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I have a question in calculus.

Let $F(x)=\int_1^\sqrt{x} t^2\cos( \pi t)dt$ Find $F'(4)$

I know $F'(X$) $=\int_1^\sqrt{x} x^2\cos( \pi x)dx$

So I made $u=x^{\frac{1}{2}}$

and I got

$u^2\cos\pi(u)$

which is

$F'(x)=x^{1/4}\cdot \left(\frac{1}{2\sqrt{x}}\right)\cdot \cos\pi\sqrt{x}$

when I plugged in four I got

$F'(4)=\sqrt{2}\frac{1}{\sqrt{2}}$

but did I do this correctly?

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1  
No, does not look right. You get $F'(x) = x^{1/4} \cos ( \pi/2) = 0$, basically. Which is incorrect. Currently you seem to have too many typos. Please fix those before someone can help. As it is currently, it will be quite hard for someone to help you. –  Aryabhata Apr 12 '13 at 18:57
    
I do not know if this makes sense but I used chain rule Ill explain. –  Fernando Martinez Apr 12 '13 at 19:05
    
I made u=x^(1/2) so that I take derivative of x^(1/2) because of F'(x)=f(u)(x)times(u'(x) –  Fernando Martinez Apr 12 '13 at 19:09

3 Answers 3

up vote 2 down vote accepted

Your post is very confusing: but I think I finally follow:

Note we have substituted $u = \sqrt x$ and $du = \dfrac 1{2\sqrt x }dx$

giving us $u^2\cos\pi(u)\,du$

So your expression for $F'(x)$ is off and needs to be, no in terms of $x$: $$F'(x)=(\sqrt x)^2\cdot \cos\pi\sqrt{x}\cdot \left(\frac{1}{2\sqrt{x}}\right)$$

Now evalate $F'(4)$:

The procedure is a little mixed up and the exposition was a little confusing.

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Since the integrand is $t^2 \cos(\pi t)$, shouldn't the first factor be $(\sqrt{x})^2$, rather than $\sqrt{\sqrt{x}}$ ? –  RecklessReckoner Apr 12 '13 at 19:59
    
@Reckless Yes indeed, corrected: I just found the exposition really difficult to follow and confused myself in the process! –  amWhy Apr 12 '13 at 20:02
    
you are right it was confusing because I was confused myself. Thanks for the help I think my mistake was that in my substitution of the value especially u^2. –  Fernando Martinez Apr 12 '13 at 21:16
    
hhm at F'(4) I got 4(1/4) for my answer –  Fernando Martinez Apr 12 '13 at 22:25
1  
@Amzoti - Yes, Fernando is a very responsive, hard working student. –  amWhy Apr 12 '13 at 22:38

The integral bound is from 1 to 2 because, if you have gotten to u-substitution, u becomes sqrt(x) and you evaluate u(4)=sqrt(4)=2 and then use integration by parts. Here is a link to integration by parts:

http://tutorial.math.lamar.edu/Classes/CalcII/IntegrationByParts.aspx

And the derivative should be x*cos(pi*sqrt(x)) because of Leibniz's rule.

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I've not learned integration by parts yet... –  Fernando Martinez Apr 12 '13 at 19:06
    
Neither have I. –  user72311 Apr 12 '13 at 19:20
    
Oops, I screwed up Leibniz rule. Multiply by the derivative of sqrt(x) and it should be fine. –  user72311 Apr 12 '13 at 19:25

This is an indirect application of the Fundamental Theorem of Calculus which is quickly passed through in many introductory texts. You have an integral function $F(x) = \int^{x}_{1} f(t) dt$; since $t^2 \cos(\pi t )$ is continuous everywhere, we can safely say that $F'(x) = f(x)$ .

When the upper limit is a function of $x$, the FTC will let us write $\int^{u}_{1} f(t) dt = F(u) - F(1)$. When we differentiate this with respect to $x$, the Chain Rule gives us $$\frac{d}{dx}\int^{u}_{1} f(t) dt = \frac{d}{dx} [F(u) - F(1)] = \frac{dF}{du} \cdot \frac{du}{dx} = f(u(x)) \cdot \frac{du}{dx}$$

So for this problem, $$\frac{d}{dx}\int^{\sqrt{x}}_{1} t^2 \cos(\pi t ) dt = (\sqrt{x})^2 \cos(\pi \sqrt{x} )) \cdot \frac{d}{dx}(\sqrt{x}).$$

To answer your question, you would complete the differentiation and evaluate the result at x = 4 .

(I'll mention, incidentally, that a problem of this type is a favorite final exam question.)

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thanks for the help and I will be sure to study it for my final which will be soon..... –  Fernando Martinez Apr 12 '13 at 21:16

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