# Definite Integrals of Definite Integrals

## Homework Statement

Suppose that f is a continuous function.

a) If g is a differentiable function and if

F(x) = $$\int_{0}^{x}$$g(x)f(t)dt

find F ' (x).

b) Show that

$$\int_{0}^{x}$$(x - t)f(t)dt = $$\int_{0}^{x}$$ [$$\int_{0}^{u}$$f(t)dt] du

Suggestion: Use a) and the racetrack theorem.

## Homework Equations

The fundamental theorems of calculus.

The racetrack theorem states that if

f(a) = g(a) and if f ' (x) = g ' (x) for all x, then f(x) = g(x) for all x.

## The Attempt at a Solution

a)

I took d/dx of both sides and then ended up with

F ' (x) = g(x)f(x) by the first fundamental theorem.

b)

I have to prove that the left side and right side are equal at some point a, and that their derivatives are equal everywhere. I've never come across the integral of an integral before so I'm not quite sure what to do.

## Answers and Replies

First you need to calculate F'(x). Since F(x) is a product of two functions - you should have no problem.

This is to warm you up.

Then you do 2). You first check that LHS=RHS at x=0. Then you calculate the derivative of LHS and of RHS. To calculate the derivative of the LHS it will be handy to split it into two terms by doing the multiplication under the integral and then use 1) for one of these terms. You must be careful - calculate the derivatives at x=a on both sides.

split it into two terms by doing the multiplication under the integral

What does that mean? Do I split it up into

$$\int_{0}^{x}}$$[xf(t)] dt - $$\int_{0}^{x}}$$[tf(t)] dt

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What does that mean? Do I split it up into

$$\int_{0}^{x}\,[xf(t)]\, dt - \int_{0}^{x}[tf(t)]\, dt$$

Yes, that's good. Then the first term is like in 1). Of course you are supposed to notice that with respect to the integration over t, x behaves like a constant.

Do functions behave like constants if you're integrating with respect to another variable as well?

Like in a), if I have

$$\int_{0}^{x}$$ g(x)f(t)dt

I'm integrating with respect to t, but then if I take

d/dx [$$\int_{0}^{x}$$ g(x)f(t)dt]

Can I treat that as:

d/dx [g(x)$$\int_{0}^{x}$$ f(t)dt]

Which would become

g(x)f(x)?

Integration variable in your integral is t. g(x) is independent of of, therefore with respect to the integral it behaves like a constant. Not so with differentiation with respect to x. When you differentiate with respect to x, you have a product of two functions of x. One is g(x), the other one is, say,

$$h(x)=\int_0^x f(t)\,dt$$

OK, here's what I have (by the way, thank you very much for helping)

a)

$$F(x) = \int_0^x g(x)f(t)\,dt$$

$$F(x) = g(x)\int_0^x f(t)\,dt$$

Let $$h(x) = \int_0^x f(t)\,dt$$

$$F(x) = g(x)h(x)$$

$$\frac{d}{dx}\ F(x) = \frac{d}{dx}\ g(x)h(x)$$

$$F'(x) = g'(x)h(x) + g(x)h'(x)$$

$$F'(x) = g'(x)\int_0^x f(t)\,dt + g(x)f(x)$$

b)

$$\int_0^x (x - t)f(t)\,dt = \int_0^x (\int_0^u f(t)\,dt)\,du$$

$$\int_{0}^{x}[xf(t)]\,dt - \int_{0}^{x}[tf(t)]\,dt = \int_0^x (\int_0^u f(t)\,dt)\,du$$

Let $$q(u) = \int_{0}^{u}f(t)\,dt$$

h(x) same as above

$$xh(x) - \int_0^x [t]f(t)\,dt = \int_0^x q(u)\,du$$

$$\frac{d}{dx}\ xh(x) - \frac{d}{dx}\ \int_0^x [t]f(t)\,dt = \frac{d}{dx}\ \int_0^x q(u)\,du$$

$$h(x) + xh'(x) - xf(x) = q(x)$$

$$\int_0^x f(t)\,dt + xf(x) - xf(x) = q(x)$$

$$\int_0^x f(t)\,dt = q(x)$$

$$\int_0^x f(t)\,dt = \int_0^x f(t)\,dt$$

How does that look?

Looks good to me.