MATH 250 TOPIC 13 INTEGRATION. 13B. Constant, Sum, and Difference Rules

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1 Math 5 Integration Topic 3 Page MATH 5 TOPIC 3 INTEGRATION 3A. Integration of Common Functions Practice Problems 3B. Constant, Sum, and Difference Rules Practice Problems 3C. Substitution Practice Problems 3D. Area Under a Curve Practice Problems 3E. Integration Summary s to Exercises and Practice Problems

2 Math 5 Review Topic 3A Page A. Integration of Common Functions There are two types of integrals, indefinite and definite. An indefinite integral is written as f(x) dx, and a definite integral is written as b a f(x) dx. Evaluating an indefinite integral requires finding a function F (x) whose derivative is the integrand f(x). That is, f(x) dx = F (x) where F (x) =f(x). F(x) is called an antiderivative of f(x). For example xdx= x + c (c representing a constant) since d dx (x + c) =x. Similarly cos xdx=sinx + c since d (sin x + c) = cos x. dx The process for evaluating a definite integral is slightly different. First, find an antiderivative. Then evaluate the antiderivative at its limits (a and b) and subtract the result. Thus, b b f(x) dx = F (x) = F (b) F (a). a a Example: 3 xdx=(x + c) 3 3 =8. =(3 + c) ( + c) Note: It is customary to leave out the c with definite integrals because the c always cancels itself out in the subtraction.

3 Math 5 Review Topic 3A Page 3 Exercise A.: Find the antiderivatives of the following basic functions. What is interesting is that in Calculus II you will cover five-six different methods of integration. In the end, however, all of the methods reduce to evaluating one of these basic integrals. f(x) dx x n dx, n dx x dx = x e x dx sin xdx cos xdx sec xdx sec x tan xdx csc xdx csc x cot xdx F s

4 Math 5 Review Topic 3B Page 4 Rewriting Integrals to Fit Basic Forms Sometimes an integral must be rewritten before it can be integrated. For example, x dx = x dx = x + c or x + c or xdx= x / dx = 3 x3/ + c or or e x e dx = e x dx = e x + c. x ( + tan x) dx = sec xdx= tan x + c. Each integrand was rewritten in one of the basic forms covered in Exercise A.. This is our first integration strategy. Always try to rewrite an integral in one of the basic forms. (Additional strategies will be discussed later.) Practice Problems 3A.. x xdx 3A.. 3A.3. 3A.4. π/ π/4 dx e x cos θdθ t 3 dt B. Constant, Sum, and Difference Rules We have a few basic rules for integration that are similar to the constant, sum, and difference rules for differentiation. Unfortunately though, there is

5 Math 5 Review Topic 3B Page 5 no product, quotient or chain rule analog for integration. The first rule is the constant rule. cf(x) dx = c f(x) dx. This means you can ignore the multiplicative constant when finding an antiderivative. Examples: 3sinxdx=3 sin xdx=3( cos x)+c = 3cosx + c 8 πx /3 dx = π 8 x /3 dx = π x+/3 3 + = π 3 5 x5/3 8 8 = 3π 5 [85/3 5/3 ]= 96π 5 Exercises: B.. Can you find sec xdx, B.. B.3. π/ π/4 4 sin x dx, e x+ dx? Hint: e x+ = e x e Another important rule involves the integral of the sum or difference of two functions. (f(x) ± g(x)) dx = f(x) dx ± g(x) dx Here you find the antiderivatives of the individual functions and add or subtract the result.

6 Math 5 Review Topic 3B Page 6 Examples:.. 3. (x / + e x ) dx = x / dx + =x / + e x e x dx = [ / / ]+[e e] = +e e ( ) dx x + tan x = x + tan xdx =ln x + ln cos x + c x(x )dx = x 5/ dx x / dx = 7 x7/ 4 3 x3/ + c Practice Problems. Can you find the following? 3B.. 3B.. 3B.3. 3B.4. π/4 π/6 (cos x tan x +)dx ( x 3 + x ) + dx x π/3 π/6 (cot x +)dx tan xdx 3B.5. 3B.6. (e x cos x) dx (x ) dx

7 Math 5 Review Topic 3C Page 7 3B.7. e x e x e x dx C. Substitution Why Substitution Works You ve already seen how to use the method of substitution to evaluate integrals. This method of integration compensates for the chain rule in differentiation. Let s start our review of substitution by looking at some derivatives involving the chain rule. (Review Topic E) d dx ex + = e x + (x) d sin(ln x) = cos(ln x) dx x d dx (x3 +3x ) / = (x3 +3x ) / (3x +3) Since integration can be considered as the reverse of differentiation, if you were asked to evaluate e x + (x) dx, you can see from above the answer is e x + + c. Similarly,. and 3. above imply that cos(ln x) dx = sin(ln x)+c, x and (x3 +3x ) / (3x +3)dx =(x 3 +3x ) / + c. If we didn t know the answer though, how could we work these problems? We make a substitution.

8 Math 5 Review Topic 3C Page 8 In algebra you probably made substitutions to simplify expressions. example, solving the following equation x /3 x /3 = appears pretty hard. If we let u = x /3 though, the problem becomes u u =; Substitution helped us recognize that the equation was quadratic and can be solved. This same idea of substitution applies to integration. Method of Substitution Let s go back to e x + (x) dx. Suppose u(x) =x +. Then du =x du =xdx. This means dx u {}}{ du {}}{ e x + xdx= e u du =e u + c = e x + + c. For WARNING! When using substitution it is important that your substitution accounts for all of the x-terms and dx in the starting integral. The new integral should only be in terms of u and du (no x s!). Now try cos(ln x) dx x. Let u(x) =lnx. Then du = dx and you end up integrating x cos udu=sinu + c = sin(ln x )+c. Exercise C.. What substitution do you think you would use with (x3 +3x ) / (3x +3)dx?

9 Math 5 Review Topic 3C Page 9 The method involved with substitution is:. Choose a substitution u = f(x).. Compute du dx. 3. Rewrite integral in terms of u and du. (Nox-terms left) 4. Evaluate integral in terms of u. 5. Replace u by f(x). The result is the antiderivative in terms of x. Note: As integrals get more complicated, we encourage you to check them by differentiating. This also allows you to practice derivatives. Each of the following examples has e f(x) in the integral. Let u(x) =f(x), determine du, substitute and integrate. Don t forget to put the final answer in terms of x! Exercises: C.. e sin x cos xdx C.3. C.4. e x/ dx x / e 3x +x (6x +)dx Now consider e x + xdx. Again we let u = x +. But now du =xdx and we don t have the in the integral. Recall cf(x) dx = c f(x) dx.

10 Math 5 Review Topic 3C Page So, solving for xdx= du we have e x + xdx= e u du = e u du = eu + c = ex + + c Note: Since the integral contained only xdx instead of xdx, wouldn t you expect the final answer to be as much. Each of the following examples involves a trigonometric function with an argument f(x). i.e., sin(f(x)), tan(f(x)), etc. Again let u = f(x), determine du, substitute, and integrate. Exercises: C.5. x sin(x +)dx C.6. C.7. (e x +)sec (e x + x) dx (x +)cos(x 3 +3x ) dx Consider a more complicated example. Evaluate x x +dx. Let u = x +. Then du = dx and x = u. (Remember, all x-terms must change to u-terms!) Thus x x +dx = (u )u / du = (u 3/ u / )du. Since everything to do with x was changed to u, this means you substituted correctly.

11 Math 5 Review Topic 3C Page Exercise C.8. Finish the preceding problem. How can we recognize when substitution won t help? Consider the problem x e x + dx, when x>. If we let u = x +, then du =xdx or du = xdx. x e x + dx = e u {}}{ {}}{ du x e x + xdx To get rid of x, we write u =x, or x = u. (We take u instead of u since x>.). This means x e x + dx = u e u du. We have substituted correctly in the sense that all x terms have been changed to u. However, the new integral is just as complicated as the old one. The main goal of substitution is to change the starting integral into another one that you can actually integrate. In this case, our substitution u = x + did not help us. We must either try a different substitution or use another method of integration (that you will learn in Calculus II). Substitution also effects the limits of integration. An integral can t be written in terms of u with limits intended for x. We must use the substitution to change the x-limits to u-limits. Example: Evaluate 3 e x + xdx. Let u = x +. Then if x =, u = +=, and if x =3, u =3 +=. Thus 3 e x + xdx= e u du = e u = e e.

12 Math 5 Review Topic 3C Page How do you know what to substitute? Suppose your integrand has a term of the form: e f(x), f(x) dx, trig(f(x)), (f(x))p. A general rule of thumb is let u = f(x), determine du, and get rid of all the x s and dx. As a final note, observe that tan xdxcan be evaluated by a substitution. sin xdx tan xdx= cos x. Let u =cosx du = sin xdx = du = ln u + c u = ln cos x + c =ln sec x + c } These are both correct forms. Why are they equivalent? Practice Problems. Evaluate the following. Change limits when necessary. Not all problems require substitution. dx 3C.. x 3C.. 3C.3. 3 xdx x x x dx 3C.4. 3C.5. x(x ) dx π/ cos ( ) x 3 dx

13 Math 5 Review Topic 3C Page 3 3C.6. e x dx e x + 3C.7. π/6 tan (x) dx To finish up this section, let s try to make your life a little easier with some integrations you should be able to do quickly and (hopefully) without substitution. For example e 3x dx dx, sin xdx,. These integrals x are of the form e ax dx dx, trig(ax), x + k The substitution for the first two is u = ax and du = adx. This leads to du. For the last integral, let u = x + k which gives a result a ln x + k + c.

14 Math 5 Review Topic 3C Page 4 Exercise C.9. Without making a formal substitution, try completing the following table. sin xdx e 3x dx e x/ dx sec 3x tan 3xdx cos πx dx dx x + dx dx x dx a= F(x) = cos x + c

15 Math 5 Review Topic 3D Page 5 D. Area Under a Curve One of the first applications covered in your previous study of definite integrals was finding the area under a curve. Actually, the integral b a f(x) dx can be defined as the area between the curve y = f(x) and the x-axis for a x b. y = f(x) Area = b a f(x) dx a b Fig. 3D.. As you (hopefully!) remember, we first evaluated b equal to the limit of Riemann sums. The idea is as follows. Arguing intuitively, we formed rectangles as below. a f(x) dx by setting it y = f(x) height = f(x i ) width = x a b a x i b This is one rectangle from the picture on the left. Fig. 3D.. The exact area, i.e. the integral, is approximated by adding the area of the rectangles (in this example there are four rectangles). The area of each rectangle = (height)(width). The height = y = f(x i ), where x i is some x-value in the base of the rectangle, and the width = x. For the drawing on the left in Fig. 3D., we write b 4 Area = f(x) dx = (approximately) f(x i ) x. a i=

16 Math 5 Review Topic 3D Page 6 The expression on the right in the line above is called a Riemann sum with n = 4. Instead of forming a Riemann sum with 4 rectangles, we could have used 6,, 6 rectangles, etc. The more rectangles we form, the better our approximation will be. If we take the limit as n, where n is the number of rectangles, we obtain the exact area. Intuitively then, the integral b a f(x) dx can be thought of as the sum of an infinite number of rectangles. The integral sign represents the sum, f(x) represents the height of the rectangle, and dx corresponds to the width of the rectangle. The dx also indicates that the variable of integration or independent variable is x. Fortunately, the Fundamental Theorem of Calculus allows us to find the area (evaluate the integral) by using antiderivatives instead of trying to add an infinite number of rectangles. Example 3D.. Find the area under f(x) =sinx and the x-axis from to π. f(x) =sinx π A typical approximating rectangle is shown. Fig. 3D.3. Area = (intuitively) the sum of the rectangles from x =tox = π π ht width {}}{{}}{ π = sin x dx = cos x = cos π ( cos ) = ( ) + =. What happens if the curve dips below the x-axis? For example, consider

17 Math 5 Review Topic 3D Page 7 f(x) =sinx, π x π. f(x) =sinx π/ π Fig. 3D.4. First, notice that on the interval π x, π/ f(x) dx = π/ sin xdx= cos x π/ =. A negative answer is not surprising since f(x i ), the height of an approximating rectangle, is negative when π x. Thus, to find the area enclosed by f(x) =sinx, π x π, wemust: (i) split the x interval into subintervals where f(x) or where f(x) ; (ii) (iii) integrate f(x) over each subinterval; take the absolute value of each resulting integral. In Fig. 3D.4, the area is: Area = π/ = +=3. π sin xdx + sin xdx Warning: If you are asked simply to evaluate an integral, then just evaluate it and don t worry about whether f(x) is positive or negative. For example, π π sin xdx= cos x = ( ) () =. π/ π/

18 Math 5 Review Topic 3D Page 8 Notice from Fig. 3D.4 that the integration process yields the net value. That is, the negative part, π/ positive, so that we only have left sin xdx, cancels π π/ sin xdx. π/ sin xdx,whichis Bottom line, if you are asked to evaluate an integral, just plunge ahead. (That s why in problem 3A.4, t 3 dt =.) If you are asked to find the area, you must worry about whether f(x) or and proceed accordingly. It is easy to extend these ideas to finding the area between two curves. Example 3D.. Find the area bounded by the curves f(x) =x and g(x) =x. First, determine where the curves intersect. Do this by setting f(x) =g(x). Thus, x =x x x = x(x ) = the curves intersect at x =andx = as indicated below. (, 4) Fig. 3D.5. You can use two approaches to find the desired area. First, find the area under the upper curve g(x) =x from x =tox = and then subtract the area under the lower curve f(x) =x from x =tox =.

19 Math 5 Review Topic 3D Page 9 g(x) =x f(x) =x (, 4) = Fig. 3D.6. Using integrals to represent the above areas, we have xdx x dx = x x3 3 =4 8 3 = 4 3. For the second approach, using Fig. 3D.5 you can draw one of the rectangles that would approximate the area and then write the corresponding integral. g(x) Height of rectangle = g(x i ) f(x i ) =x i x i Width of rectangle = dx f(x) Area of rectangle = (x i x i )dx Fig. 3D.7. The rectangles are formed when x. Thus, area = [g(x) f(x)] dx = (x x ) dx = x x3 3 which, of course, gives the same answer as the first approach. = 4 3, A little hint: Notice how much it helps to sketch the functions no matter which approach you use.

20 Math 5 Review Topic 3D Page Another hint: In the second approach above, you are finding the height of the rectangle by saying the height = upper curve lower curve. This idea of upper curve lower curve works no matter where the curves are. That is, the curves could both be below the x-axis, one curve could be above and the other below, or both could be above the x-axis like Fig. 3D.7. It doesn t matter. The height of the rectangle is always the upper curve lower curve and is always positive. Consider the following example. Example 3D.3. g(x) =x 3. Find the area bounded by the curves f(x) =x and Setting f(x) =g(x) we see that x = x 3 x 3 x = x(x ) = the curves intersect at x =,,. Fig. 3D.8. Notice that for <x<, g(x) =x 3 is above f(x) =x. However, for <x<, f(x) =x is above g(x) =x 3. This means we should divide the problem into two parts: finding the area when <x< and finding the area when <x<. Looking at Fig. 3D.8 and remembering that the height of an approximating rectangle is always the upper curve lower curve, we see that the area when <x< is given by: (x 3 x) dx = x4 4 x ( = 4 ) = 4. For <x<, the area is: (x x 3 ) dx = x x4 4 = ( ) = 4 4.

21 Math 5 Review Topic 3D Page Adding the two results: =. Of [ course, we could have used symmetry and said that the area is ] (x x 3 ) dx to arrive at. There will be times in Calc II when it will be necessary to consider integrals where y is the independent variable and x is the dependent variable. Example 3D.4. Find the area in the first quadrant bounded by x = y, the y-axis, and y =. x = y ; y Fig. 3D.9. Now our approximating rectangles are horizontal. The height or length is some function x = h(y), and the width is dy. In Fig. 3D.9, the area of an approximating rectangle is (length)(width) = y dy. Now add up all the rectangles by integrating. Since dy varies from y =toy =, area = We offer one final example to you. y dy = y3 3 = 3. Example 3D.5. Find the area bounded by the curves y = x, y = 3 x, and y = x 3.

22 Math 5 Review Topic 3D Page This problem is definitely more challenging. First, draw a picture. y = x y = 3 x A B y = x Fig. 3D.. To find the point of intersection A, set x =4,y =3. ForB, set x = x 3 x = 3 x 8=x x = x 3/ x =6,y =. Now draw an arbitrary rectangle with width dx. In trying to do this, we see that there are two possible rectangles which can be described, as below. y = x y = 3 x (4, 3) (6, ) y = x Fig. 3D..

23 Math 5 Review Topic 3D Page 3 The ( rectangle ) on the left has height upper curve-lower curve = 3 x x, for x 4 and the rectangle on the right has height ( 3 = x x ) for 4 x 6. Since we have two different descriptions, we 3 need two integrals. I II Fig. 3D.. 4 ( 3 x Area = I + II = x 3 = 3 x 3/ x ( = 8 8 ) + ( ln 6 6) 3 ( ) 3 =+ln. ) 6 ( dx + 4 x x ) dx 3 ) ( ln x x ( ln Exercise D.. Find the area enclosed by the three curves in Example 3D.5 using horizontal rectangles. ) Practice Problems. curves. Find the area of the regions bounded by the given 3D.. f(x) = cos x, x-axis, x = π 6,andx = π 3

24 Math 5 Review Topic 3E Page 4 3D.. f(x) = x,g(x) =x and x = e 3D.3. f(x) = cos x, g(x) =sinx, x π 3D.4. Area of the region to the right of the y-axis and bounded by x =y y. E. Integration Summary If you feel that integration is not as straightforward as differentiation, you are correct. Complex processes (like limits and integration) are always more difficult to master. Here are a few guidelines that can help when evaluating integrals. ) Memorize the basic list of integration formulas (Review Topic 3A, Exercise A.); this includes power, exponential, log, and trigonometric rules. ) Try to make a substitution u = g(x) that enables the integrand to transform to a basic formula. Learning what to choose for u comes after much time and practice. 3) If u substitution doesn t apply, try to alter the integrand. This may involve trig identities or various algebraic manipulations (see Exercise E. below). Imagination helps but good algebra/trig skills help more. 4) CHECK more answers to integrals by differentiating (rather than by looking in the back of the book). The more you analyze differentiation, the better you get at integration. A real challenge in integration involves evaluating integrals that are similar in appearance yet require different methods. Consider the following example.

25 Math 5 Review Topic 3E Page 5 Example : Evaluate the following: x(x a) ) dx b) x x dx c) Solution: x x dx a) Expand, then integrate using the power rule. x(x ) dx = (x 5/ x / ) dx = 7 x7/ 3 x3/ + c b) Substitution. Let u = x. x x dx = u / du = 3 (x ) 3/ + c c) Substitution. Let u = x. x (u+) u {}}{{}} / { du {}}{ x dx = x x dx = (u +) u / du = (u 5/ +u 3/ + u / ) du In any substitution, you must acount for the entire integrand. = 7 (x )7/ (x )5/ + 3 (x )3/ + c Now it s time to see if you are ready to begin Calc II. Exercise E.: Evaluate each set of integrals. x x I. dx and x x dx II. xe x dx and xe x dx III. e x dx and +ex dx +ex

26 Math 5 Review Topic 3E Page 6 IV. cos x dx and +sinx cos x +sinx dx Conclusion In Calc II you will learn more ways to evaluate an integral. Regardless of the method, the intent (strategy) is always the same; manipulate and find a form that can be integrated. We ve done all we can. The rest is up to you. Practice doesn t make one perfect, but it sure helps. Good luck in Calc II. Beginning of Topic Review Topics 5 Skills Assessment

27 Math 5 T3 Exercise A s Page 7 s: f(x) dx x n dx x dx = e x dx sin xdx cos xdx sec xdx dx x sec x tan xdx csc xdx csc x cot xdx F (x) x n+ + c, n n + ln x + c e x + c cos x + c sin x + c tan x + c sec x + c cot x + c csc x + c Question: Why is c included in the antiderivative of an indefinite integral? a) My instructor told me to put it there. b) c stands for correct. I want my instructor to know I ve checked my answer. c) Since the derivative of any constant is, the inclusion of c suggests the antiderivative is not a single function but rather a family of functions that differ only by a constant. Return to Review Topic

28 Math 5 Problems 3A s Page 8 3A.. x xdx= xx / dx = x 3/ dx = x5/ 5 + c = 5 x5/ + c 3A.. π/ π/4 cos θdθ =sinθ π/ π/4 =sin π sin π 4 = 3A.3. dx e x = e x dx = e x + c 3A.4. t3 dt = t4 4 = ()4 4 ( )4 4 = 4 4 = Question: If a definite integral measures the area under the curve, why does it appear there is no area under t 3 over [, ]? (See Review Topic 3D.)

29 Math 5 T3 Exercise B s Page 9 Can you find B.. sec xdx B.. B.3. π/ π/4 4 sin x dx e x+ dx s: B.. sec xdx= sec xdx= tan x + c Return to Review Topic B.. π/ π/4 4 π/ sin dx =4 csc xdx x π/4 π/ =4( cot x) =4 [ cot π ( cot π ] 4 ) π/4 =4[ +]=4 Return to Review Topic B.3. e x+ dx = e x edx= e e x dx = e x+ + c Return to Review Topic

30 Math 5 Problems 3B s Page 3 3B.. π/4 π/6 4(cos x tan x +)dx =4 =4[ cos x + x] π/4 π/6 π/4 π/6 =4 (sin x +)dx [ cos π 4 + π ( 4 cos π 6 + π )] 6 = + 3+ π 3 3B.. x 3 + x + dx = x ( x + x + ) x dx = x3 3 + x +ln x + c 3B.3. π/3 π/6 (cot x +)dx = cot x π/3 π/6 π/3 π/6 csc xdx = cot π ( 3 cot π ) 6 = B.4. tan xdx= (sec x ) dx = tan x x + c

31 Math 5 Problems 3B (Continued) s Page 3 3B.5. (e x cos x) dx = e x sin x + c 3B.6. (x ) dx = (x 4 x +)dx = x5 5 x3 3 + x + c 3B.7. e x e x dx = e x (e x ) dx = e x x + c

32 Math 5 T3 Exercise C s Page 3 C.. C.. What substitution do you think you would use with (x3 +3x ) / (3x +3)dx? e sin x cos xdx C.3. e x/ dx x / C.4. e 3x +x (6x +)dx s: C.. u = x 3 +3x anddu =(3x +3)dx u / du = u / + c = x 3 +3x +c Return to Review Topic C.. u =sinx, du =cosxdx e u du = e u + c = e sin x + c Return to Review Topic C.3. u =x /, du = dx x / e u du = e u + c = e x/ + c Return to Review Topic C.4. u =3x +x, du =(6x +)dx e u du = e u + c = e 3x +x + c Return to Review Topic

33 Math 5 T3 Exercise C (Continued) s Page 33 C.5. sin(x +)xdx C.6. sec (e x + x)(e x +)dx C.7. cos(x 3 +3x )(x +)dx s: C.5. u = x +,du =xdx sin u du = sin udu= ( cos u)+c = cos(x +)+c Return to Review Topic C.6. u = e x + x, du =(e x +)dx sec udu= tan u + c = tan(e x + x)+c Return to Review Topic C.7. u = x 3 +3x, du =(3x +3)dx =3(x +)dx cos u du 3 = cos udu= 3 3 sin u + c = 3 sin(x3 +3x ) + c Return to Review Topic

34 Math 5 T3 Exercise C (Continued) s Page 34 C.8. x x +dx = 5 (x +)5/ 3 (x +)3/ + c Return to Review Topic C.9. sin xdx a = e 3x dx a =3 e x/ dx a = sec 3x tan 3xdx a =3 cos πx dx a= F(x) = a = π cos x + c 3 e3x + c e x/ + c sec 3x + c 3 sin πx + c π dx dx ln x + + c x + dx dx ln x + c x Return to Review Topic

35 Math 5 Problems 3C s Page 35 3C.. u = x, du = dx du u = u / du =u / + c =(x ) / + c / 3C.. u = x, du = dx. Limits: x = u =;x =3 u =. u + du = (u / + u / )du u/ = 3 u3/ +u / = 3 ( ) + = = 3 ( 8) ( ) 3 + 3C.3. u = x, du =xdx u /du = u / du = 3 u3/ + c = 3 (x ) 3/ + c 3C.4. No substitution needed! Rewrite as (x 3/ x / ) dx = 5 x5/ 3 x3/ + c

36 Math 5 Problems 3C (Continued) s Page 36 3C.5. u = x 3 ; du = 3 dx. Limits: x = π u = π ; x = u =; 3 3 π/3 cos udu= 3 π/3 sin u = 3 (sin π ) 3 sin = C.6. u = e x +,du = e x dx du u =ln u + c = ln ex + + c 3C.7. u =x, du =dx. Limits: x = π 6 u = π ; x = u =. 3 π/3 tan u du = π/3 π/3 tan udu = (sec u ) du = π/3 (tan u u) = [( tan π 3 π ) ] tan 3 = ( π ) 3 = 3 6 (3 3 π)

37 Math 5 T3 Exercise D. s Page 37 D.. The area is described by Fig. 3D., but now we must use horizontal rectangles. Again, there are two possible descriptions. Also, as in Example 3D.4, we must solve for x in terms of y for all the curves. y = x or x = y (4, 3) (6, ) y = 3 4 x or x = 9 y y = x 3 or x =3y 4 6 ( The length of the upper rectangle is right curve left curve = y 4 ) 9 y for y 3, and the length of the lower rectangle is (3y 49 ) y for y. Thus, Area = (3y 49 ) 3 ( y dy + y 4 9 y ( ) 3y = 4y3 ) + ( ln y 4y ( = 6 3 ) ( + ( ln 3 4) 7 ( ) 3 =+ln, ) dy ln 3 7 which is the same answer obtained in Example 3D.5 using vertical rectangles. ) Return to Review Topic

38 Math 5 Problems 3D s Page 38 3D.. Area = π/3 π/6 cos xdx=sinx π/3 π/6 =sin π 3 sin π 6 = 3 3D.. g(x) =x f(x) = x A x = e To find the point A, setf(x) =g(x) x = x x3 = x =, y =. Area = sum of rectangles e = (x x ) dx = x + x e = (e ) + e ( = e4 + e 3 + )

39 Math 5 Problems 3D (Continued) s Page 39 3D.3. A sin x cos x π π B To find the x coordinates of the points A and B, set sin x =cosx x = π 4, 5π. Since the area involves the upper curve lower curve, 4 three integrals are necessary (see Example 3D.3). Thus, area = π/4 (cos x sin x) dx + + π 5π/4 =(sinx +cosx) 5π/4 π/4 (cos x sin x) dx π/4 +( cos x sin x) (sin x cos x) dx 5π/4 π/4 =sin π 4 +cosπ 4 (sin + cos ) [cos 5π4 ( +sin5π4 cos π ) ] 4 +sinπ 4 +(sinπ ( +cosπ) sin 5π ) 4 +cos5π 4 = [ ( )] + ( ) = 4 +(sinx +cosx) π 5π/4

40 Math 5 Problems 3D (Continued) s Page 4 3D.4. In this problem, it is easier to use horizontal rectangles like Example 3D.4. x=y y First, find where the curve intersects the y-axis. y y = y( y) = y =,y= Area = [(y y ) ]dy = y y3 3 =4 8 ( ) 3 = 4 3

41 Math 5 T3 Exercise E. s Page 4 E.-I. ( x x dx = x x ) dx x ( ) = x x dx =ln x + x + c and x ( x dx = x ++ ) x dx = x + x +ln x + c for help with long division, see Review Topic. E.-II. xe x dx = e u du = + c ex and u = x, du =xdx xe x dx This integral can t be evaluated until you learn (in Calc II) a method called Integration by Parts.

42 Math 5 T3 Exercise E. (Continued) s Page 4 E.-III. and e x du +e dx = x u du =ln +e x + c =ln(+e x )+c u =+e x, du = e x dx Why can the absolute value symbol be dropped? +e dx = x +e e x e x dx = x e x e x + dx du = u = e x + u = ln e x + + c = ln(e x +)+c For more examples on multiplying by a form of, see Review Topic. E.-IV. cos x du +sinx dx = u =ln +sinx + c =ln(+sinx)+c u =+sinx, du =cosxdx and cos x sin +sinx dx = x +sinx dx = = x +cosx + c ( sin x) dx