Science:Math Exam Resources/Courses/MATH101/April 2015/Question 01 (c)/Solution 1

Note that by integration rule of power functions we have

$\int x^{q}dx={\begin{cases}{\frac {1}{q+1}}x^{q+1}+C&q\neq -1\\\log x+C&q=-1.\end{cases}}$.

In the case of $q=-1$, the given integral can be calculated as

$\int _{0}^{1}{\frac {1}{x}}dx=\lim _{\delta \to 0+}\int _{\delta }^{1}{\frac {1}{x}}dx=\lim _{\delta \to 0+}\log x{\bigg |}_{\delta }^{1}=\lim _{\delta \to 0+}(-\log \delta )=\infty .$

This implies that we lose the convergence of the integral, so that options K,M and P are not the answer, because they include $q=-1$.

Now assume $q+1<0$. Then, we have

$\int _{0}^{1}x^{q}dx=\lim _{\delta \to 0+}\int _{\delta }^{1}x^{q}dx=\lim _{\delta \to 0+}{\frac {1}{q+1}}x^{q+1}{\bigg |}_{\delta }^{1}=\lim _{\delta \to 0+}{\frac {1}{q+1}}(1-\delta ^{q+1})=-\infty .$

We used the assumption in the last inequality. Therefore the integral diverges also when $q+1<0$ i.e when $q<-1$, so option J is also wrong.

However, when $q+1>0$, $x^{q+1}$ is a power function with positive exponent (polynomial) and hence

$\int _{0}^{1}x^{q}dx={\frac {1}{q+1}}x^{q+1}|_{0}^{1}={\frac {1}{q+1}}(1-0)={\frac {1}{q+1}}$.

Correct choice is $\color {blue}L$.