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

MATH103 April 2005

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Question 05 (a)
Consider the curve given by $\displaystyle y=f(x)={\frac {1}{x^{p}}}$ for $\displaystyle 1\leq x\leq B$ where $\displaystyle p,B$ are constants with $\displaystyle B>1$ and $\displaystyle p>0$. (a) Compute the area of the region between this curve and the x-axis for $\displaystyle 1\leq x\leq B$. (Later we will refer to this area as A.)

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
The area A under the graph of a function y = f(x) on the interval [a,b] is $A=\int _{a}^{b}f(x)dx$

Hint 2
Note the subtlety when p = 1, this is a special case that needs to be treated separately!

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

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