# Derivative of an Integral

• Oct 24th 2009, 05:58 AM
synapse
Derivative of an Integral
Hi everyone! I was wondering if anyone here could kindly point me in the direction of how to get around solving for the derivative of $\displaystyle \int_x^{2x} 2^{t^2} \, dt$ w.r.t. $\displaystyle x$. Thank you very much in advance.

Also, I'm new here, so if I'm flouting any rules please do let me know and I will gladly make changes as necessary. Thanks again! (Hi)
• Oct 24th 2009, 06:17 AM
tonio
Quote:

Originally Posted by synapse
Hi everyone! I was wondering if anyone here could kindly point me in the direction of how to get around solving for the derivative of $\displaystyle \int_x^{2x} 2^{t^2} \, dt$ w.r.t. $\displaystyle x$. Thank you very much in advance.

Also, I'm new here, so if I'm flouting any rules please do let me know and I will gladly make changes as necessary. Thanks again! (Hi)

The fundamental theorem of integral calculus tells us that:

$\displaystyle \int\limits_x^{2x} 2^{t^2} \, dt\,=F(2x)-F(x)\,,\,\,with\,\,\,F'(x)=2^{x^2}$ and thus $\displaystyle \frac{d}{dx}\int\limits_x^{2x} 2^{t^2} \, dt=2F'(2x)-F'(x)$ and there you go./

Tonio
• Oct 24th 2009, 06:22 AM
Scott H
There are two theorems we may use to our advantage here:

\displaystyle \begin{aligned} \frac{d}{dx}\int_0^x f(t)\,dt&=f(x)\\ \int_x^0 f(t)\,dt&=-\int_0^x f(t)\,dt. \end{aligned}

The first implies that as $\displaystyle x$ moves to the right, the area under a curve changes at a rate equal to the height of the function at $\displaystyle x$. The second is defined for convenience in calculus.