# Thread: Fundamental Theorem Of Calculus

1. ## Fundamental Theorem Of Calculus

It makes sense when I read the statements but I can't seem to apply it without getting totally lost.
If I have a problem that looks something like:

d/dx of the integral of some function f(x) evaluated from -1 to sin(x) then how do I apply FToC to solve it?
Also what if the bounds are 2x to x^2 or something along those lines?

I also have a related question that I am dead stuck on but an answer to the above would be appreciated more than a solution to the below if you don't have time or something.

2 plus the integral of (f(t)/(t^3))dt evaluated from a to x equals 6x^-5. Find a value for a and a function f(t) so that this is true.

2. Originally Posted by Vashti
It makes sense when I read the statements but I can't seem to apply it without getting totally lost.
If I have a problem that looks something like:

d/dx of the integral of some function f(x) evaluated from -1 to sin(x) then how do I apply FToC to solve it?

[snip]
From the chain rule and the FTC it can be shown that:

$\displaystyle \frac{d}{dx} \left [ \int_{a}^{g(x)} f(t) \, dt \right] = f(g(x)) \, g'(x)$.

In your case a = -1 and $\displaystyle g(x) = \sin x$.

3. Originally Posted by Vashti
[snip]d/dx of the integral of some function f(x) evaluated from -1 to sin(x) then how do I apply FToC to solve it?
Also what if the bounds are 2x to x^2 or something along those lines?

[snip]
From the chain rule and the FTC it can be shown that

$\displaystyle \frac{d}{dx} \left [ \int_{h(x)}^{g(x)} f(t) \, dt \right] = f(g(x)) \, g'(x) - f(h(x)) \, h'(x)$.

In your case h(x) = 2x and g(x) = x^2.

4. if it's not too much trouble could someone prove that for me? You say to use the chain rule but I'm not sure how...

5. p.s. thanks so much you've helped me a lot already.

6. Originally Posted by Vashti
[snip]

2 plus the integral of (f(t)/(t^3))dt evaluated from a to x equals 6x^-5. Find a value for a and a function f(t) so that this is true.
$\displaystyle 2 + \int_{a}^{x} \frac{f(t)}{t^3} \, dt = \frac{6}{x^5}$ .... (1)

Differentiate both sides of (1) with respect to x using the rule in post #2:

$\displaystyle \frac{f(x)}{x^3} = -\frac{30}{x^6} \Rightarrow f(x) = .....$

Now substitute your answer for f(x) into (1), do the required integration and solve for a.

7. Originally Posted by mr fantastic
From the chain rule and the FTC it can be shown that

$\displaystyle \frac{d}{dx} \left [ \int_{h(x)}^{g(x)} f(t) \, dt \right] = f(g(x)) \, g'(x) - f(h(x)) \, h'(x)$.

[snip]
$\displaystyle \frac{d}{dx} \left [ \int_{h(x)}^{g(x)} f(t) \, dt \right] = \frac{d}{dx} \left [ \int_{h(x)}^{a} f(t) \, dt + \int_{a}^{g(x)} f(t) \, dt \right]$

$\displaystyle = \frac{d}{dx} \left [ -\int_{a}^{h(x)} f(t) \, dt + \int_{a}^{g(x)} f(t) \, dt \right]$

$\displaystyle = - \frac{d}{dx} \left [\int_{a}^{h(x)} f(t) \, dt \right] + \frac{d}{dx} \left[ \int_{a}^{g(x)} f(t) \, dt \right]$

Now apply the result given in post #2 to each term. To prove the result in post #2:

Let w = g(x). Then you have:

$\displaystyle \frac{d}{dx} \left [ \int_{a}^{g(x)} f(t) \, dt \right] = \frac{d}{dw} \left [ \int_{a}^{w} f(t) \, dt \right] \, \times \, \frac{dw}{dx} = f(w) \, g'(x) = f(g(x)) \, g'(x)$.

8. you are my hero! thank you so so much!

9. Originally Posted by Vashti
you are my hero!
Of course... Did you ignore his job is to save the world ?