Spot the mistake in Fundamental Theorem of Calculus question

yoyobarn · **Joined:** Fri, 12 Nov 2010 07:00:12 UTC **Posts:** 48

Hi all,

Can anyone tell me which step did I do wrong in this FTC question?

Question: Let h be the function defined by $h(x)=\int_0^{x^2} e^{x+t} dt$ for all real numbers x. Find the value of h'(1)

.

My solution:
Let u=x^2

.

$h(u)=\int_0^u e^{\sqrt{u}+t} dt$

$\frac{dh}{du}=e^{\sqrt{u}+u$ , by FTC.

$\frac{du}{dx}=2x$

$\therefore h'(x)=\frac{dh}{du}\cdot \frac{du}{dx}$

$h'(1)=e^2\cdot 2$ .

Correct answer is: h'(1)=3e^2-e

Thanks a lot!

aswoods · **Posted:** Sun, 21 Aug 2011 13:06:52 UTC

yoyobarn wrote:

$h(u)=\int_0^u e^{\sqrt{u}+t} dt$

$\frac{dh}{du}=e^{\sqrt{u}+u$ , by FTC. ← ERROR

$\frac{du}{dx}=2x$

The integrand is not a function solely of t, but of u and t.

To see this, bring $e^{\sqrt{u}}$ outside the integral, and apply the product rule:
$\\h(u)=e^{\sqrt{u}}\int_0^ue^t\mathrm{d}t\\h'(u)=e^{\sqrt{u}}\frac{1}{2\sqrt{u}}\int_0^ue^t\mathrm{d}t+e^{\sqrt{u}}e^u$
The final e^u is provided by the FTC.

yoyobarn · **Joined:** Fri, 12 Nov 2010 07:00:12 UTC **Posts:** 48

thanks a lot.

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Spot the mistake in Fundamental Theorem of Calculus question

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