Fundamental Theorem Of Calculus

**Vashti** · June 18th 2008, 03:25 PM

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.

**mr fantastic** · June 18th 2008, 03:32 PM

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:

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

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

**mr fantastic** · June 18th 2008, 03:39 PM

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

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

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

**Vashti** · June 18th 2008, 03:43 PM

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...

**Vashti** · June 18th 2008, 03:44 PM

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

**mr fantastic** · June 18th 2008, 03:48 PM

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.

$2 + \int_{a}^{x} \frac{f(t)}{t^3} \, dt = \frac{6}{x^5}$ .... (1)

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

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

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

**mr fantastic** · June 18th 2008, 03:58 PM

Originally Posted by mr fantastic

From the chain rule and the FTC it can be shown that

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

[snip]

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

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

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

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:

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

**Vashti** · June 18th 2008, 04:02 PM

you are my hero! thank you so so much!

**Moo** · June 19th 2008, 12:22 AM

Originally Posted by Vashti

you are my hero!

Of course... Did you ignore his job is to save the world ?

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