How Do You Derive the Integral from g(x) to h(x)?

[mateomy]

Here's the problem: I've solved it (sorta), but I still have a question...

If f is continuous and g and h are differentiable functions, find a formula for

$$\frac{d}{dx}\,\int_{g(x)}^{h(x)} f(t)\,dt$$

Breaking it apart further...

$$\frac{d}{dx}\, \bigg(\int_0^{h(x)} f(t)dt ,\ - \int_0^{g(x)} f(t)dt \, \bigg)$$

Now working with the individual pieces...

setting h(x) to s and h'(x) to ds\dx

$$\frac{d}{ds} \, \int_0^s f(t)\,dt \, \frac{ds}{dx} \, --> \, f(s)\frac{ds}{dx} \, --> \, f(h(x))h'(x)$$

I do the same operation with g(x) but I have to reverse the limits of integration so the integral is a negative one at the end of all the work, looking ultimately like so...

$$-f(g(x))g'(x)$$

Then showing the end result...
$$\frac{d}{dx}\,\int_{g(x)}^{h(x)} f(t)\,dt \, = f(h(x))h'(x)-f(g(x))g'(x)$$

Thats how it is in the answer column anyway, I got the same result but rather than subtracting the g(x) portion FROM the h(x) portion I came up with an addition because of the f(g(x))g'(x) being a negative, so the (b-a) of the integration would turn into h(x)-(-g(x)) which would turn the subtraction to an addition. I can't figure out why it isnt. Can somebody please explain this to me.

 

[lippyka]

The reason why you got an addition instead of a subtraction is because you are taking the derivative of the integral with respect to x, not with respect to the integration variable t. Taking the derivative with respect to x changes the sign of the g(x) term, so the subtraction becomes an addition. This is because when you take the derivative of the integral with respect to x, you need to use the chain rule. The chain rule states that if you have a composite function, such as the integral from g(x) to h(x), then you need to take the derivative of the outside function (in this case, the integral) with respect to the inside function (in this case, x). This means that the derivative of the integral with respect to x is equal to the derivative of the integral with respect to t multiplied by the derivative of x with respect to t. In this case, the derivative of x with respect to t is negative, so it changes the sign of the g(x) term.