Science:Math Exam Resources/Courses/MATH103/April 2017/Question 06 (a)

MATH103 April 2017

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Question 06 (a)
Consider the function $y=f(x)=x^{-{\frac {3}{2}}}-1$. (a) Calculate the area, $A$, bounded by the $x$-axis, $y$-axis and $y=f(x)$ (shaded area in figure) or show that it does not exist

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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
First, set up the integral. Then, as the integral could be improper, integrate the integrand first.

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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. The area that we wish to determine is the left shaded area, which equals to the following integral $\int _{0}^{1}x^{-3/2}-1{\text{ }}dx.$ Integrating the integrand gives, by the Power Rule of Differentiation, $\int x^{-3/2}-1{\text{ }}dx=-2x^{-1/2}-x$ We see that $-2x^{-1/2}-x$ is defined everywhere on $(0,1]$ but undefined at $x=0$. Hence, we have an improper integral, and it equals to $\int _{0}^{1}x^{-3/2}-1{\text{ }}dx=\lim _{\epsilon \to 0^{+}}\int _{\epsilon }^{1}x^{-3/2}-1{\text{ }}dx$ Evaluating this using the Fundamental Theorem of Calculus now gives $\lim _{\epsilon \to 0^{+}}\int _{\epsilon }^{1}x^{-3/2}-1{\text{ }}dx=\lim _{\epsilon \to 0^{+}}((-2-1)-(-2\epsilon ^{-1/2}-\epsilon ))$ We note that $\lim _{\epsilon \to 0^{+}}\epsilon =0$ and that $\lim _{\epsilon \to 0^{+}}\epsilon ^{-1/2}=\infty$. Hence, the integral diverges and the area is $\color {blue}\int _{0}^{1}x^{-3/2}-1{\text{ }}dx=\infty$