Assignment 11 Assigned Mon Sept 27

Size: px

Start display at page:

Download "Assignment 11 Assigned Mon Sept 27"

Lizbeth Williams
5 years ago
Views:

1 Assignment Assigned Mon Sept 7 Section 7., Problem 4. x sin x dx = x cos x + x cos x dx ( = x cos x + x sin x ) sin x dx u = x dv = sin x dx du = x dx v = cos x u = x dv = cos x dx du = dx v = sin x = x cos x + x sin x + cos x + C. Section 7., Problem 5. x ln x dx = x ln x x= x= x dx = ( ln ln ) x= 4 x x= u = ln x du = dx/x dv = x dx v = x = ln Section 7., Problem 7. tan y dy = y tan y y y + dy u = tan y dv = dy du = dy v = y y + dy = y tan y ln y + + C. 3

2 Section 7., Problem. x 3 e x dx = x 3 e x 3x e x dx = x 3 e x 3x e x + 6xe x dx = x 3 e x 3x e x + 6xe x 6e x dx u = x 3 dv = e x dx du = 3x dx v = e x u = 3x dv = e x dx du = 6x dx v = e x u = 6x dv = e x dx du = 6 dx v = e x = x 3 e x 3x e x + 6xe x 6e x + C. Section 7., Problem 3. First we compute the indefinite integral. x sin x dx = x cos x + cos x dx = x cos x + sin x + C. u = x du = dx dv = sin x dx v = cos x In order to compute the area, we need to be sure that we count regions below the x-axis positively. (a) (Area for x π) = π x sin x dx = ( x cos x + sin x) x=π x= = π. (Note that cos π =.) (b) Here we need a negative sign because the region is entirely below the x-axis. (Area for π x π) = (c) We re back above the x-axis. (Area for π x 3π) = π π x sin x dx = ( x cos x + sin x) x=π x=π = (π cos π sin π) (π cos π sin π) = 3π. 3π π x sin x dx = ( x cos x + sin x) x=3π x=π = ( 3π cos 3π + sin 3π) ( π cos π + sin π) = 5π. 3

3 (d) Ignoring the sign for the moment, we have (n+)π nπ x sin x dx = ( x cos x + sin x) x=(n+)π x=nπ = ( (n + )π cos(n + )π + sin(n + )π) ( nπ cos nπ + sin nπ) = ±(n + + n)π = ±(n + )π. So (Area for nπ x (n + )π) = (n + )π. Section 7., Problem 37. First we compute the indefinite integral. (This is almost the same as Example 4 in the text.) e t cos t dt = e t sin t + = e t sin t e t cos t e t sin t dt e t cos t dt u = e t du = e t dt u = e t du = e t dt Now we can solve for the integral that we want, e t cos t dt = e t sin t e t cos t. The average A is (note that the integrand has an extra in it) A = π π e t cos t dt = π ( ) e t sin t e t cos t π = π e π + e The numerical value is approximately.588. dv = cos t dt v = sin t dv = sin t dt v = cos t = e π. π Section 7., Problem 4. x n e ax dx = xn e ax n a a x n e ax dx u = x n du = nx n dx dv = e ax dx v = a eax 33

4 Assignment Assigned Fri Oct Section 7.4, Problem. We set 5x 3 (x 3)(x ) = A x 3 + B x. So we need to solve 5x 3 = A(x ) + B(x 3) = (A + B)x (A + 3B). A + B = 5, A + 3B = 3. Doubling the first equation and subtracting it from second equation eliminates AS and gives B = 3. Then the first equation gives A =. We then check that A = and B = 3 is a solution to the second equation, too. So 5x 3 (x 3)(x ) = x x. Section 7.4, Problem 3. We set x + 4 (x + ) = A x + + B (x + ). x + 4 = A(x + ) + B = Ax + (A + B). So we need A =, A + B = 4. So B = 3 and we obtain x + 4 (x + ) = x (x + ). 34

5 Section 7.4, Problem 6. (If you re clever, you ll notice that you can cancel a factor of z from the top and the bottom, so the problem is really to find a partial fraction expansion for /(z z 6). But I won t be clever and will do the problem without first canceling the z.) First we factor the denominator z 3 z 6z = z(z + )(z 3). (It s obvious to pull off a factor of z, and then you can either factor the polynomial z z 6 directly by inspection, or use the quadratic formula to find its roots.) We set z z(z + )(z 3) = A z + So we need to solve the equations B z + + C z 3. z = A(z + )(z 3) + Bz(z 3) + Cz(z + ) z = A(z z 6) + B(z 3z) + C(z + z) z = (A + B + C)z + ( A 3B + C)z + ( 6A). A + B + C =, A 3B + C =, 6A =. The last equation gives A =, so the first two equations become B + C =, 3B + C =. Adding three times the first equation to the second gives 5C =, so C = 5. The first equation gives B = C = 5. So we have z z 3 z 6z = 5z + + 5z 5 35

6 Section 7.4, Problem. First we need the partial fraction expansion. x + 4 x + 5x 6 = x + 4 (x )(x + 6) = A x + B x + 6. x + 4 = A(x + 6) + B(x ) = (A + B)x + (6A B). We solve A + B =, 6A B = 4, and find that A = 5 7 and B = 7. Then x + 4 x + 5x 6 dx = 5/7 x + /7 x + 6 dx = 5 7 x dx + 7 x + 6 dx = 5 7 ln x + ln x C. 7 36

7 Section 7.4, Problem 7. First we have to do long division to get the degree of the numerator to be smaller than the degree of the denominator. So x + x + ) x 3 x x 3 + x + x x x x 4x 3x + x 3 x + x + = x + 3x + x + x + = x + 3x + (x + ). Next we compute the partial fraction expansion. 3x + (x + ) = A x + + B.3x + = A(x + ) + B = Ax + (A + B). (x + ) So A = 3, and then from A + B = we find that B =. Thus Now we can integrate x 3 x + x + dx = 3x + (x + ) = 3 x + (x + ). x + 3 x + (x + ) dx = x x + 3 ln x + + x + + C. Finally, the problem asks for a definite integral, so x 3 ( x + x + dx = x x + 3 ln x + + ) x= x + x= ( = + 3 ln + ) ( + 3 ln + ) = + 3 ln. The numerical value is approximately

8 Section 7.4, Problem. First we compute the partial fraction expansion. So we need to solve (x + )(x + ) = A x + + Bx + C x +. = A(x + ) + (Bx + C)(x + ) = (A + B)x + (B + C)x + (A + C). A + B =, B + C =, A + C =. The first two equations give B = A and C = B = A, so the last equation A =, so A =. Then B = and C =, so we obtain (x + )(x + ) = x + + x + x + = ( x + x ) x. + Now we can integrate, (x + )(x + ) dx = ( x + x x + = dx x + Finally, we can evaluate the definite integral ) dx x x + dx + dx x + = ln x + 4 ln x + + tan (x) + C. ( (x + )(x + ) dx = ln x + 4 ln x + + ) x= tan (x) x= ( = ln 4 ln + ) tan () = 4 ln + π 8. The numerical value of the integral is approximately

9 Section 7.4, Problem 49. We need to integrate kx(n x) dx = The partial fraction expansion of x(n x) is dt. x(n x) = A x + B N x. = A(N x) + Bx = ( A + B)x + AN. So A + B =, which gives A = B, and also AN =, which gives A = /N. So we can integrate t = kx(n x) dx = Nk x + N x dx = ( ) ln x ln N x + C Nk = Nk ln x N x + C. We are told that k = /5, N =, and that x() =. Plugging these values into our formula, we can find the value of C, = /5 ln + C, so C = ln So our complete formula, with the given values of k and N, is t = 4 ln x x + 4 ln 499 = 4 ln 499x x. (a) We can partially solve for x as Then some algebra gives 499x x = e4t. x = e4t e 4t. (b) We want to know the value of t when x = N = 5. It s easiest to use our earlier formula that expresses t in terms of x. Thus t = ( ) x=5 4 ln = ln days

(x + 1)(x 2) = 4. x

(x + 1)(x 2) = 4. x dvanced Integration Techniques: Partial Fractions The method of partial fractions can occasionally make it possible to find the integral of a quotient of rational functions. Partial fractions gives us