Derivative of a definite integral (FTIC)
By John Thompson
$\displaystyle \left.\frac{d}{dx}\right|_{x=\pi} \int_{t=0}^{x} \frac{\cos 3t}{\sqrt{1+t}} \,dt$
What would be the correct way to solve this? I'd really like to understand the reasoning behind it as well please?
Would something like this be going in the correct direction?: $\displaystyle \int_0^{\pi} \frac{d}{dt}\!\left( \frac{\cos 3t}{\sqrt{1+t}} \right) dt$
$\endgroup$1 Answer
$\begingroup$The notation $$ \frac{d}{dx}f\,\Bigr|_{x=a}\text{ means }f'(a), $$ that is, derivate and then evaluate at $x=a$.
The fundamental theorem of calculus tells you that $$ \frac{d}{dx}\int_{0}^{x} \frac{\cos 3\,t}{\sqrt{1+t}} \,dt=\frac{\cos 3\,x}{\sqrt{1+x}}. $$ Then $$ \frac{d}{dx}\Bigr|_{x=\pi}\int_{0}^{x} \frac{\cos 3\,t}{\sqrt{1+t}} \,dt=\frac{\cos 3\,\pi}{\sqrt{1+\pi}}=-\frac{1}{\sqrt{1+\pi}}.$$
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