Homework Help: Question about the fundamental theorem of Calculus

1. Jan 9, 2010

James889

Hi,

Suppose we're asked to find the derivative of the integral
$$f(x)~=~\int_{-13}^{sin~x} \sqrt{1+t^2}~dt$$

Now, the solution apparently looks like this:
$$f(x)^{\prime} = \sqrt{1+sin^2(x)}\cdot~cos(x)$$

Why?

Why does the solution contain the upper limit plugged in ?

A more sensible answer (to me) would be $$\sqrt{1+x^2}$$

2. Jan 9, 2010

phsopher

Denote the integrand as a derivative of some unknown function:

$$\sqrt{1+t^2} = F'(t)$$

Now calculate the integral and then take a derivative of f(x).

Last edited: Jan 9, 2010
3. Jan 9, 2010

JSuarez

Because of the composition rule for derivatives. You have:

$$F\left(y\right)=\int_{-13}^{y}\sqrt{1+t^{2}}dt$$

And $$f(x) = F(sin x)$$. What is the derivative of that?

4. Jan 9, 2010

rochfor1

Your answer would be sensible if the upper limit of integration were simply x, which is the situation the FTC deals explicitly with. Not to worry though...the chain rule comes to the rescue. Perhaps you could try defining another function $$F(x)=\int_{-13}^x \sqrt{ 1 + t^2 }\ dt$$. Then think about what F'(x) is and what F(sin x) is equal to.

EDIT: beaten to the punch by jsuarez

Last edited: Jan 9, 2010
5. Jan 9, 2010

James889

$$f(x)^{\prime}=\int_{-13}^{sin~x}\sqrt{1+x^2} - \sqrt{1+(-13)^2}$$

$$== \sqrt{1+sin^2x} \cdot cos~x$$ because the second part is just zero.

But what would happen if we had sin x in the upper limit and another trig function in the lower limit like cos x ?

6. Jan 9, 2010

rochfor1

? You're f(x)' makes no sense to me. If you had functions as both limits of integration, simply rewrite it as two integrals where each has one limit that's a function and one limit that's a number. (Think about how you rewrite one integral as two.)

7. Jan 9, 2010

Bohrok

Then you would have a second term with cos x plugged in times its derivative, -sin x:

$$\frac{d}{dx} \int_{\cos x}^{\sin x} \sqrt{1 + t^2} ~dt = \sqrt{1 + \sin^2x} ~\cdot ~\cos x ~+ ~\sqrt{1 + \cos^2x} ~\cdot ~\sin x$$

8. Jan 9, 2010

JSuarez

If you have something like:

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

Where $$f(t)$$ is continuous (not strictly necessary, I'll assume it just to guarantee integrability) $$g(x), h(x)$$ differentiable, then you have:

$$F(x) = \int_{g(x)}^{h(x)}f(t)dt = \int_{a}^{h(x)}f(t)dt + \int_{g(x)}^{a}f(t)dt = \int_{a}^{h(x)}f(t)dt - \int_{a}^{g(x)}f(t)dt$$

For a suitable a, so:

$$F'(x) = f(h(x))h'(x) - f(g(x))g'(x)$$