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

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

${\begin{aligned}g(x)&=\int _{0}^{1}(xe^{t}-t)^{2}dt\\&=\int _{0}^{1}(x^{2}e^{2t}-2xe^{t}t+t^{2})dt\\&=x^{2}\int _{0}^{1}e^{2t}dt-2x\int _{0}^{1}e^{t}tdt+\int _{0}^{1}t^{2}dt\\\end{aligned}}$

The first integral in the sum is

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

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

${\begin{aligned}\int _{0}^{1}e^{t}tdt&=te^{t}{\big |}_{0}^{1}-\int _{0}^{1}e^{t}dt\\&=(e-0)-(e-1)=1\end{aligned}}$

The last integral in the sum is:

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

So

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

The derivative is directly computed to be:

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

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

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