MATH105 April 2013
• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q1 (f) • Q1 (g) • Q1 (h) • Q1 (i) • Q1 (j) • Q1 (k) • Q1 (l) • Q1 (m) • Q1 (n) • Q2 (a) • Q2 (b) • Q2 (c) • Q3 (a) • Q3 (b) • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) • Q6 (a) • Q6 (b) •
Question 01 (c)

ShortAnswer Questions. Put your answer in the box provided but show your
work also. Each question is worth 3 marks, but not all questions are of equal difficulty.
Evaluate $\displaystyle \int _{0}^{5/2}{\frac {dx}{\sqrt {25x^{2}}}}$.

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

The identity $\displaystyle {}\sin ^{2}\theta +\cos ^{2}\theta =1$ may be useful here.

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Solution

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We start by noting that the denominator is in the form
 $\displaystyle {}a^{2}x^{2}$
with a = 5. This is a good indicator that we want to do a trig substitution with
 $\displaystyle {}x=a\sin \theta$
because
 $\displaystyle {}a^{2}\sin ^{2}\theta +a^{2}\cos ^{2}\theta =a^{2}$
and hence
 $\displaystyle {}a^{2}a^{2}\sin ^{2}\theta =a^{2}\cos ^{2}\theta$
We will use this identity to handle the square root in the denominator. So, let
 ${\begin{aligned}x&=5\sin \theta \\dx&=5\cos \theta d\theta .\end{aligned}}$
We must also modify the limits of integration since we have a new variable. When
 ${\begin{aligned}x&=0=5\sin \theta \end{aligned}}$
solving for $\theta$ gives $\displaystyle {}\theta =0$. Solving
 ${\begin{aligned}x&=5/2=5\sin \theta \end{aligned}}$
for $\theta$ gives $\displaystyle {}\theta ={\frac {\pi }{6}}$. Substituting these values into our integral we obtain
 $\int _{0}^{5/2}{\frac {dx}{\sqrt {25x^{2}}}}=\int _{0}^{\pi /6}{\frac {5\cos \theta d\theta }{\sqrt {25(5\sin \theta )^{2}}}}=\int _{0}^{\pi /6}{\frac {5\cos \theta d\theta }{\sqrt {(5)^{2}(5)^{2}(\sin \theta )^{2}}}}.$
We can use the identity above we get
 $\int _{0}^{\pi /6}{\frac {5\cos \theta d\theta }{\sqrt {(5)^{2}(5)^{2}(\sin \theta )^{2}}}}=\int _{0}^{\pi /6}{\frac {5\cos \theta d\theta }{\sqrt {(5)^{2}(\cos \theta )^{2}}}}=\int _{0}^{\pi /6}1d\theta ={\frac {\pi }{6}}$
Since this is a definite integral, the result is a number and we do not need to return to the previous variables.

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