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While revising, I came across this question:

Let $h(x)=\int_0^{x^2}e^{x+t}dt$.

Find $h'(1)$.

I tried using the substitution $t=u^2$, then $dt=2u du$. The integral becomes $h(x)=\int_0^x{e^{x+u^2}}\cdot 2udu$.

Then by the Fundamental Theorem of Calculus, $h '(x)=e^{x+x^2}\cdot 2x$, and hence $h'(1)=2e^2$.

However, the correct answer is $3e^2-e$.

May I know what is wrong with the above approach?

Thanks a lot.

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1  
Notice that the theorem can't give out $F'(x)=f(x,x)$ when applying to $F(x)=\int_a^x f(x,t)dt$ where $f(x,t)$ is related to $x$. –  Frank Science Jun 28 '12 at 9:23
    
The function $h(x)$ depends on $x$ in two different ways: 1) because the upper bound of the interval of the definite integral depends on $x$, 2) because the integrand itself also depends on $x$. You have calculated the contribution from 1), but have not done anything about 2). –  Jyrki Lahtonen Jun 28 '12 at 9:28

3 Answers 3

up vote 3 down vote accepted

$\frac{d({\int_{a(x)}^{b(x)}} f(x,t)dt)}{dx}= f(x,b(x)).b'(x)-f(x,a(x)).a'(x)$+$\int_{a(x)}^{b(x)}(\frac{\partial}{\partial x}f(x,t))dt$. Putting $f(x,t)=e^{x+t}$,this gives $h'(x)=(e^{x+x^2}2x)-(0)+(e^{x+x^2}-e^x) \implies h'(x)=e^{x+x^2}(2x+1)-e^x\implies h'(1)=3e^2-e $.

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+1 Looks good! Siminore's way is simpler, but doesn't really explain, where the OP went wrong. –  Jyrki Lahtonen Jun 28 '12 at 11:23
    
My fear is that @yoyostein does not know how to differentiate functions of two variables. The function $h$ is $x \mapsto F(x^2,x)$, where $$F(v,w) = \int_0^v e^{w+t}\, dt.$$ To use the chain rule, he needs to compute the differential $dF$. –  Siminore Jun 28 '12 at 11:37
    
But, if he is doing this kind of problems he needs to know this.He can't always split $f(x,t)$ into $g_1(x).g_2(t)$. –  Aang Jun 28 '12 at 11:39
    
I can't know that. This might be a problem in calculus of one variable. Indeed, it is ;-) –  Siminore Jun 28 '12 at 12:28

Observe that $$ h(x) = e^x \int_0^{x^2} e^t \, dt = e^x \left( e^{x^2}-1 \right). $$ Now you can compute $h'(x)$ for every $x$, I guess.

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As Siminore has pointed out $$ h(x) = e^x \int_0^{x^2} e^t \, dt = e^x \left( e^{x^2}-1 \right). $$

Now

$$ h'(x) = e^{x^2+x} \left( 2x+1 \right) -e^x. $$

Replacing x with 1 we get $$ h'(x) = 3e^{2} -e. $$

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