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Where, $F(x) = \int^x_0 t\cosh(t^4)\,dt$

This is a 2 part question:

(a) calculate $\int^1_0 xF(x)\,dx$ (the answer will involve $F(1)$)

(b) $ \lim_{x\to0}\frac{F(x)}{x^2}$

For part (a)

I am given that:

$F(x) = \int^x_0 t\cosh(t^4)\,dt$

To calculate $\int^1_0 xF(x)\,dx$ I use integration by parts and take advantage of the Fundamental Theorem of Calculus (FTC):

Let $u = F(x)$, $du = \frac{d}{dx}\int^x_0 t\cosh(t^4)\,dt = x\cosh(x^4) \, dx$ by the FTC

Let $dv = x$, $v = \dfrac{x^2}{2}$

Therefore, I have:

$$ \begin{align} & {}\quad \int^x_0 t\cosh(t^4)\,dt = \dfrac{x^2}{2} F(x)|^1 _0 - \dfrac{1}{2}\int^1_0x^3\cosh(x^4) \, dx \\[6pt] & = \dfrac{1}{2}F(1) - \dfrac{1}{2}\cdot\dfrac{1}{4}[\sinh(x^4)|^1 _0 ] \\[6pt] & = \dfrac{1}{2}F(1) - \dfrac{1}{8}\sinh(1) \end{align} $$

Is this correct? Can it or should it be broken down further?

For part (b):

$\lim_{x\to0}\frac{F(x)}{x^2}$ is 0/0, since $\lim_{x\to0}F(x)$ = 0 by the FTC.

Therefore, $\lim_{x\to0}\frac{F(x)}{x^2} = \lim_{x\to0}\frac{x\cosh(x^4)}{2x} = \dfrac{1}{2}$

Have I made any errors? Thanks so much in advance!

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Looks OK to me. –  Gerry Myerson Sep 25 '12 at 2:58
1  
I did the first part another way (double integral), got the same answer. –  André Nicolas Sep 25 '12 at 2:59
    
Thanks to you both! –  JackReacher Sep 25 '12 at 3:36
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