Science:Math Exam Resources/Courses/MATH103/April 2005/Question 05 (a)/Solution 1

There are actually two cases we have to consider here, if p=1 and if $p\neq 1$. We will see why below:

First assume that $p\neq 1$. Given the function $y=f(x)={\frac {1}{x^{p}}}$ on the interval [1,B], the area between this curve and the x-axis is

${\begin{aligned}A&=\int _{1}^{B}{\frac {1}{x^{p}}}dx=\int _{1}^{B}x^{-p}dx\\&=\left[{\frac {x^{-p+1}}{-p+1}}\right]_{1}^{B}={\frac {1}{1-p}}\left(B^{1-p}-1\right)\end{aligned}}$

Note that, if p=1, then the above does not work because we would divide by 1-p=0.

Hence we treat p=1 separately. The function becomes $y=f(x)={\frac {1}{x}}$ and on the interval [1,B], the area between this curve and the x-axis is

${\begin{aligned}A&=\int _{1}^{B}{\frac {1}{x}}dx\\&=\left[\ln {x}\right]_{1}^{B}=\ln {B}\end{aligned}}$

So our final answer is

$A=\int _{1}^{B}{\frac {1}{x^{p}}}\,dx={\begin{cases}{\frac {1}{1-p}}\left(B^{1-p}-1\right),&p\neq {1}\\\ln {B},&p=1\end{cases}}$