Science:Math Exam Resources/Courses/MATH101/April 2010/Question 09/Solution 1

First we solve the integral considering $x$ as a constant:

$\begin{array}{rl}g(x)& =\underset{0}{\overset{1}{\int }}(x{e}^t-t{)}^{2}dt\\ & =\underset{0}{\overset{1}{\int }}(x^2{e}^{2t}-2x{e}^{t}t+t^2)dt\\ & =x^2\underset{0}{\overset{1}{\int }}{e}^{2t}dt-2x\underset{0}{\overset{1}{\int }}{e}^{t}tdt+\underset{0}{\overset{1}{\int }}{t}^{2}dt\\ \end{array}$

The first integral in the sum is

$\underset{0}{\overset{1}{\int }}{e}^{2t}dt=\frac{1}{2}(e^2-e^0)=\frac{1}{2}(e^2-1)$

Now we solve $\underset{0}{\overset{1}{\int }}{e}^{t}tdt$ by parts letting $u=t$ and $dv=e^{t}dt$ so that $du=dt$ and $v=e^t$. Hence

$\begin{array}{rl}\underset{0}{\overset{1}{\int }}{e}^{t}tdt& =t{e}^t{|}_0^1-\underset{0}{\overset{1}{\int }}{e}^{t}dt\\ & =(e-0)-(e-1)=1\end{array}$

The last integral in the sum is:

$\underset{0}{\overset{1}{\int }}{t}^{2}dt=\frac{1}{3}(1^3-0^3)=\frac{1}{3}$

So

$g(x)=\frac{x^2}{2}(e^2-1)-2x+\frac{1}{3}$

The derivative is directly computed to be:

$g^{\prime }(x)=x(e^2-1)-2$

Which is zero at $x=\frac{2}{e^2-1}$.

To make sure it is a minimum we use the second derivative test. The second derivative is:

$g^{″}(x)=e^2-1$

Which is always positive (Remember that $e>2$). This means $x=\frac{2}{e^2-1}$ is the minimum for $g$. Plugging in this value gives

$\begin{array}{rl}g\left(\frac{2}{e^2-1}\right)& ={\left(\frac{2}{e^2-1}\right)}^2\frac{e^2-1}{2}-2\left(\frac{2}{e^2-1}\right)+\frac{1}{3}\\ & =\left(\frac{4}{(e^2-1{)}^2}\right)\frac{e^2-1}{2}-\frac{4}{e^2-1}+\frac{1}{3}\\ & =\frac{2}{e^2-1}-\frac{4}{e^2-1}+\frac{1}{3}\\ & =-\frac{2}{e^2-1}+\frac{1}{3}\\ & =-\frac{6+e^2+1}{3(e^2-1)}\\ & =\frac{e^2-7}{3(e^2-1)}\end{array}$