Derivative of an Integral

**synapse** · October 24th 2009, 05:58 AM

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 $\int_x^{2x} 2^{t^2} \, dt$ w.r.t. $x$ . 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!

**tonio** · October 24th 2009, 06:17 AM

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 $\int_x^{2x} 2^{t^2} \, dt$ w.r.t. $x$ . 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!

The fundamental theorem of integral calculus tells us that:

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

Tonio

**Scott H** · October 24th 2009, 06:22 AM

There are two theorems we may use to our advantage here:

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

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

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