Let H(x)=integral 2 to x^3 cos t dt?

Question

evaluate the integral to find an equation for h(x) that does not involve the integral sign

Answer 1

The integral of cos(t) with respect to t is simply sin(t). So when you plug in the bounds, you get
H(x) = sin(x^3) - sin(2)

Answer 2

H(x) = Integral (2 to x^3, cos(t) dt )

By the Fundamental Theorem of Calculus, all we have to do is plug in our upper bound into the function, use the chain rule, plug in the lower bound, use the chain rule.

H'(x) = cos(x^3) {3x^2} - cos(2^3) {0}

I demonstrated the chain rule using the { } brackets. Note that the derivative of x^3 is 3x^2, and the derivative of 2 is 0. Simplifying, we get

H'(x) = 3x^2 cos(x^3) - cos(8) (0)
H'(x) = 3x^2 cos(x^3)

Edit: My mistake; I thought you were trying to find H'(x) (given the similarity with this question and the Fundamental Theorem of Calculus questions).

Answer 3

H(x) = sin(x^3)-sin(2)