solve There is no need to evaluate the sum. x dx/(x2 - 5)(1/3) sin(6x)cos(6x) dx

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There is no need to evaluate the sum. x dx/(x2 - 5)(1/3) sin(6x)cos(6x) dx 5dx tan6(8x - 2)sec2(8x - 2)dx Apply the Mean value Theorem to f(x) = x2(3 - x)2 on the interval [-1, 2] and determine the average value of the function. Find the maximum product of 2 numbers whose difference is 36. Then find the minimum product. Be sure to verify using first or second derivative tests. A rectangle is inscribed inside the parabola y = 9 - x2 and the x axis. Find the dimensions of the rectangle yielding the maximum area. Again be sure to use the appropriate derivative tests.

Explanation / Answer

12. ?sin(6x)cos(6x)dx

u=sin(6x)

du=6cos(6x)dx

1/6?udu

1/6*(u^2/2)+C

[sin^2(6x)]/12+C

13. ?5xdx/[sqrt(x)*(2+sqrt(x))^4]

5?sqrt(x)/[2+sqrt(x))^4]

u=2+sqrt(x)

du=[1/2*x^-(1/2)]dx

10?xdu/u^4

x=(u-2)^2

10?[(u-2)^2]du/u^4

10?(u^2-4u+4)du/u^4

10(1/u^2-4/u^3+4/u^4)du

10[-1/u+2/u^2-4/(3u^3)]+C

-10/u+20/u^2-40/(3u^3)+C

-10/(2+sqrt(x))+20/(2+sqrt(x)^2)-40/[(3(2+sqrt(x)^3)]+C

14.?tan^6(8x-2)sec^2(8x-2)dx

u=tan(8x-2)

du=8sec^2(8x-2)dx

1/8*?u^6du

u^7/56+C

tan^7(8x-2)/56+C

hope this helps

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