Functions Let f(x) = P(x) Q(x) = a nx n +a n x n + +a x+a b m x m +b m x m + +b x+b ; a n, b m be a rational function. Suppose that the polynomial functions P(x) an Q(x) have no common factors. - If a is real zero of Q(x), then x = a is a vertical asymptote for the graph of f. - If n = m, then y = a n b m is a horizontal asymptote for the graph of f. 3- If n < m, then y = is a horizontal asymptote for the graph of f. 4- If n > m, then the graph of f has no horizontal asymptote. 5- If n = m +, then the quotient y = mx + b of P(x) an Q(x) is a slant asymptote for the graph of f. 6- Determine whether the graph will intersect its nonvertical asymptote y = b or y = mx + b by solving f(x) = b or f(x) = mx + b.

Page 6 Example: Analyze an sketch the rational function f(x) = x x - We first observe that the numerator P(x) = x an the enominator Q(x) = x have no common factors. - Also since f( x) = f(x) f(x) is o function (graph of f is symmetric with respect to the origin. 3- Since f() = implies the y intercept is (, ). 4- Solving P(x) = x = implies x = an so the only intercept is (, ). Since the egree of the numerator x is an the egree of the enominator x is an ( < ) it follows that y = is a horizontal asymptote. 5- Zeroes of the enominator Q(x) = x are x = ±. Therefore the lines x = an x = are vertical asymptotes. lim x lim x x = lim x (x )(x+) x + = (x )(x+) x (x )(x+) = lim x + x (x )(x+) =

Page 7 Example: Analyze an sketch the rational function f(x) = x x 6 x 5 - Note that the numerator P(x) = x x 6 an the enominator Q(x) = x 5 have no common factors. - f(x) is neither even nor o. 3- Since f() = 6 5 implies the y intercept is (, 6 5 ). 4- Solving P(x) = (x + )(x 3) = implies x = an x = 3, so the x intercepts are (, ) an (3, ). 5- The zeros of Q(x) = x 5 is obviously x = 5. So x = 5 is a vertical asymptote. lim x 5 x = lim x (x )(x+) x 5 + = (x )(x+) 6- Since egree of P(x) = > egree of Q(x) =. Hence f(x) has no horizontal asymptote. But it has an oblique asymptote. To fin it we use long ivision or synthetic ivision x x 6 x 5 = (x + 4) + 4 x 5. f has a slant asymptote y = x + 4. Type equation here.

Page 8 Example: Analyze an sketch the rational function f(x) = (x 9) x 4 - Note that the numerator P(x) = (x 9) an the enominator Q(x) = x 4 have no common factors. - Since f( x) = f(x), f(x)is an even function (graph of f is symmetric with respect to the y axis. 3- Since f() = 9 implies the y intercept is (, 9 ). 4- Solving P(x) = (x 9) = implies x = 3 an x = 3, so the x intercepts are ( 3, ) an (3, ). 5- The zeros of Q(x) = x 4 are obviously x = an x = So the lines x = an x = are vertical asymptotes. lim (x 3)(x+3) x = lim (x 3)(x+3) (x )(x+) x + = (x )(x+) lim (x 3)(x+3) x = lim (x 3)(x+3) (x )(x+) x + = (x )(x+) 6- Since egree of P(x) = equal egree of Q(x) =. Hence f(x) has a horizontal asymptote which is y = y = is a horizontal asymptote. Type equation here.

Page 9 Tutorial Exercise. Page 6,3,5,7,9,,3,5,9, 3, 5,7,9,3,39 48,59 74. Exercise.- Page 4 9,,3,7,9,. Exercise.5- Page 3,3,5,9,,,3,5,6,9,3. Assignment () () Express the given limit as a number, as, or as. a) lim x 5 c) lim x 4 + x 5 b) lim x ( x 5 ) 4 ( x +4 ) 4 ) lim x x 4 e) lim x x 3x 4x + 5 f) lim x x 4x + () Graph the rational functions. Inclue the graphs an equations of the asymptotes a) f(x) = x x b) f(x) = x (x ) c) f(x) = x x + e) f(x) = x ( x ) (x 3). ) f(x) = x + x 8 x 4 (3) Fin the set of values of k for which the line x = y + intersects the curve y = x + 3x + k at two istinct points. (4) Fin the omain of the given function f. a) f(x) = x x x + b) f(x) = 3 x x +. (5) Fin the x an y intercepts, if any, of the function f. a) f(x) = x 4 5x + 4 b) f(x) = 4 x. (6) ABC is a right-angle triangle with AB perpenicular to BC. The coorinates of the points A, B an C are ( 5,), (x, 5) an (3, ) respectively. Fin the possible values of x.

Lecture [ Inverse Functions / Transformation of Graphs ] Composition of functions Page Definition: If f an g are functions, the composite function (f g) ( f compose with g ) is efine by (f g)(x) = f(g(x)). The omain of (f g) consists of the numbers x in the omain of g for which g(x) lies in the omain of f. Example: If f(x) = x an g(x) = x +, fin (a) (f g)(x) (b) (g f)(x) (c) (f f)(x) () (g g)(x) Composite Domain (f g)(x) = f(g(x)) = g(x) = x + [, [ (g f)(x) = g(f(x)) = f(x) + = x + [, [ (f f)(x) = f(f(x)) = f(x) = x = x 4 [, [ (g g)(x) = g(g(x)) = g(x) + = x + ], [ Note that: the functions g f an f g are usually quite ifferent.

Definition: - Functions Page A function f is sai to be one to one if each y value in its range correspons to only one x value in its omain. Horizontal line Test That is f(x ) f(x ) whenever x x Or equivalently f(x ) = f(x ) then x = x A function is one to one if an only if no horizontal line intersects its graph more than once. f(x) = x 3 is one to one f(x) = x is not one to one Inverse functions: If y = f(x) is - function, then there exist a unique inverse function f (x) with the following properties Properties of the inverse functions - The omain of f = range of f. - The range of f = omain of f. 3- y = f(x) is equivalent to x = f(y). 4- The graph of f an f are reflections in the line y = x. 5- f(f (x)) = x for x in the omain of f. 6- f (f(x)) = x for x in the omain of f. CAUTION The symbol f (x) oes not mean f(x).

To get f (x) Page Algebrically a) Let y = f(x), b) Write x in terms of y, c) Interchange x an y, ) Replace y by f (x). Geometrically Reflecting the graph of y = f(x) in the line y = x. Example: Fin the inverse function f (x) when f(x) = 5x + 4, x [, [. Fin the omain an range of the inverse function. We write, y = 5x + 4 x = y 4 5 x = + y 4 5 Interchanging x an y y = x 4 5 Replace y by f (x) f (x) = x 4 5 The omain of f(x) is x [, [ an the range of f(x) is y [ 4, [ So, the omain of f (x) is x [ 4, [ an the range of f (x) is y [, [.

Page 3 Example: Fin the inverse function an its omain an range if f(x) = 4 x 3, for x R {3}. Make x the subject of y = 4 x 3 y (x 3) = 4 xy = 3y + 4 x = 3y + 4 y Interchanging x an y an replace y by f (x) y = 3x + 4 x f (x) = 3x + 4 x f (x) exists for all real values except x =, So, the omain of f (x) is x R {}. The range of f (x) is the omain of f(x) giving y R {3}. Transformations of graphs of functions y = f(x) ± a translate y = f(x) parallel to the y axis by ±a units y = f(x ± a) translate y = f(x) parallel to the x axis by a units y = f(x) reflect y = f(x) in the x axis y = f( x ) reflect y = f(x) in the y axis y = a f(x) stretch y = f(x) parallel to the y axis by a scale factor of a y = f( ax ) stretch y = f(x) parallel to the x axis by a scale factor of a

Page 4 Example: Describe the transformations neee to transform the graph of y = x into the graph of y = 4 x. First, a reflection in x axis followe by a translation 4 units upwars parallel to the y axis. Example: Complete the square an explain how to transform the graph of y = x into the graph of y = 3x 6x +. y = 3x 6x + Completing the square y = 3[ x x ] + y = 3[ (x ) ] + y = 3(x ) This is the graph of y = x translate unit to the right, stretche parallel to the y axis by a scale factor 3 then translate parallel to the y axis by unit. Example: Sketch the graph of f(x) = x + 5. This is the graph of f(x) = x translate parallel to the x axis by 5, then stretche parallel to the y axis by a scale factor of.

Tutorial Page 5 Exercise.3 Page 9,3,4,3,4,7,9,,3 6,37 4,44,45,46. Exercise.6 Page 59,,5,6,7,9,,,3,7,9,3 3, 39 43,45,47. Assignment () () Fin the inverse for each of the following functions: a) f(x) = x, x R, x (b) f(x) = c) f(x) = x 3, x R, x 3. () Use the metho of completing the square to sketch: x + 5 a) y = x + x + 4 (b) y = 6 + x x., x R, x 5 (3) The given function f is one-to-one. Without fining the inverse, fin the omain an range of f. a) f(x) = 3 + x (b) f(x) = x x 4. (4) Let f(x) = 5 x an g(x) = x. Fin the omain of the given function. a) f(g(x)) (b) g(f(x)). (5) Express F(x) = 9x 3 + 4 as the composition of two functions f an g. (6) Use the graph of y = f(x) to match the function with its graph. (i) y = f(x + 5) (j) y = f(x) 5 (k) y = f( x) (l) y = f(x 4) (m) y = f(x + 6) + (n) y = f(x ) + 3 (7) Challenge For the given function f(x) = x +, x [, [. Fin x such that f(x) = f (x).

Page 6 Lecture 3 [ The Derivative ] Tangent lines an rates of change Consier the curve y = f(x) an a secant line intersecting the curve at the points P(a, f(a)) an Q(a + h, f(a + h)) The slope of the secant line PQ is f(a + h) f(a) m sec = h = The average rate of change of f on the interval [a, a + h] As h approaches zero (h ) m tan = lim h f(a+h) f(a) h (provie existence) ( ) = The slope of the tangent line at x = a = The instantaneous rate of change of f at x = a = f (a) (the erivative of y = f(x) at x = a) Equation of the tangent line through (a, f(a)) with slope m tan = f (a) is y f(a) = f (a) (x a) Replacing a by x in the expression ( ) gives the efinition of the erivative. The erivative f (x) = lim h f(x + h) f(x) h f (x), when it exists, is the slope of the tangent line (an the instantaneous rate of change) at the variable point (x, f(x)).

The most common notations use to enote the erivative of y = f(x) Page 7 f (x), y, f, x x x (f(x)), y (x), D x (f(x)). Example: Let y = f(x) = x ; () Compute y x () Fin an equation of the line tangent to the graph of f at (4, ). () f (x) = lim h f( x + h ) f(x) h = lim h x + h x h x + h x = lim h h ( x + h + x ). ( x + h + x ) = lim h x + h x h ( x+h + x ) = lim h h h ( x + h + x ) = lim h x + h + x = =. x + x x () The slope of the tangent line at (4, ) is y x = x=4 4 = 4 Therefore, an equation of the tangent line is y = 4 (x 4) or y = 4 x + Rules of ifferentiation x C =, C is any real number. x xr = r x r, for any real number r. x C f(x) = C f (x) x [f(x) ± g(x)] = f (x) ± g (x) x [f(x) g(x)] = f(x) g (x) ± g(x) f (x) x [ f(x) g(x) ] = g(x) f (x) f(x) g (x) [ g(x) ] ; g(x).

Example: Page 8 x 5 = x x = x x = x x = x = x x 3 = 3 x 5 x 5x = x = x x x ( x ) = x x = ( ) x = x x ( x ) = x 3 x ( x4 + 3x 3 x ) = x 3 + 9x x [(3x 5x)(x 3 + x + 9)] = (3x 5x)(6x + 4x) + (x 3 + x + 9)(6x 5) x ( 3 ( x + 4 ) ( x x + 4 ) = x ) ( 3 x )() ( x + 4 ) The Chain Rule Let u = g(x) an y = f(u) The erivative of the composite function y = f( g(x) ) can be expresse as y = y u x u x (Version ) OR x f(g(x)) = f (g(x)) g (x) (Version )

Example: Fin y. If y = (x 3 + x ) 5 Let u = x 3 + x y = u 5 an y = y u = x u x 5u4 u = 5(x 3 + x ) (3x + ). Page 9 Example: Fin y. If y = 3x x Let u = 3x x y = u an y = y u = x u x u u = u u = 3x x (3 x). The General Power Rule If y = [ u(x) ] n Then y x = n [ u(x) ]n u (x). Example: Fin y. If y = ( + 4x 4x )3 y = 3 ( + 4x 4x ) ( 4x)(4) (+4x)( 4) ( ). ( 4x ) Repeate use of the Chain Rule Example: Fin y. If y = ( x 3 + 4 3x 4 3 ). y = 3 ( x 3 + 4 3x 4 ) ( 3x x 3 + 4 x3 ). Example: Fin y. If y = + + x. y = + + x ( ). ( + x x ).

Higher Orer Derivatives Page Higher erivatives are obtaine by repeately ifferentiating a function y = f(x). If f (x) is ifferentiable, then the secon erivative, enote f or y, is the erivative f (x) = x (f (x)) The secon erivative is the rate of change of f (x). The process of ifferentiation can be continue, provie that the erivatives exist. The thir erivative, enote f (x) is the erivative of f (x). More generally the n th erivative f (n) (x) is the erivative of the (n ) st erivative. Example: Calculate f ( ) for f(x) = 3x 5 x + 7x. We must calculate the first three erivatives: f (x) = 5x 4 4x 4x 3 f (x) = x (5x4 4x 4x 3 ) = 6x 3 4 + 4x 4 f (x) = x (6x3 4 + 4x 4 ) = 8x 68x 5 At x =, f ( ) = 8 + 68 = 348. Tutorial 3 Exercise. Page 34,3,5,7,9,,3,5,3. Exercise.3 Page 5,3,5,7,9,,3,5,7,9,7,9,37,39,4,45,47. Exercise.5 Page 65,3,5,7,9,,3,5,7,3,35,37,4 Exercise. Page 43,3,5,7,3. Exercise.4 Page 59,3,5,7,9,,3,5,7,9,,3 Assignment (3) () Use the efinition of the erivative to fin f (x): (a) f(x) = x + (b) f(x) = ( x + ).

Page () Fin the first erivative of the following functions: (i) y = (4x 3 + 5)(x 4 x) 3 (ii) y = 3x + 7x (iii) y = 4 x + 8 6 3 + 3π + x π. x x (3) Fin the points on the curve y = x 3 + 3x 4 where the tangent is parallel to the line y = 9x. (4) Fin the secon erivative of the following functions: (i) y = ( 5x ) 3 (ii) y = x3 (x+). (5) Fin the points on the graph of f(x) = x 3 + 3x + where the tangent is horizontal? (6) Match the graph of each function in (a) () with the graph of its erivative in (i) (iv).

Lecture 4 [ Trigonometric Functions ] Page Angles are measure in raians or egrees θ(raians) = S( units of length ) r ( units of length ) π raians (ra) = 36 ra = ( 8 π ) 57.3 = π 8 ra.7 ra Graphs of The Trigonometric Functions A graph of y = sin θ is generate as a point moves aroun the unit circle. Graphs of y = sin θ an y = cos θ over one perio of length π. Perioic Function: A function f(x) is calle perioic with perio T if f(x + T) = f(x) (for all x) an T is the smallest positive number with this property. The sine an cosine functions are perioic with perio T = π since angles that iffer by an integer multiple πk correspon to the same point on the unit circle. - The functions sin θ an cos θ are efine for all real numbers θ. We often write sin x an cos x instea of θ epening on the application.

Page 3 Recall that there are four other stanar trigonometric functions, each efine in terms of sin x an cos x. sin x cos x tan x = cos x = b a cot x = sin x = a b sec x = cos x = c a csc x = sin x = c b Trigonometric Ientities cos x + sin x = sin(x + y) = sin x cos y + cos x sin y () + tan x = sec x sin(x y) = sin x cos y cos x sin y () cot x + = csc x cos(x + y) = cos x cos y sin x sin y (3) cos(x y) = cos x cos y + sin x sin y (4) From () sin(x) = sin x cos x. (5) From (3) cos(x) = cos x sin x cos(x) = cos x ( cos x) = ( sin x) sin x cos(x) = cos x = sin x. (6) From (6) cos x = [ + cos(x) ] an sin x = [ cos(x) ]

Page 4 Derivatives of sine an cosine functions x sin x = cos x x cos x = sin x The erivatives of the other Trigonometric functions are x tan x = sec x x sec x = sec x tan x x cot x = csc x x csc x = csc x cot x Verify the formula x tan x = sec x Use the Quotient Rule an the ientity cos x + sin x = : x tan x = x sin x cos x (sin x) sin x ( sin x) ( ) = cos x cos x = cos x + sin x cos x = cos x = sec x. Example: Fin the erivative of y = sec x tan x y = sec x ( csc x cot x ) + csc x (sec x tan x ). Example: Fin the erivative of y = sec x +tan x y = ( + tan x)(sec x tan x) sec x (sec x) ( + tan x) = sec x (tan x + tan x sec x) ( + tan x) = sec x (tan x ) ( +tan x).

Page 5 If u = g(x) is a ifferentiable function, then applying the chain rule: x sin u = cos u u x tan u = sec u u x sec u = sec u tan u u x cos u = sin u u x cot u = csc u u x csc u = csc u cot u u Example: Fin y if, ) y = cos(x 3 ) y = sin(x 3 ) 3x. ) y = csc 4 (x ) y = 4 csc 3 (x ) ( csc(x )cot (x )) x. 3) y = + tan 4 x y = +tan 4 x (4 tan3 x sec x ). 4) y = sin (cos (sin x)) y = cos(cos(sin x)) ( sin(sin x)) (cos x). 5) y = cos (+x) +cos x y = (+cos x) ( sin(+x)) cos (+x) ( sin x) ( +cos x) 6) y = [ + cot 5 (x 4 + )] 9 y = 9[ + cot 5 (x 4 + )] 8 5 cot 4 (x 4 + ) ( csc (x 4 + )) 4x 3. Example: Fin the equation of the tangent line to y = x cos(x ) + 3 at the point (, 3). y = x cos(x ) + 3 y = x ( sin(x )) + cos(x ) The slope at x = is m = y x= = An equation of the tangent line is (y 3) = ()(x ) or y = x + 3. Example: Sketch the graph of y = 3 sin(x π 3 ). The function y = sinx has been shifte π 3 units to the right then stretche parallel to the y axis by a factor of 3.

Tutorial 4 Page 6 Exercise.4 - Page 39 5,7,9,,3,5,6,7, 9,. Exercise.6 - Page 73 5 (o). Assignment (4) () Solve the equations (a) cos x = sin x for x π. (b) cos x = cos x for x π. (c) sin x = cos x for x π. () Fin the first erivative of the following functions; (a) y = sec(tan (x 4 )). (b) y = sec 3 (sec x + ). (c) () y = ( + cos 3 x) cot 3 x. y = tan x + x csc 3 x. (f) y = tan (cos 3 ( + x )). (g) y = sin 3x 4 + cos x. (h) y = x ( cos x) sin x cos x. (i) y = cot x ( cos x sin x ). cos 3 x (3) If y = a cos(ωx) + b sin(ωx), show that y + ω y =. (4) Sketch the graph of the function ientifying the range an omain. (i) y = + cos x (ii) y = sin x (iii) y = tan x (iv) y = 3 3 sin(x + π ). (5) Fin an equation of the tangent to the function f(x) = sin ( π x) at x =.

Page 7 Lecture 5 [Inverse Trigonometric Functions / Implicit ifferentiation] Inverse Trigonometric Functions The six trigonometric functions are not -. However, we can restrict their omains to intervals on which they are one-to-one. So that they can have inverse functions. Domain restrictions that make the sin, cos, an tan functions - reflection in the line y = x y = sin x or y = arcsin x x = sin y where x π y π y = cos x or y = arccos x x = cos y where x y π y = tan x or y = arctan x x = tan y where < x < π < y < π Examples: () sin ( 3 ) = π 3 (3) cos ( ) = π 4 (5) tan ( 3 ) = π 6 () sin ( ) = π 4 (4) cos ( ) = π 3 (6) tan ( 3) = π. 3

Page 8 Examples: (a) sin (sin π 6 ) = π 6. (b) cos (cos 3 ) = 3. (c) tan (tan 3π 4 ) = tan ( ) = π 4. Example: Fin tan(sin ( 3 4 )). Metho : You nee to fin tan θ given that θ = sin ( 3 4 ) sin θ = 3 4 cos θ + sin θ = cos θ = ( 3 4 ) = 7 4 (taking +ve root for cosine in the st quarant) Hence, tan θ = 3 sin θ = 4 cos θ 7 4 = 3 7 an tan(sin ( 3 4 )) = 3 7. Metho : Draw a triangle to show sin θ = 3 4 Use Pythagoras Theorem: x + 3 = 4 x = 6 9 = 7 So tan θ = 3 7 an tan(sin ( 3 4 )) = 3 7

Derivatives of Inverse trigonometric functions Page 9 x sin x =?? Let y = sin x x = sin y π y π = cos y y x y = x cos y an cos y > since π y π, so cos y = sin y = x Therefore y = x cos y = x x sin x = x x The general formulas of the erivative of inverse trigonometric functions. x sin u = u. u u <. x cos u = u. u u < 3. x tan u = + u. u 4. x cot u = + u. u 5. x sec u = u u. u u > 6. x csc u = u u. u u > Example: Fin y x ) y = x sin x y = x for each the following : x + sin x (). ) y = x tan (x ) + (csc 3x) y = x x + (x ) + tan (x ) + (csc 3 3x) ( 3x (3x) ). 3) y = cot(cos (x )) + cot (cosx) y = csc ( (cos (x )) x (x ) + ( 4) y = x tan ( x) + sec (4x) y = x + (cos x) +( x) + x tan ( x ) x + ( sin x ) ) 4 4x (4x).

Page 3 5) y = tan ( x) cos (x) 6) y = y = cos (x) ( + x )( x ) tan ( x ) ( x ) ( cos x) + sin (x) cot ( x3 + ) x 3. 7) y = ( x ) ( sin x) + ( ) +( x3 + x 3 ) [ ( x3 ) ( 3x ) ( x 3 + ) (3x ) ( x 3 ) ]. Implicit Differentiation Fortunately, we on t nee to solve an equation for y in terms of x in orer to fin the erivative of y. Instea we can use the metho of Implicit ifferentiation. This consists of ifferentiating both sies of the equation with respect to x an then solving the resulting equation for y. Example: Fin y x for 3y + y + x y = x 3 Differentiating both sies with respect to x 6 y y + y + x y + xy = 3x y [ 6y + + x ] = 3x xy y = [ 3x xy 6y + + x ]. Example: Fin y x for sin y cos(xy) = x y. Differentiating both sies with respect to x cos y y + sin(xy) [ x y + y] = y y [ cos y + x sin(xy) + ] = y sin(xy) y = [ y sin(xy) cos y + x sin(xy) + ].

Tutorial 5 Page 3 Exercise.4 - Page 39 57 64. Exercise.8 - Page 9,3,5,7,9,3,9,,3,5,9,3,39,47,49. Assignment (5) () Write the expression in algebraic form. (a) cos (sin (x)) (b) sec( sin (x )) (c) sin( cos x ) (Hint use : sin θ = sin θ cos θ). () Fin the st erivative of (a) tan 3x + ( tan x ) (b) csc ( x ) + cot x (c) sin x sin x + x + sec x () tan ( cos x) + cos (sin x). (3) If y = sin x x, show that ( x ) y = 3xy + y. (4) If f(x) = sin ( x x + ) an g(x) = tan ( x ), show that f (x) = g (x). (5) If y = x cos x + 4 sin x x 4 x. Show that y = x cos x. (6) Fin an equation of the tangent line to the graph of the equation tan (xy) = sin (x + y) at the point (, ). (7) Fin the point(s) on the graph of x + y = 5 at which the slope of the tangent is.

Page 3 Lecture 6 [ Exponential Functions an Logarithmic Functions ] Definition: Exponential functions The function f(x) = a x base a. ; x is a real number, a >, a, is calle the exponential function, f(x) = a x, < a < f(x) = a x, a > The omain is x ], [ The omain is x ], [ The range is y ], [ The range is y ], [ Horizontal asymptote y = Horizontal asymptote y = ecreasing on its omain. increasing on its omain. The Natural Exponential Function f(x) = e x which has the base, e.78888459 Among all exponential functions y = a x, the base e is the unique base for which the slope of the tangent line to the graph of f(x) = e x at the point (,) is equal to. ( i.e. f () = ). We shall show later that e = lim ( + n n n ).

Logarithmic functions f(x) = a x has an inverse f, which is calle the logarithmic function with base a an is enote by log a. f (x) = y x = f(y) Page 33 Then we have log a x = y x = a y Thus, if x >, then log a x is the exponent to which the base a must be raise to give x. For x >, a >, a. The logarithmic function has omain ], [ an range ], [ Inverse Properties for a x an log a x log a a x = x for every x R a log a x = x for every x >

Natural Logarithms Page 34 log e x = ln x ( a = e ) - ln x = y x = e y - ln e = - ln e x = x, x R - e ln x = x, x > - lim x ln x = - lim x + ln x = Change of base formula Every logarithmic function is a constant multiple of the natural logarithm log a x = ln x ln a For a, b > an c any real number - ln(ab) = ln a + ln b - ln ( a ) = ln a ln b b - ln a b = b ln a Derivative of the exponential function (f(x) = e x ) f (x) = lim h f( x + h ) f(x) h = lim h ex+h e x h = lim h eh e x e x h = lim h ex (e h ) h = e x lim (eh ) h h = e x f () = e x () = e x x ex = e x

Derivative of the exponential function (f(x) = a x ) Page 35 f(x) = a x = (e ln a ) x = e x ln a f (x) = e x ln a ln a ( Chain Rule ) x ax = a x ln a Ex: x 5x = 5 x ln 5. Derivative of the logarithmic function (f(x) = ln x) y = ln x e y = x e y y x = y x = e y = x x ln x = x Derivative of the logarithmic function (f(x) = log a x) y = log a x a y = x a y ln a y x = y x = a y ln a = x ln a x log a x = x ln a In general: ) x eu = e u u ) x au = a u u ln a 3) x ln u = u u 4) x log a u = u ln a u Example: Fin y x if: - y = xe x + x e + ex + 7 e y = x e x () + e x + ex e x x ln. - y = tan(e tan x ) + log 5 (x + ) y = sec (e tan x ) e tan x sec x + 3- y = cos( e x ) + sin(3 x ) x ( x + ) ln 5. y = sin( e x ) e x + cos(3 x ) 3 x ln 3.

Page 36 4- y = ln(sec(ln x)) + e tan x y = sec(ln x) sec(ln x) tan(ln x) x + etan x +x = tan(ln x) x + etan x +x. To fin x [ f(x)g(x) ], Logarithmic ifferentiation coul be use, as in the next example. Example: Differentiate y = x x. Take ln both sies, ln y = ln x x ln y = x ln x y y = x x + ln x x y = y[ x x ln x + ] x y = x x [ x x ln x + ]. x Or y = e x ln x y = e x ln x [ x + ln x ] = x x x x [ x x ln x + ]. x Taking the erivative of some complicate functions can be simplifie by using logarithms. Example: Fin y if y = x x + (x+) 5 (3x + ) ln y = ln (x x + ) ln((x + ) 5 (3x + ) ) ln y = ln x + ln x + [ln(x + ) 5 + ln(3x + ) ] ln y = ln x + ln(x + ) 5 ln(x + ) ln(3x + ) y y = x + x x + 5 x + ( 6x 3x + ) y = x x + (x+) 5 (3x + ) [ x + x x + 5 x+ x 3x + ].

Page 37 Example: Fin the inverse function f (x) for the function f(x) = + e x. Sketch the graphs of f(x) an f (x) on the same iagram, labelling any intersections with the coorinate axes. Give the omain an range of f (x). y = + e x ln(y ) = ln(e x ) = x Interchanging x an y. ln(x ) = y y = ln(x ) f (x) = ln(x ) Domain f (x) is x ], [ Range f (x) is y ], [ Tutorial 6 Exercise.5 - Page 39,3,5,7,9,3,33,35,36,37,4,43,45. Exercise.7 - Page 8 3 (o), 39,4,43. Assignment (6) () Fin the erivative of the given function. (a) y = ln x + ( ln x ) (b) y = ln(ln(ln x)) + e x tan(e x ) (c) () (e) (f) y = csc(e tan x ) + ln x y = x sin x cos x y = [ln(sin x)] y = sin (e x ) + cos (ln x) (g) y = cot3 (x) ln(+x ). x x cos x

() Fin an equation of the tangent line to the graph of y = ln(e 3x + x) at x =. Page 38 (3) Explain why there is no point on the graph of y = e x at which the tangent line is parallel to x + y =. (4) Show that the given function satisfies the given ifferential equation, where c an c are arbitrary real constants. (a) y = c e 3x + c e x ; y + y 6y = (b) y = c x + c x ln x ; 4x y + 8xy + y =, for x >. (5) If x y + y x = 4. Fin y. (6) Sketch the graph of each function an state the horizontal asymptote. i) f(x) = e x ii) f(x) = e x iii) f(x) = + e x iv) f(x) = e x + 3. v) f(x) = e x (7) Fin the omain of the given function. Fin the x intercept an the vertical asymptote of the graph. Use transformations to sketch the graph. (a) f(x) = ln(x + ) (b) f(x) = ln( x).

Page 39 Lecture 7 [Hyperbolic an Inverse Hyperbolic Functions] Hyperbolic Functions The hyperbolic functions are forme by taking combinations of the two exponential functions e x an e x. The hyperbolic functions simplify many mathematical expressions an occur frequently in mathematical an engineering applications. Definition Hyperbolic sine an cosine Hyperbolic sine Hyperbolic cosine sinh x = ex e x cosh x = ex + e x Definition Other hyperbolic sine an cosine Hyperbolic tangent tanh x = sinh x cosh x = ex e x e x + e x Hyperbolic cotangent coth x = cosh x sinh x = ex + e x e x e x Hyperbolic secant sech x = cosh x = e x + e x Hyperbolic cosecant csch x = sinh x = e x e x

Hyperbolic Ientities Page 4 cosh x sinh x = cosh( x) = cosh x tanh x = sech x sinh( x) = sinh x coth x = csch x cosh(x ± y) = cosh x cosh y ± sinh x sinh y sinh(x ± y) = sinh x cosh y ± cosh x sinh y sinh x = sinh x cosh x tanh( x) = tanh x cosh x = cosh x + sinh x cosh x = cosh x + sinh x = cosh x cosh x sinh x = ( ex + e x ) ( ex e x ) sinh x cosh x = ( ex + e x )( ex e x ) = ex + + e x ( e x + e x ) 4 = ex e x = 4 4 =. = sinh x. Example: Solve the equation cosh(x) + sinh(x) = 5 ( ex + e x ) + ( ex e x ) = 5 6 e x 4e x 5 = 6 (e x ) 5e x 4 = e x = ( 5) ± 5 4 (6) ( 4) (6) e x = 5 + (e x > ) x = ln ( 4 3 ) x = ln (4 3 ).

Derivatives of Hyperbolic Functions Page 4 x [sinh x] = x [ ex e x ] = ex + e x = cosh x x [cosh x] = x [ ex + e x ] = ex e x = sinh x x [tanh x] = x sinh x cosh x (cosh x) sinh x (sinh x) [ ] = cosh x cosh = x cosh x = sech x. In general, let u be a ifferentiable function of x. x x [sinh u] = cosh u u [cosh u] = sinh u u x [tanh u] = sech u u x [coth u] = csch u u x x [sech u] = sech u tanh u u [csch u] = csch u coth u u Fin the first erivative of each function Example: If y = tan x + x tanh x. Fin y. y = tan x tanh x ln x + e y = tan x +x ln + etanh x ln x [ tanh x ( x ) + ln x sech x Example: If y = (ln(cosh e x )) tanh x. Fin y. ln y = tanh x ln[ln(cosh e x )] y y = tanh x ln(cosh e x ) sinh(ex ) e x + ln[ln(cosh e x )] sech x y = y( tanh x ln(cosh e x ) sinh(ex ) e x + ln[ln(cosh e x )] sech x).

Inverse Hyperbolic Functions Page 4 Because the hyperbolic functions are efine in terms of exponential functions, it is not surprising to fin that the inverse hyperbolic functions can be written in terms of logarithmic functions Example: Show that. sinh x = ln( x + x + ). tanh x = ln( + x x ) ) ) y = sinh x x = sinh y y = tanh x x = tanh y x = ey e y x = ey e y e y + e y e y x e y = xe y + x e y = e y e y e y = ( x) ± 4x 4 () ( ) () xe y + x = e y e y = x ± x + e y ( x ) = + x e y = x + x +, e y > e y = +x y = ln( x + x + ) y = ln( +x x ) sinh x = ln( x + x + ). tanh x = +x ln ( ). x x

Page 43 x sinh x = x ln( x + x + ) Alternativly, we may procee as follows: = = [ + ( x+ x + x + ) (x) y = sinh x x = sinh y [ x + + x ] = cosh y y x+ x + x + = x + x sinh x = = = cosh y sinh y + + x as before In general x sinh u = + u u x cosh u = x tanh u = u u, u u, u > u < x coth u = u u u > x sech u = x csch u = u < u < u u u u u + u Example: Fin the erivative of each function ) y = cosh (x + 5 ) y = (x +5) (x) ) y = e x sech x y = e x ( x x x ) + sech x e x x.

Page 44 3) y = sin4 x tanh x (x+) sinh x ln y = 4 ln (sin x) + ln(tanh x) sinh (x) ln (x + ) y y = 4 cos x + sin x tanh x sech x (sinh (x) x+ + ln(x + ) + x ). y = sin4 x tanh x (x+) sinh x [4 cot x + sech x tanh x sinh (x) x+ ln(x+) + x )]. Example: Solve the equations ) ln ( + tanh ex tanh e x ) = 8 ln ( + tanh ex tanh e x ) = 4 tanh (tanh(e x )) = 4 e x = 4 x = ln 4 x = ln ) ln (sinh( x) + sinh (x) + ) = 3x 6 sinh (sinh x) = 3x 6 x = 3x 6 x = 3

Tutorial 7 Page 45 Exercise.9 - Page 97 7. Assignment (7) () If sinh x =, fin the values of the remaining hyperbolic functions. () Show that (a) tanh(ln x) = x x + (c) cosh x = cosh x + sinh x. (3) Fin the erivative of the given function (a) y = coth(cosh 3x) + tanh(sinh x 3 ) (b) y = (x cosh x) 3 + (ln(sech x)) (c) y = sinh (sin x ) + csch (e tan x ) () y = ln(sech x) + x tanh ( x ) (e) y = (f) y = (tanh x ) 3 e x + cosh x + (cosh 6 x) + (x + ) sech(ln x ) (g) y = coth (csc x) + (tanh( x + )) (h) y = ln ( sin 4 x (x +) tanhx coshx 3 ). +e x (4) Show that if y = cosh (ln e cosh( x +) ), then yy = x. (5) Fin the point(s) on the graph of y = sinh x at which the tangent line has slope.

Page 46 Lecture 8 [L Hopital s Rule] L Hopital s Rule Suppose that f an g are ifferentiable on the interval ]a, b[, except possibly at the point c ] a, b [ an that g (x) on ]a, b[, except possibly at c. Suppose further that f(x) lim = or x c g(x) an that, lim f (x) x c g (x) = L ( or ± ). Then, f(x) lim x c g(x) = lim x c f (x) g (x) Remark : f(x) The conclusion also hols if lim x c g(x) is replace with any of the limits lim x c + f(x) g(x), lim x c f(x) g(x), lim x f(x) g(x), lim x f(x) g(x) Ineterminate Forms,,,,,,.

Page 47 = Apply L Hopital s rule = lim x cos x = =. = Apply L Hopital s rule = lim x x+ = lim x+4 x = = x+ x+4 = Apply L Hopital s rule = lim t π sin t cos t sin t = lim t π cos t =. = Apply L Hopital s rule = lim x cos x 3x = ( ) h = lim x sin x 6x = ( ) h = lim x cos x 6 = 6.

Page 48 = Apply L Hopital s rule = lim x x = lim x x =. = Apply L Hopital s rule = lim x 4x3 e x = ( ) h = lim x x 4 e x = ( ) h = lim x 4x 8 e x = ( ) h = lim x 4 6 e x = 4 6 = 3. = Apply L Hopital s rule h = lim π+ t h sec t = 3sec 3 t ( ) = lim t π h = ( ) = lim 3cos t + cos 3 t t π + cos 3t ( sin 3t 3 ) 6 cos t ( sin t ) = lim π+ t sin 3t cos 3t sin t cos t = lim π+ t sin 6t sin t h = ( ) = lim 6 cos 6t = 6 t π + cos t = 3.

Page 49 = Apply L Hopital s rule = lim x + x csc x = lim x + sin x x = ( ) h = lim x + sin x cos x = = Apply L Hopital s rule lim x + cos x sin x sin x = lim cos x x + sin x = ( ) h = lim x + sin x cos x = =

Page 5 Apply L Hopital s rule = lim x + sin x x x sin x = ( ) h = lim x + = cos x h x cos x + sin x = ( ) = lim x + sin x x sin x + cos x + cos x = =. = Apply L Hopital s rule = lim x x cot( πx ) = ( h ) = lim x csc ( πx ) π = π. = ( ) Apply L Hopital s rule = lim x + ln x = ( x h ) = lim x + x x 3 = lim x + x =.

Page 5 = Let y = ( cos x ) x (then take ln for both sies) ln y = ln(cos x) lim x ln y = lim x ln ( cos x) x = ( ) h = lim x sin x cos x x x = lim x tan x x = ( ) h = lim x sec x = Hence, lim x + ( cos x ) x = e. = Let y = ( x + 3 x ) x+ (then take ln for both sies) ln y = (x + ) ln( x + 3 x ) lim x ln y = lim x ln( x + 3 x ) (x + ) = ( ) h = lim x x (x ) ( x +3) ( x+3 (x ) ) = lim 4 (x+) x = (x + ) (x+3) (x ) h = lim x 8(x + ) x + = h 8 = lim x = 4 Hence, lim ( cos x ) x = e 4. x

Page 5 = Let y = ( sin x) x (then take ln for both sies) ln y = x ln(sin x) lim x + ln y = lim x ln(sin x) = ( ) = lim x + x + ln(sin x) x = ( ) h = lim x + cos x sin x x = = lim x + cos x sin x x h x = lim = tan x () x = lim = = sec x x + x + Hence, lim ( sin x ) x = e =. x + = Let y = ( + x ln x ) (then take ln for both sies) ln y = ln( + ln x x ) h ln( lim ln y = lim +x ) = ( ) = lim x x ln x x x +x x = lim x h x = +x ( ) 4x = lim = x x Hence, lim x ( + x ln x ) = e. Tutorial 8 Exercise 3. - Page 9 43 (o).

Assignment (8) Page 53 Fin the inicate Limits (a) lim x x + x 6 x 4 (b) lim x π + sin x x π (c) lim x x sin ( 4x ) () lim ln( sin x ) x + ln(tan x ) (e) lim x + x ln x (f) lim x x tan x cos x (g) lim x (sec 3 x) cot x (h) lim x ( ) x sin x x (i) lim x tan x x x sin x (j) lim x e x x cos x (k) Lim x + ln(cot x ) (l) lim e csc x x + ln(x) ln (sin x) (m) lim π tan x x x (n) lim x x ln x x + ln x (o) lim π+ t tan t tan 3t (p) tan x lim x ( sinh x ) (q) lim x ( + e x ) e x ( cos x ) (r) lim x ( x ) (s) πx tan( lim x ( x ) ) (t) lim x + sin x ln x (u) lim x π sec x tan x (v) lim x ( x x + )x.

Page 54 Lecture 9 [ Antierivatives / The Definite Integral ] Antierivatives F(x) is antierivative of the function f(x) means, x F(x) = f(x) Ex: F(x) = 3x 4 is antierivative of f(x) = x 3 because x 3x4 = x 3 Note: If F(x) is an antierivative of f(x) then all other antierivatives of f(x) have the form F(x) + c, where c is constant. The process of fining antierivatives is calle antiifferentiation or integration, an is enote by integration sign Inefinite integral notation

Page 55 Differentiation Formula x [ xn+ n + ] = xn Integration Formula x n x = xn+ + c, n n + x [ln x] = x x x = ln x + c x x x [sin x ] = cos x [cos x ] = sin x x [tan x ] = sec x [sec x ] = sec x tan x cos x x = sin x + c sin x x = cos x + c sec x x = tan x + c sec x tan x x = sec x + c x [sin x ] = x x x = sin x + c x [tan x ] = + x + x x = tan x + c x [sec x ] = x [ex ] = e x x x x [ax ] = a x ln a x x [sinh x] = cosh x [cosh x] = sinh x x x x = sec x + c e x x = e x + c a x x = ax + c, a >, a ln a cosh x x = sinh x + c sinh x x = cosh x + c Example: x 5 x = x 5 x = x 4 4 + c = 4x 4 + c. Example: x x = x x = x3 3 + c = 3 x3 + c.

Properties of the inefinite integral Page 56 Let F (x) = f(x) an G (x) = g(x) then, (i) (ii) k f(x) x = k f(x) x = k F(x) + c. [ f(x) ± g(x) ] x = f(x) x ± g(x) x = F(x) ± G(x) + c. Example: I = ( 4x + 5 sin x ) x x = 4 x x x x + 5 sin x x = 4 x ln x 5 cos x + c Example: I = 4x3 7 x x = ( 4x3 7 ) x = ( 4x5 7x ) x x x = 4 x7 7 7 x = 8 7 x7 4 x + c Example: I = x x + x = x + x + x = x + x + x x + x = x tan x + c. Differential equations A ifferential equation is an equation that involves erivatives of an unknown function. An n th orer ifferential equation of the form n y x n = g(x) can be solve by integrating the function g(x) in succession n times. In this case the family of solutions will contain n constants of integration. Example: Fin a function f such that f (x) = x + for which the slope of te tangent line to its graph at (, ) is 3. solution f (x) = x +

Page 57 f (x) = x + x f (x) = 4 x 3 + x + c m = f () = 3 3 = 4 () 3 + () + c c = 3 f (x) = 4 x 3 + x 3 f(x) = (4x 3 + x 3) x f(x) = x 4 + x 3x + c f() = = () 4 + () 3() + c c = f(x) = x 4 + x 3x +. The Definite Integral Let f(x) be a continuous function efine on [a, b]. Consier the following four steps Divie the interval [a, b] into n subintervals [x k, x k ] of withs x k = x k x k, where a = x < x < < x n < x n = b. Let P enote the largest with of the n subintervals x, x,, x n. Choose a number x k in each subinterval [x k, x k ] Then the efinite integral of f(x) from a to b is enote b by f(x) x, a is efine to be b f(x) x = a n lim f(x k ) P k= x k

Geometric Interpretation of the Definite integral Page 58 If f(x), on [a, b], then on [a, b], If f(x) is both positive an negative on [a, b], b f(x) x = a b the area of the region uner then f(x) x = A a + A 3 A the graph of y = f(x), above = Area above[a, b] the x axis from x = a to x = b. Area below [a, b]. (See Fig) The Funamental Theorem of Calculus Let f be a continuous function efine on [a, b] an F is any antierivative of f on the interval, then b f(x) x = F(b) F(a) a Properties of the efinite integral a. f(x) x = a b a a b. f(x) x = f(x) x b a b a 3. c f(x) x = c f(x) x b a b a 4. [ f(x) ± g(x) ] x = f(x) x b a c a 5. f(x) x = f(x) x + Note (c, is a constant) b f(x) x c If f is an even integrable function on [ a, a], then a ± b g(x) x a (a < c < b) f(x) x = f(x) x a a

If f is an o integrable function on [ a, a], then Page 59 a f(x) x = a Example: Evaluate (x + x ) x (x + x ) x = [ x3 + 3 x () ] = [ 3 + 3 () ] [ ()3 + 3 () ] = ( 8 3 + 4) ( 3 + ) = 6 3. Example: Evaluate π 4 3 cos x x π 4 3 cos x x π 4 = 3 sec x x π 4 = 3 [tan x] = 3 [ tan (π 4) tan() ] = 3. Example: Evaluate 4 + x x = 4 + x x 4 + x x = 8 + x x = 8 [tan x] = 8 [tan () tan ()] = 8( π ) = π. 4 Example: Evaluate e x x e x x = [ e x ] = e e = e. e π Example: Evaluate sin x x π π π sin x x =

Tutorial 9 Page 6 Exercise 4. - Page 37 43(o), 5 58. Exercise 4.5 - Page 34 3(o), 5 54. Assignment (9) () Fin constants c an c such that F(x) = c x sin x + c cos x is an antierivative of f(x) = x cos x. () Evaluate. (a) x ( x + ) x (b) x x x 3 x x (c) x + x + x () ( x + ) x x 4 (e) x 5 + x 4 5 x (f) x x (g) sin x sin x x (h) tan x x (h) cos x cos x sin x x (i) x3 x + x + 4 + x x (j) x (Hint: use sin x cos x cos x + sin x = ). (3) Verify by ifferentiation. (a) sin x cos x x = sin x + c (b) x e x x = x e x e x + c (c) ln x x = x ln x x + c. (4) Fin a function f such that f (x) = x 3 an the line x + y = is tangent to the graph of f. (5) Fin a function f such that f (x) = 6, f ( ) =, f( ) =. (6) Given that the graph of f passes through the point (, 6) an that the slope of its tangent line at (x, f(x)) is x +, fin f().

Page 6 Lecture [ Integration by Substitution ] Integration by the u substitution Consier a composite function F(g(x)), where F is an antierivative of f ; That is F = f By the chain rule x F(g(x)) = F (g(x)) g (x) = f(g(x)) g (x) f(g(x)) g (x) x = F(g(x)) + c In the composite function f(g(x)), ientify the inner function as, u = g(x) u = g (x) x f(g(x)) g (x) x = f(u) u = F(u) + c PROCEDURE. Given an inefinite integral involving a composite function f(g(x)), ientify an inner function u = g(x) such that a constant multiple of u (x)( equivalently, g (x)) appears in the integral.. Substitute u = g(x) an u = u (x) x in the integral. 3. Evaluate the new inefinite integral with respect to u. 4. Write the result in terms of x using u = g(x). Disclaimer: Not all integrals yiel to the substitution Rule

Perfect substitution Page 6 Example: Evaluate I = ( 5x + ) 6 5 x. Let u = 5x + u = 5 x Substituting SOLUTION I = ( 5x + ) 6 5 x = u 6 u = u7 + c = ( 5x+) 7 7 7 + c. Example: Evaluate I = sin 3 x cosx x. SOLUTION We rewrite the integral as (sin x) 3 cosx x Let u = sin x u = cos x x Substituting I = sin 3 x cosx x = u 3 u = u4 + c = sin 4 x 4 4 + c. Example: Evaluate I = (tan x ) + x x. SOLUTION We rewrite the integral as I = (tan x ) + x x Let u = tan x u = + x x Substituting I = (tan x ) x = u u = u3 + c = (tan x ) 3 + x 3 3 + c.

Introucing a constant Page 63 Example: Evaluate I = x 5 ( x 6 + ) x. SOLUTION Let u = x 6 + u = 6 x 5 x Substituting x 5 x = u 6 I = x 5 ( x 6 + ) x = u u 6 = 6 u u I = u + c = (x6 +) + c. 6 66 csc ( x) cot ( x ) Example: Evaluate I = x x. SOLUTION We rewrite the integral as I = csc ( x) cot ( x ) x x Let u = x u = x x x x = u Substituting I = csc ( x) cot ( x ) x x = csc (u) cot (u) u I = csc(u) + c = csc( x ) + c Example: e4 x Evaluate I = x. x SOLUTION We rewrite the integral as I = e4 x x x

Page 64 Let u = 4 x u = 4 x x x x = 4 u Substituting 4 x I = e x x = 4 eu u = 4 eu + c = 4 x e4 + c. Example: Evaluate I = x 6 + x 6 x. SOLUTION We rewrite the integral as I = x 6 + ( x 3 ) Let u = x 3 u = 3x x x u 3 = x x Substituting I = x 6 + ( x 3 ) x = 3 4 + u u = 3 4 tan ( u 4 ) + c I = tan ( x3 4 ) + c Example: Evaluate I = x 5 + x 3 x. SOLUTION Let u = x 3 + 5 u = 3x x u 3 = x x Substituting x I = x = u = 5 + x 3 3 u 3 ln u + c I = 3 ln 5 + x3 + c.

Variation on the substitution metho Page 65 Example: Evaluate I = x x + x. SOLUTION Substitution : The composite function x + suggests that u = x +. You might oubt whether this choice will work because u = x an the x in the numenator of the integral is unaccounte. But lets procce. Let u = x + x = u & u = x Substituting I = x x+ x = u u u = ( u u u ) u = ( u u ) u = u3 3 u + c = 3 (x + )3 (x + ) + c Substitution : Another possible substitution is Let u = x + u = x+ x u = x+ x & x = u Substituting I = x x = x x x+ x+ = ( u ) u = ( u3 3 u ) + c I = 3 (x + )3 (x + ) + c

Page 66 Example: Evaluate I = + x x SOLUTION Let u = x x = u & x = u u Substituting I = + x x = u + u u = ( u + ) + u u = ( + u ) u I = ( u ln + u ) + c = x ln( + x ) + c OR using the substitution Let u = + x x = (u ) & x = (u ) u Substituting I = + x ( u ) x = u u = ( u ) u = ( u ln u ) + c I = u ln + u + c = x ln( + x) + c Completing the square Example: Evaluate I = x + 4 x + 6x + 8 x SOLUTION I = x + 4 x + 6x + 8 x = x + 4 ( x + 3 ) + 9 x Let u = x + 3, then x = u 3 & x = u Substituting I = x + 4 x + 6x + 8 x = u + u + 9 u = u u + 9 u + u + 9 u

Page 67 I = u u + 9 u + u + 9 u = ln( u + 9 ) + 3 tan ( u 3 ) + c = ln( ( x + 3 ) + 9 ) + 3 tan ( x + 3 ) + c. 3 Substitution Rule for Definite Integrals Let u = g(x) be a function that has continuous erivative on the interval [a, b], an let f be a function that is continuous on the range of g. If F (u) = f(u), then b f(g( a x)) g (x) x = g(b) f(u) u = F(g(b)) g(a) F(g(a)) Example: Evaluate x ( x + 3 ) 3 SOLUTION Let u = x + 3, then x = u 3 & x = u Because we have change the variable of integration from x to u, the limits of integration must also be expresse in terms of u. In this case x = implies u = + 3 = 3 x = implies u = + 3 = 5 lower limit upper limit Substituting x ( x + 3 ) 3 5 = u 3 3 u = u 5 3 = ( 5 3 ) = 8 5.

Special Trigonometric Integrals Page 68. tan x x = sin x cos x sin x x = cos x x = ln cos x + c = ln sec x + c cos x. cot x x = sin x x = ln sin x + c sec x (sec x+tan x ) 3. sec x x = (sec x+tan x ) x = sec x + sec x tan x (sec x + tan x ) x = ln sec x + tan x ) + c csc x (csc x cot x ) 4. csc x x = (csc x cot x ) x = csc x csc x cot x (csc x cot x ) x = ln csc x cot x ) + c 5. sin x x = ( cos x ) x = x sin x 4 + c 6. cos x x = ( + cos x ) x = x + sin x 4 + c 7. tan x x = ( sec x ) x = tan x x + c 8. cot x x = ( csc x ) x = cot x x + c 9. csc x x = cot x + c. sec x x = tan x + c

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Tutorial Page 7 Exercise 4.6 - Page 35 4 (o). Assignment () () Evaluate the given inefinite integral using an appropriate u- substitution. (a) x + x 3 + 3x + 6 x (b) x + 3 x x 3 + 3x + 6 (c) x (ln x) x () (x ) x + x (e) csch( x ) coth( x ) x (g) (i) (k) (m) x + ( x ) 3 3 x ( + x ) x 9 + 8x x 4 ln x x (o) x x + (q) cos x + sin x x (f) sin x x 4 + cos x x (h) cos x x (j) x x x (l) x x ln x x 3 x x (n) e (x+) (x + ) x x (p) x + x 6 x x (r) x x (s) x 3 ( x + ) 5 x (t) ( x + ) ( x ).5 x (u) (w) sec x tan x (sec x ) x (v) ( x + x )7 x 9 x sin x cos x x (x) x 5 + sin x + x (y) x sec (x 3 ) x (z) x x x+ + 5x + () Use the Funamental Theorem of Calculus to evaluate the given efinite integral ln (a) tanh x x π π (c) x sin x + x (e) ( + x )3.5 x x (g) + e x x () x x. ln (b) e x cosh x x (f) x (h) 4 e cos x x x sec(lnx) tan (lnx) x x x. (tan x ) ( + x )

Page 7 Lecture [ Area Between Curves ] Applications of the efinite integral Area of Region Between curves () If f(x) on [ a, b ], then the area uner the curve y = f(x) over [ a, b ] is, b A = f(x) x a () If f(x) an g(x) are continuous with f(x) g(x) on [ a, b ], then the area of the region between the curves y = f(x) an y = g(x) from a to b is b A = [ f(x) g(x) ] x a Example: Fin the area boune by the graph of y = 3 x an y = x 9 To get points of intersection we solve 3 x = x 9 (x + 4)(x 3) = Hence x = 4 an x = 3. 3 A = [(3 x) (x 9)] 4 x = [ x x x3 ] 3 = 343. 4 6

Page 7 Example: Fin the area of the region boune by the graphs of y = cos x, y = x + for x. A = [ y top y bottom ] x A = [ (x + ) cos x ] x A = [ x3 + x sin x ] = sin(). 3 3 Sometimes, the upper an lower bounary is not efine by a single rule as in the following example Example: Fin the area boune by the graphs of y = x an y = x for x. To fin the points of intersection we solve, x = x x = x = ± A = [( x ) x ]x + [x ( x )] x = [ x x3 3 ] + [ x 3 3 x ] = 4.

Page 73 Some regions are best treate by regaring x as a function of y. If a region is boune by curves with equations x = f(y), x = g(y), y = c an y = where f(y) g(y) on c y then this area is given by A = [ f(y) g(y) ] y c Example: Fin the area enclose by the line y = x an the parabola y = x + 6 To get the intersection points we solve x = y + an x = y 3 y y 8 = (y 4)(y + ) = y =, y = 4 4 A = [ x right x left ] y 4 A = [ (y + ) ( y 3) ] y A = [ y y3 + 4y ] 4 6 = 8.

Page 74 Note: In the last example we coul foun the area by integrating with respect to x instea of y, but the calculation is much more involve. It woul have meant splitting the region in two an computing the areas labelle A an A. A = A + A A = [ x + 6 ( x + 6) ] x 3 5 + [ x + 6 (x ) ] x = 8 Example: Fin the area of the region in the st quarant that is boune above by y = x an below by the x axis an the line y = x. To get the intersection points we solve x = y an x = y + y y = (y + )(y ) = y =, y = A = [ x right x left ] y A = [ (y + ) ( y ) ] y A = [ y + y y3 3 ] = + 4 8 3 = 3.

Page 75 Example: Fin the area of the region boune by the curves y = sin x, y = cos x, x = an x = π. To get the intersection points we solve sin x = cos x x π tan x = x = π 4 Observe that cos x sin x when x π 4, but sin x cos x when π 4 x π Therefore the require area is A = A + A π 4 A = (cos x sin x) x = [sin x + cos x ] π 4 = ( + ) ( + ) = π A = (sin x cos x) x = [ cos x sin x ] π 4 π 4 π = ( ) ( ) A = + A = A + A = ( ) + ( + ) =.

Tutorial Page 76 Exercise 5. - Page 383,,3,4,5,6,7,8,,,,9 5. Assignment () () Use calculus to fin the area of the triangle with vertices (, ), (,), (, ). () Sketch an fin the area boune by the graphs of. (a) y = 4 x, y = x +. (b) y = e x, y = e x +, x =. (c) y = x, y = x, x =, x =. () y = cos x, y = cos x, x =, x = π. (e) x = y, y = x. (f) y = 4 x, y = 4, x =. (g) y = sin x, y = cos x, x =, x = π. (h) y = x, y = x 3, x =, x =. (i) y = x + x, y = x + 4, x =, x = 4. (j) y = tan x, y =, x = π 4, x = π 3. (k) y = x, x =, x = 4. (l) y = x x, y = x 4.

Page 77 Lecture [ Volume of Revolution ] Applications of The Definite Integral The volume of a soli of revolution If a region in the plane is revolve about a line, the resulting soli is a soli of revolution, an the line is calle the axis of revolution. I) The Disk Metho As long as the rotational soli resulting from your graph has no hollow space in it, you can use the isk metho to calculate its volume.

Page 78 Example: Let R be the region boune by the curve y = f(x) = (x + ), the x axis, an the lines x = an x =. Fin the volume of the soli of revolution obtaine by revolving R about the x axis. Volume = π ((x + ) ) x = π (x + ) 4 x = π [ ( x + )5 5 ] = 4 π 5. Revolving About a Line That Is Not a Coorinate Axis Example: Fin the volume of the soli forme by revolving the region boune by f(x) = x an g(x) = about the line y =. Solve x = to etermine that the limits limits of integration are ±, an R(x) = ( x ) = x The volume is given by V = π ( x ) x = 6 π 5.

Example: Fin the volume of the soli generate by revolving the region between the parabola x = y + an the line x = 3 about the line. R(y) = 3 ( y + ) = y The volume is given by V = π ( y ) y V = π ( 4 4y + y 4 ) y V = π [ 4y 4y3 + y5 ] 64 π =. 3 5 5 Page 79 II) The Washer Metho If a soli of revolution has any hollow spots in it, you ll have to use a moifie version of the isk metho, calle the washer metho. It gets its name from the fact that a thin slap of the soli resembles a circular washer of outer raius R an inner raius r. Horizontal axis of revolution Horizontal axis of revolution Volume = π ( R (x) r (x) ) x b a

Page 8 Vertical axis of revolution Vertical axis of revolution Volume = π ( R (y) r (y) ) y c Integrating with respect to y, two integral case Example: Fin the volume of the soli forme by revolving the region boune by the graphs of y = x +, y =, x =, an x = about the y axis. For the region shown, the outer raius is R(y) =. There is however, no convenient formula that represents the inner raius. When y, r(y) =, but when y, r(y) is etermine by the equation y = x +, which implies that r(y) = y. r(y) = {, y y, y The volume is given by, V = π ( ) y + π ( ( y ) ) y V = π y + π ( y) y = 3π. Note that the first integral π y represents the volume of a right circular cyliner of raius an height. This portion of the volume coul have been etermine without using calculus.