All right, last integral question before bed...?

Question

So let's get back to that interval thing. Let's say that the interval is, instead, [x^2, pi]. And the integral is S(sin2t dt). How do I take the derivative of that one?

Answer 1

F(x) = Integral (x^2 to pi, sin(2t) dt )

To solve this, use the Fundamental Theorem of Calculus once again; when taking the derivative, plug in the upper bound, use the chain rule on that bound, plug in the lower bound, use the chain rule on that bound.

F'(x) = sin(2[pi]) {0} - sin(2x^2) {2x}, OR

(To make things clear, I used { } brackets to denote how I used the chain rule. The derivative of PI is 0, and the derivative of x^2 is 2x)

F'(x) = 0 - sin(2x^2) (2x)

F'(x) = -2x sin(2x^2)

Answer 2

π
∫sin(2t)dt =
x^2

f(x) = (1/2)[-cos(2π) + cos(2x^2)]
f(x) = (1/2)cos(2x^2) - 1/2
f(x) = - (1/2)(4x)sin(2x^2)
f(x) = - 2xsin(2x^2)

Answer 3

doody