Patrick's question at Yahoo Answers (First fundamental theorem of Calculus)

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SUMMARY

The discussion centers on applying the First Fundamental Theorem of Calculus to solve for the second derivative of a function defined by an integral. The function in question is \( f(x) = \int_1^x \sin(\pi t^2) \, dt \). By differentiating twice, it is established that \( f'(x) = \sin(\pi x^2) \) and \( f''(x) = 2\pi x \cos(\pi x^2) \). The final result for \( \frac{f''(2)}{\pi} \) is calculated as 4.

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Fernando Revilla
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Here is the question:

don't know how to type the question so here's the link to the image.

http://goo.gl/vQhhs

Thanks in advance :)

Here is a link to the question:

Integration by parts? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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Hello Patrick,

The problem is:

Find $\dfrac{f''(2)}{\pi}$ if $f(x)=\displaystyle\int_1^x\sin (\pi t^2)\;dt$. Enter your answer as an integer.

Solution. Using the First fundamental theorem of Calculus, $f'(x)=\sin (\pi x^2)$. Deriving again, $f''(x)=2\pi x\cos (\pi x^2)$ so, $$\dfrac{f''(2)}{\pi}=\frac{4\pi\cos(4\pi)}{\pi}=4\cos(4\pi)=4$$
 

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