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

MATH101 April 2010

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Question 09
For any real number $x$, define $g(x)=\int _{0}^{1}(xe^{t}-t)^{2}dt$. Find the minimum value of $g(x)$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
We can consider $x$ as a constant inside the integral

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 ${\begin{aligned}g\left({\frac {2}{e^{2}-1}}\right)&=\left({\frac {2}{e^{2}-1}}\right)^{2}{\frac {e^{2}-1}{2}}-2\left({\frac {2}{e^{2}-1}}\right)+{\frac {1}{3}}\\&=\left({\frac {4}{(e^{2}-1)^{2}}}\right){\frac {e^{2}-1}{2}}-{\frac {4}{e^{2}-1}}+{\frac {1}{3}}\\&={\frac {2}{e^{2}-1}}-{\frac {4}{e^{2}-1}}+{\frac {1}{3}}\\&=-{\frac {2}{e^{2}-1}}+{\frac {1}{3}}\\&=-{\frac {6+e^{2}+1}{3(e^{2}-1)}}\\&={\frac {e^{2}-7}{3(e^{2}-1)}}\end{aligned}}$

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