## Definite Integrals, extrema

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### Definite Integrals, extrema

Find and classify the relative maxima and minima of f(x), if
f(x) = integral from 0 to x of {(t^2-4)dt} /{1+(cost)^2}

I got x=2 and x=-2 for f'(x)=0, but, I don't know how to calculate the relative maxima and minima. Please help me.
Thank you
Aurica
Queen of Hearts

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Joined: Mon May 04, 2009 4:49 am

### Re: Definite Integrals, extrema

Here, we may use the fact that

$\frac{d}{dx}\int\,_0^x\,g(t)\,dt\,=\,g(x).$

(As you move $x$ to the right, the rate of change of the area under a curve is equal to the height of the function.)

Your solutions $x\,=\,\pm 2$ are correct. To prove that they are relative maxima and minima, we can use the Second Derivative Test:

$\begin{eqnarray*} f''(x) &>& 0\mbox{ }\rightarrow\mbox{ }x\mbox{ is a local minimum}\\ f''(x) &<& 0\mbox{ }\rightarrow\mbox{ }x\mbox{ is a local maximum} \end{eqnarray*}$
Scott
King of Diamonds

Posts: 251
Joined: Sat Dec 13, 2008 9:34 am

### Re: Definite Integrals, extrema

Thanks for your help. But, how can I calculate f(2)? is x=-2 also a solution since the domain of f(x) is x>=0 ?
Aurica
Queen of Hearts

Posts: 58
Joined: Mon May 04, 2009 4:49 am

### Re: Definite Integrals, extrema

If the domain is, as you say, $[0,\,\infty)$, then $x\,=\,2$ is the only local extremum. However, since $\cos\,x$ is squared and added to $1$ in the denominator, $f(x)$ is defined everywhere.

I just put the integral into the Wolfram Online Integrator and it gave an answer that was not expressed in terms of elementary functions. If the problem only asks to classify the extrema, then I don't think you have to worry about the values of the extrema at those points.
Scott
King of Diamonds

Posts: 251
Joined: Sat Dec 13, 2008 9:34 am