EDEXCEL NATIONAL CERTIFICATE UNIT 28 FURTHER MATHEMATICS FOR TECHNICIANS OUTCOME 4 - CALCULUS TUTORIAL 2 - INTEGRATION

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1 EDEXCEL NATIONAL CERTIFICATE UNIT 8 FURTHER MATHEMATICS FOR TECHNICIANS OUTCOME - CALCULUS TUTORIAL - INTEGRATION CONTENTS Be able to apply calculus Differentiation: review of standard derivatives, differentiation of a sum, function of a function, product and quotient rules, numerical values of differential coefficients, second derivatives, turning points (maimum and minimum) e.g. volume of a rectangular bo Integration: review of standard integrals, indefinite integrals, definite integrals e.g. area under a curve, mean and RMS values; numerical e.g. trapezoidal, mid-ordinate and Simpson s rule It is assumed that the student has completed the module MATHEMATICS FOR TECHNICIANS. In this tutorial you will learn how to differentiate more complicated epressions. D.J.Dunn

2 . REISION You should have covered integration in the core unit where you learned that integration is the reverse of differentiation. The standard integrals are: n a a d C n d ln C a a e d e C a sin a d cos C a cos a d sin C a tan d -ln C sin d - sin C cos d sin C tan d ln d n tan - C ln - C FINDING AREAS Lets consider how to find the area under the graph of y = f() = +. A graph of this function looks like this. If we solve the area by use of calculus the area would be precisely solved as follows. Over the range = to = the area is epressed as follows. A ( )d Carrying out the integration gives the following. Evaluating between limits we get the following. A ( A )d.. units A ( )d This is a precise answer and we will compare it with the results found in the following work. D.J.Dunn

3 GRAPHICAL METHODS Consider the same function again and this time more grid lines are shown. COUNTING RECTANGLES A simple but crude way to find the area under the graph is to count the rectangles. Each rectangle on the graph above has an area of unit. Count them up judging the divided ones to the nearest half. You should get an answer of about units depending on how good you are at doing it. MID-ORDINATE RULE The values of y corresponding to =, =, = and so on are called the ordinates. The values of y corresponding to =.5, =.5, =.5 and so on are called the mid-ordinates. Each column is approimately a rectangle w wide and h high. The area is approimately w h. The area under the whole graph is approimately A = w h + w h + w h +w h A = w(h + h + h + h ) Usually, as in this case, w = Putting in the mid-ordinate values we find the following. A = ( ) = units Clearly if we took more strips by say making w =.5, we would get a more accurate answer and in the limits as w becomes very small the answer will be the same as found by integrating. D.J.Dunn

4 TRAPEZOIDAL RULE Consider that each strip has a straight line joining the top corners as shown. The height at the middle is not quite the same as the mid-ordinate and is the average of the two ordinates. If h is the average then h = (A+B)/ h = (B+ C)/ h = (C+ D)/ h = (D+E)/ The area of each strip is wh = w(a+b)/ wh = w(b+ C)/ wh = w(c+ D)/ wh = w(d+e)/ The total area is A = (w/)[(a+b) + (B+C) + (C+D) + (D+E)] A = (w/)[(a+b + B+C + C+D + D+E] A = (w/)[(a + E) + (B+C+D)] Hence in our eample A = (/)[(+9) +(+7+)] = (/)(+6) = This is slightly larger than the correct figure but again, if smaller strips are used, the answer will be accurate. The above rule may be written as follows. w A First Last sum of the rest WORKED EXAMPLE No. Find the area under the graph of the function y = sin between the limits and radians using integration and the trapezoidal rule. Evaluating the ordinates and mid-ordinates at intervals of /8 produces the table and graph shown. Integration π A sinθ dθ π - cosθ cos π cos D.J.Dunn

5 Mid-ordinates w = / π A w(h h h h) ( ).68 Trapezoidal Rule w = / w A First Last sum of the rest π/ A units Note one answer is slightly large and the other slightly small. SIMPSON'S RULE The area is divided into an even number of strips. The ordinates are h, h... The area is calculated on the assumption that the curve joining neighbouring ordinates are a quadratic that passes through the mid ordinate. It follows that if the curve is a parabola, the area will be eact. The derivation is not given here as it is quite complicated but the result is as follows. I w first last sum of the even ordinates sum of the remaining odd ordinates WORKED EXAMPLE No. Find the area under the curve f() = between = and = using Simpson's Rule with eight strips and determine the error. A ( 8)d 8 () 8() 6.67 SIMPSON'S RULE () w I first last sum of the even ordinates sum of the remaining odd ordinates.5 I I The error is zero and it always is when the function is a quadratic. It follows that for a quadratic you only need two strips. D.J.Dunn 5

6 SELF ASSESSMENT EXERCISE No.. Find the area under the graph of the following functions using integration, the mid-ordinate rule and the trapezoidal rule. y = between the limits = and = 5 y = e between the limits of = and = 5. y = sin between the limits = and = 8 o.. Estimate the value of the definite integral I d by Simpson's rule using four strips. What is the error in the estimate? 5 (65. and by integration it is 6.8 giving an error of.5 units) In Engineering the area under the graph represents real things. For eample the area under a force distance graph represents the work done or energy used and the area under a pressure volume graph also represents work done during the compression or epansion of a gas. D.J.Dunn 6

7 WORKED EXAMPLE No. The pressure (p) and volume () during a gas epansion is related by the law p =. -.. Determine the work done when the volume is epanded from -6 m to -6 m. Use calculus and the trapezoidal rule to find the answer. INTEGRATION Area Work W. W Joules W pd but p. -. d. -.. TRAPEZOIDAL RULE w = -6 m w W W W 5 First Last sum of the rest ( Joules -6 D.J.Dunn 7

8 SELF ASSESSMENT EXERCISE No.. The electric current charging a capacitor is related to time by the following law. I = ( e -t/ ) Amps Calculate the charge Q (the area under the graph) between the limits t = and calculus and the trapezoidal rule. (Graph and ordinates are calculated for you) (Answer around Coulombs) t = 6 s. Use. Find the area (with units) under the following function between the limits = and = m using integration, the mid-ordinate rule and the trapezoidal rule. (Answers around 76.7 m ) y = + m. Find the area under the following function between the limits t = and t = s using integration, the mid-ordinate rule and the trapezoidal rule with steps of.s. v = t + e t m/s (Answers around. m). Find the area (with units) under the following function between the limits = and =. radian using integration, the mid-ordinate rule and the trapezoidal rule with steps of. radian T = cos (Nm) (Answers around.96 Joules) 5. Find the area (with units) under the following function between the limits = and = 5 m using integration, the mid-ordinate rule and the trapezoidal rule with steps of. p = ln N/m (Answers around 8.9 Nm or Joules) D.J.Dunn 8

9 . INTEGRATING POLYNOMIALS The rule for integrating a single polynomial is: a n n a n Note that no power shown against a variable (e.g. ), means and that this integrates as /. Anything raised to the power of zero is so a number on its own integrates e.g. could be written as and this integrates to / =. If the polynomial is a sum each term integrates separately. WORKED EXAMPLE No. Evaluate F() ( )d F() ( )d F() F() F() SELF ASSESSMENT EXERCISE No.. Integrate the following epressions. i. y = 5 / ii. y = /. Evaluate the following. i. ( )d (Answer 8.67) 5 ii. ( )d (Answer 5) D.J.Dunn 9

10 . INTEGRATING TRIGONOMETRIC EXPRESSIONS The standard integrals are given in the table and used in the following eamples. WORKED EXAMPLE No. 5 Evaluate A π sin d From the list of standard integrals we see sin = - cos so: A π cos cos π--cos -(-) - (-) WORKED EXAMPLE No. 6 Evaluate A π/ sin d From the list of standard integrals we see sin = ½ ¼ sin so: A sin A π/ sin d π/ π sinπ sin SELF ASSESSMENT EXERCISE No. Solve the following integrals. All angles are in radian... W d (Answer 5.9) π.. A sin( θ ) dθ (Answer ).5. A cos ( θ ) dθ (Answer.58).5. A sin ()d (Answer.7) D.J.Dunn

11 . INTEGRATING LOGARITHMIC AND EXPONENTIAL EXPRESSIONS The standard integrals are given in the table and used in the following eamples. WORKED EXAMPLE No. 7 Evaluate A e d From the list of standard integrals we see e a =e a /a so: e A e e - WORKED EXAMPLE No. 8 Evaluate I ln() d Using the standard integral I ln() ln.55 WORKED EXAMPLE No. 9 Evaluate I ln() d Using the standard integrals I ln() - d I ln() ln() ln() ln().8.97 I ln() ln() SELF ASSESSMENT EXERCISE No. 5. ln() I e d (-). sin() ln() I d (.) 5. e ln( ) I d () I d (.8). sin () - ln() D.J.Dunn

12 . AERAGE ALUES OF FUNCTIONS The mean value of a function can be defined accurately as the area under the graph divided by the base length of the graph. Consider a simple sine function such as used to describe a sinusoidal voltage. The function is v = sin(θ) where θ is the angle in radians and the amplitude. The plot is simply as shown. Let s find the mean value of over the range θ = to θ = π. Area π sin(θ ) dθ Area π cos( θ ) cos( π ) cos() ( ) Base length = π Mean = /π Repeat for the range to π π cos( θ ) cos( π ) cos() ( ) Area Mean = as epected for a full cycle. WORKED EXAMPLE No. Find the mean value of the function y = over the range = to = A d.667 Mean =.667/ = ROOT MEAN SQUARE ALUES (rms) This is mainly used in electrical engineering as a way of epressing alternating current and voltage as a meaningful quantity. Basically r.m.s. values are the equivalent D.C. values that will give correct power dissipation into a resistive load. The mean values of sinusoidal voltages and currents are zero and cannot be used to calculate electric power. The power formula is P = I R or /R or I so if we used the mean values of I or then the power formula will work. To get these we first find the mean value of the function for I or. For a sinusoidal voltage this would be done as follows. v = sin(θ) v = sin (θ) Over one cycle the plot is as shown. Note that all values of v are positive. The area under the graph for one complete cycle from θ = to θ =π is : Area π sin ( θ ) dθ π θ π Area cos( θ)sin(θ) cos( π ) sin( π ) cos( ) sin() π Area π The mean value is π /π =.5 remember is the amplitude. D.J.Dunn

13 If we did the same for current we would find the mean of the i is.5i. We could use this value to find the mean power using the P = I R or /R If we take the square root of these values we have the r.m.s. value. rms =.5 =.77 or I rms =.5I =.77 I or I If we use the r.m.s. value we could calculate the power with the formula P = rms I rms Amplitude For all sinusoidal quantities the r.m.s. are always given by Sometimes we have alternating quantities other than sinusoidal so the same basic process should be used. The mid ordinate rule, trapezoidal or Simpson s rule could be used to find the mean of a function. SELF ASSESSMENT EXERCISE No.6 An alternating voltage has a saw tooth form as shown. Calculate the r.m.s. value. The following is not stated as a requirement in the syllabus but students would do well to have a go at it. D.J.Dunn

14 6. INTEGRATION BY SUBSTITUTION A comple equation may be simplified with a substitution but it takes eperience to recognise these cases. WORKED EXAMPLE No. 7 Evaluate I d 8 A suitable substitution is z 8 Differentiate to get 7 Substitute back into the original equation I dz 8 Substitute for z 7 I 8 C 8 7z 8 C dz d 8 8 dz 8 d 8 WORKED EXAMPLE No. The voltage c across a capacitor when it discharges through a resistance is given by dc S C T where T is a time constant and s is the voltage at t =. dt Find the equation relating c with time t. Let S C = Differentiate and since S is a constant we find -d C = d d The equation becomes T Rearranging dt t d Integrating dt ln T The limits become clear when we substitute = S C dt T d T t T t dt ln C S C ln S C ln S d S ln S C Take antilogs and Rearrange ln e t T S C S C S - e T t t T D.J.Dunn

15 WORKED EXAMPLE No. Find I ( ) d Substitute z = and = ( z)/ Differentiate to get dz = -d and substitute d = -dz/ Substitute to get rid of all the terms - z dz z z I ( ) d (z ) dz z z z z I C C Now substitute back I C dz z - z 5 WORKED EXAMPLE No. Find I sin θ cosθ dθ Substitute sin(θ) = and noting that I sin θ cosθ dθ sin θ I C C d cos θ θ dsin d hence cos(θ) dθ = d dθ dθ SELF ASSESSMENT EXERCISE No. 7 Solve the following integrals. d. 5 I 5 5 C d. 6 d. I ln6 C sin I C sin. I sin θ cosθ θ dθ C d 5. I 6 ln( + ) + C D.J.Dunn 5

16 7. INTEGRATION BY PARTS The rule is without eplanation udv uv WORKED EXAMPLE No. 5 Find I e d Let = u and let e = dv du = d and v e e vdu and best shown with an eample. udv uv vdu e e d e e C e C WORKED EXAMPLE No. 6 Find I e d Let = u and let e e = dv du = d and v e e e e udv uv vdu d e d We must repeat the process for the integration ofe d e Let u = and dv e du = d v e e e e e e d d Now put the two parts together including the constant of integration. e e e e I e d C C e C e I C SELF ASSESSMENT EXERCISE No. 8 Solve the following integrals.. I ln() d I ln( ) C 6. I sin() d I cos() sin(). I e cos() d e I sin() cos() C e I e sin() d I cos() sin() C cos() C D.J.Dunn 6