LESSON 5: INTEGRATION BY PARTS (II) MATH FALL 2018

Transcription

1 LESSON 5: INTEGRATION BY PARTS (II MATH 6 FALL 8 ELLEN WELD. Solutions to In-Class Examples The following are solutions to examples done in class. Example. Suppose a turtle is moing at a speed of 8(t+ 3 ln(t+ / miles/hour. How far does the turtle trael in half an hour? Round your answer to the nearest thousandth. Solution: Here, we want to integrate the function 8(t + 3 ln(t + / because we need to know the distance accumulated by the turtle between and hours. This is described by / 8(t + 3 ln(t + / dt. u-substitution will not work here and so we need to use integration by parts. Howeer, we should first simplify our integral. By our rules about ln x, We write / 8(t+ 3 ln(t+ / dt = ln(t + / = ln(t +. / 8 (t+3 ln(t+ dt = / (t+ 3 ln(t+ dt. Next, by LIATE, we take u = ln(t +, which means d = (t + 3 dt. Obsere that u=t+ ( (t + 3 du=dt dt = u 3 dt = 4 u4 = 4 (t + 4. Hence, integrating d, ( (t + 3 dt = (t + 3 dt = ( (t + 4 = 4 (t + 4. Thus, our table becomes u = ln(t + du = t + dt d = (t + 3 dt ( = (t + 4

2 ELLEN WELD So we write / (t + 3 ln(t + dt ( / = ln(t + (t + 4 u = / (t + 4 ln(t + = / (t + 4 ln(t + / / ( (t + 4 dt } {{} t + du (t + 3 dt ( (t by ( = (t + 4 ln(t + 8 (t + 4 / = ( + = ( 4 3 ln ( = 34 3 ln 5.5 miles 4 ( ln ( / + ( 4 ( 8 + ( 4 ( 3 ln( 8 ( + 4 ln( + ( ln(7t + Example. A factory produces pollution at a rate of tons/week. How (7t + 3 much pollution does the factory produce in a day? Round your answer to the nearest hundredth. Solution: Our function measures output in terms of weeks but we are asked about the pollution produced in a day. Hence, the integral we must compute is /7 4 ln(7t + (7t + 3 dt. Howeer, this form looks somewhat unwieldy, so we introduce a small cosmetic change to our integral. Set x = 7t +, then dx = 7dt dx 7 = dt with Thus, /7 x( = 7( + = and x(/7 = 7(/7 + =. 4 ln(7t + 4 ln(x dt = (7t + 3 x 3 ( dx ln(x = dx. 7 x 3 So now, we need only compute the far right integral and we hae soled the problem.

3 AN UNOFFICIAL GUIDE TO MATH 6 FALL 8 3 Our integral cannot be computed using u-substitution, so we apply integration by parts. By LIATE, we take u = ln(x, and so d = dx. Since x3 = x dt = x 3 dt = x3+ = x = x = x, our table becomes u = ln(x d = x 3 dx Therefore, we write ln(x x 3 du = x dx = x dx = ln(x ( x u = ln(x x = ln(x x } {{ } = ln(x x = ln(x + x = ln( + ( = ln + ( + x dx 3 x x ( x ( }{{ x dx } du ( ln( + ( + ln( + ( = 8 ( ln( + + = ln(. tons Example 3. Suppose the number of Emerald Ash Borers (an inasie species in Indiana is increasing at a rate of E(t = 3t e t members/month where t is the number of months from the start of. (t = corresponds to Jan. and, for our purposes, we assume each month is of equal duration. What is the aerage population per month of Emerald Ash Borers in Indiana between March and May of? Round your answer to the nearest integer.

4 4 ELLEN WELD Solution: Because we are asked to find the aerage number of Emerald Ash Borers between March and May of, we will need to set up a definite integral. First, let s determine our bounds. Since t = is Jan., consider t <, Jan t <, Feb t < 3, March 3 t < 4 April 4 t < 5 May Hence, we ought to integrate from t = to t = 5. Our formula for aerage alue is then gien by 5 3t e t dt. Second, we integrate. Simplifying, our integral is 5 3t e t dx = 3 3t e t dx = t e t dt. This is an integration by parts integral where, by LIATE, we take u = t and thus our d = e t dt. We write Continuing, we hae u = t du = t t e t dt = t (e t u d = e t dt = e t 5 = t e t 5 te t dt which inoles another integration by parts problem. ( (e t (t dt du We find (. Let u = t and d = e t dt. Our table is then u = t du = d = e t dt = e t dt Hence, te t dt = t e t u 5 = te t 5 e t e t dt ( dt du = te t 5 et 5 = te t e t 5

5 AN UNOFFICIAL GUIDE TO MATH 6 FALL 8 5 Returning to our original problem, we hae t e t dt = t e t 5 te t dt ( = t e t 5 (te t e t 5 = t e t te t + e t 5 = (5 e t (5e 5 + e 5 [ ( e (e + e ] = 5e 5 e 5, 8 Emerald Ash Borers. Additional Examples Example 4. Find the area under the cure of f(x = x(x 3 6 oer the interal x 3. Solution: The area under a cure is gien by a definite integral. In this case, the definite integral is 3 x(x 3 6 dx. This can be computed using integration by parts, but it is easier to use u-substitution. Take u = x 3, then du = dx. Note u = x 3 u + 3 = x and so we can write 3 x(x 3 6 dx = u(3 u( (u + 3u 6 du. Since u = x 3, ealuating at x = and x = 3, we find u( = 3 = 3 and u(3 = 3 3 =. Hence, u(3 (u + 3u 6 du = u( 3 (u + 3u 6 du.

6 6 ELLEN WELD Finally, we ealuate: 3 (u + 3u 6 du = 3 (u 7 + 3u 6 du = 8 u u7 3 = 8 (8 + 3 ( 7 (7 8 ( (37 ( 656 = ( = ( = 656 = Example 5. Suppose the probability of a gold necklace haing a gold purity of x percent (so x is gien by P (x = e3 e 3 4 xe3x. Find the probability that a gold necklace has a purity of at least 75%. Round your answer to the nearest percent. Solution: We want our gold necklace to hae a purity of at least 75%. Hence, 75 x x. Thus, the question comes down to computing e 3 e 3 4 xe3x dx = e3 xe 3x dx. e 3 4 For the moment, let s focus on the integral and forget about the constant out front. We want to sole xe 3x dx. This is an integration by parts problem, which means it comes down to choosing the correct u and d. By LIATE, let u = x and d = e 3x dx. Integrating d, we get ( = d = e 3x dx u=3x du=3 dx = 3 eu du = 3 eu = 3 e3x. So our table becomes u = x du = dx d = e 3x dx = 3 e3x

7 We write AN UNOFFICIAL GUIDE TO MATH 6 FALL 8 7 xe 3x dx = x ( 3 e3x u = 3 xe3x + e 3x dx 3 = 3 xe3x e3x = 3 xe3x e3x ( ( 3 e3x } {{ } dx du = 3 e3 e3 [ 3 (e3( ] e3( Hence, e 3 e 3 4 = 4 e3( + e3( 3 e3 e3 ( = 4 + ( e 3( 3 + e 3 xe 3x dx = So the probability is 8%. [( e3 e ( e 3( 3 + ] e 3.8