Derivative of integral bounded by functions

[springo]

Homework Statement

What's the derivative of the following two:
\int_{a}^{h(x)}f(t)\,\mathrm{d}t
\int_{u(x)}^{v(x)}f(t)\,\mathrm{d}t

Homework Equations

The Attempt at a Solution

I thought of doing the following:
\int_{h(a)}^{h(x)}f(t)\,\mathrm{d}t = \int_{a}^{x}f\circ h(u)\cdot h'(u)\,\mathrm{d}u
(with t = h(u) )
But then I don't know how to continue.

Thanks for your help.

 

[Dick]

If the antiderivative of f(t) is F(t), then the integral of the second one is F(v(x))-F(u(x)), right? Differentiate that using the chain rule.

 

[Mark44]

These problems are applications of the Fundamental Theorem of Calculus. As a hint, I'll work a different, but related, problem.

\int_a^{x^2} f(t) dt
Suppose an antiderivative of f(t) is F(t). I.e., F'(t) = f(t).
So
\int_a^{x^2} f(t) dt = F(x^2) - F(a)
If we differentiate both sides of this equation with respect to x, we get
d/dx \int_a^{x^2} f(t) dt = d/dx(F(x^2) - F(a))
= F'(x^2) * d/dx(x^2) - 0 (F(a) is a constant, so its derivative is zero)
= f'(x^2) * 2x

Is that enough to get you going?

 

[springo]

OK, so I guess then...
\frac{\mathrm{d}}{\mathrm{d}x}\left(\int_{u(x)}^{v(x)}f(t)\,\mathrm{d}t \right)=v'(x) \cdot f \circ v(x) - u'(x) \cdot f \circ u(x)
As for the other one, since u'(x) = da/dx = 0, we have:
\frac{\mathrm{d}}{\mathrm{d}x}\left(\int_{a}^{h(x)}f(t)\,\mathrm{d}t \right)= h'(x)\cdot f\circ h(x)

Is that all OK?
Thanks.

 

[Dick]

Looks ok to me.