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Problem:

Let: $F(x)=\int_{\sin x}^{\cos x}e^{t^{2}+xt}\,dt$. Compute F^{'}(0).

I tried the following: $F(x)=\int_{a}^{\cos x}e^{t^{2}+xt}\,dt - \int_{a}^{\sin x}e^{t^{2}+xt}\,dt$ where $a$ is between $\cos x$ and $\sin x$. Then I applied the second Fundamental theorem of Calculus to get: F^{'}(x)=e^{\cos^{2}x+x\cos x}-e^{\sin^{2}x+x\sin x} and Hence F^{'}(0)=e-1. But in the book, the given solution is : $\frac{e-3}{2}$. Can anyone tell me what's wrong with my proof and give me the correct solution?

  • 0
    Do you mean to have the $x$ in the exponent in the integrand?2012-01-26

2 Answers 2

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$F(x)=\int_{\sin x}^{\cos x}e^{t^{2}+xt}\,dt$

To compute $F^{'}(0)$ let's use my latest derivation (applied to $a(x)=\sin x$, $b(x)=\cos x$, $h(x,t)=t^2+xt$) :

$F'(x)=\int_{a(x)}^{b(x)}\frac{\partial h(x,t)}{\partial x}.e^{h(x,t)}dt+b'(x)e^{h(x,b(x))}-a'(x)e^{h(x,a(x))}$

So that $F'(0)=\int_{sin(0)}^{cos(0)}t.e^{t^2+0t}dt-\sin(0)e^{cos(0)^2+0\cos(0)}-\cos(0)e^{\sin(0)^2+0\sin0))}$

or $F'(0)=\int_0^1 t.e^{t^2}dt-1=\frac{e^{t^2}}2|_0^1-1=\frac{e-3}2$

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For convenience I use $x$ as the variable of integration. Let $ F(t)=I(u(t),v(t),t)=\int_{u(t)}^{v(t)}f(x,t)dx $

You have variable limits and need to differentiate under the integral sign. By the Leibniz integral rule, the derivative is

$\begin{eqnarray*} F^{\prime }(t) &=&\frac{\partial I}{\partial t}\frac{dt}{dt}+\frac{\partial I }{\partial v}\frac{dv}{dt}+\frac{\partial I}{\partial u}\frac{du}{dt} \\ &=&\int_{u(t)}^{v(t)}\frac{\partial f(x,t)}{\partial t}dx+f(v(t),t)v^{\prime }(t)-f(u(t),t)u^{\prime }(t). \end{eqnarray*}$

Hence $ F^{\prime }(0)=\int_{u(0)}^{v(0)}\left. \frac{\partial f(x,t)}{\partial t} \right\vert _{t=0}dx+f(v(0),0)v^{\prime }(0)-f(u(0),0)u^{\prime }(0). $

In the present case $ \begin{eqnarray*} u(t) &=&\sin t,\qquad u(0)=0 \\ u^{\prime }(t) &=&\cos t,\qquad u^{\prime }(0)=1 \\ v(t) &=&\cos t,\qquad v(0)=1 \\ v^{\prime }(t) &=&-\sin t,\qquad v^{\prime }(0)=0. \end{eqnarray*}$

and $ \begin{eqnarray*} f(x,t) =e^{x^{2}+xt}, \qquad f(1,0) =e, \qquad f(0,0) =1. \end{eqnarray*}$

So $ F^{\prime }(0)=\int_{0}^{1}\left. \frac{\partial f(x,t)}{\partial t} \right\vert _{t=0}dx+0-1 $

Since $ \left. \frac{\partial f(x,t)}{\partial t}\right\vert _{t=0}=\left. \frac{ \partial }{\partial t}e^{x^{2}+xt}\right\vert _{t=0}=xe^{x\left( x+t\right) }\vert _{t=0}=xe^{x^{2}} $

and $ \int_{0}^{1}xe^{x^{2}}dx=\frac{1}{2}e-\frac{1}{2} $

we finally get $ F^{\prime }(0)=\frac{1}{2}e-\frac{1}{2}-1=\frac{1}{2}e-\frac{3}{2}. $

You find another example here.