Fundamental Theorem of Calculus Part 2 help?

Question

How do i calculate the derivative of
d/dx∫sin^2tdt

the top portion of the integral has a 0 and the bottom has an x.


I'd appreciate any help. Thanks.

Answer 1

Interchange the limits of integration, and multiply outside by (-1); this keeps the problem the same. Next multiply by 1/2 outside and by 2 inside. The problem is now -d/dx[integ(0,x)sin(2t)d(2t)]. By the Fundamental Theorem, the derivative is -sin(2x).

Answer 2

do no longer enable the wording of the question confuse you. It merely ability to evaluate the confident necessary in the type you have been taught to realize this. In different words, locate an anti-spinoff (call it G) and then compute G(3) - G(0). If I bear in mind my freshman Calculus wisely, the required itself is easy to evaluate: ? 2*sin(x) = 2cos(x) + C and ? 4e^x dx = 4e^x + D the place C and D are any consistent (use 0 for the two). solid success

Answer 3

d/dx∫sin^2tdt
d/dx (x/2 - sin(2x)/4) (0 to x)
d/dx (-x/2 + sin(2x)/4)
-1/2 + 1/2cos(2x)
-1/2 + 1/2(1-2sin^2(x))
-sin^2(x)

Answer 4

Answer: -sin²x