Science:Math Exam Resources/Courses/MATH200/December 2011/Question 05 (a)

MATH200 December 2011

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q2 (a)  •  Q2 (b)  •  Q3 (a)  •  Q3 (b)  •  Q4  •  Q5 (a)  •  Q5 (b)  •  Q6  •  Q7  •  Q8 (a)  •  Q8 (b) i  •  Q8 (b) ii  •  Q8 (b) iii  •  

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  •  April 2012  •  December 2011  •  

Question 05 (a)
Consider the iterated integral $\int _{-4}^{0}\int _{\sqrt {-y}}^{2}\cos(x^{3})\,dx\,dy.$ i) Draw the region of integration. ii) Evaluate the integral.

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
For evaluating the integral, change the order of integration.

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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. (i) (ii) Since $\displaystyle \cos(x^{3})$ is hard to integrate with respect to $\displaystyle x$, we change the order of integration. Using the sketch of (i) and that $\displaystyle x={\sqrt {-y}}$ is equivalent to $\displaystyle y=-x^{2}$ (and $\displaystyle x\geq 0$) we find that $\displaystyle {\begin{aligned}\int _{-4}^{0}\int _{\sqrt {-y}}^{2}\cos(x^{3})\;{\text{d}}x{\text{d}}y&=\int _{0}^{2}\int _{-x^{2}}^{0}\cos(x^{3})\;{\text{d}}y{\text{d}}x\\&=\int _{0}^{2}\cos(x^{3})[y]_{-x^{2}}^{0}\;{\text{d}}x\\&=\int _{0}^{2}\cos(x^{3})x^{2}\;{\text{d}}x\\&=[{\frac {1}{3}}\sin(x^{3})]_{0}^{2}={\frac {1}{3}}\sin(8)-{\frac {1}{3}}\sin(0)\\&={\frac {1}{3}}\sin(8)\end{aligned}}$

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