Derivative of an integral

**Hampus** · October 21st 2009, 03:35 PM

Hi, how do you solve this (btw, sorry, I don't know how to make snazzy math grafics :/):

Find G'(x) for G(x) = the integral exp(t^2)dt in the interval 1/x< a< x.

**skeeter** · October 21st 2009, 04:22 PM

Originally Posted by Hampus

Hi, how do you solve this (btw, sorry, I don't know how to make snazzy math grafics :/):

Find G'(x) for G(x) = the integral exp(t^2)dt in the interval 1/x< a< x.

$\frac{d}{dx} \int_v^u f(t) \, dt = f(u) \cdot \frac{du}{dx} - f(v) \cdot \frac{dv}{dx}$

$G(x) = \int_{\frac{1}{x}}^x e^{t^2} \, dt$

$G'(x) = e^{x^2} + \frac{e^{\frac{1}{x^2}}}{x^2}$

**Hampus** · October 21st 2009, 04:52 PM

Could you maybe explain why this is the case? (please?)

**redsoxfan325** · October 21st 2009, 04:57 PM

Originally Posted by Hampus

Could you maybe explain why this is the case? (please?)

Say that $\int e^{t^2}\,dt=F(t)$ . (It doesn't matter what $F(t)$ is.)

So $\int_{1/x}^x e^{t^2}\,dt=F(x)-F(1/x)$ by the Fundamental Theorem of Calculus.

To take the derivative of this expression, use the chain rule to get

$F'(x)\cdot1-F'(1/x)\cdot-\frac{1}{x^2}$

But because of how we defined $F$ , we know that $F'(x)=e^{x^2}$ .

So the above is $e^{x^2}+\frac{e^{1/x^2}}{x^2}$ .

**Hampus** · October 21st 2009, 05:21 PM

Feels like something I could have figured out if I gave it a thought :/ Thanks a lot though, very helpful, it all makes sense

Thread: Derivative of an integral

LinkBack

Thread Tools

Display