Science:Math Exam Resources/Courses/MATH105/April 2009/Question 01 (j)

MATH105 April 2009

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Question 01 (j)
Find the area under the graph of $\displaystyle y=e^{-2x}$ for $\displaystyle x\geq 0$

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
To proceed, we want to evaluate the integral $\displaystyle \int _{0}^{\infty }e^{-2x}\,dx$

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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. This question is asking for the value of $\displaystyle \int _{0}^{\infty }e^{-2x}\,dx$ To solve this, we solve the improper integral as follows. First change to a limit $\displaystyle \int _{0}^{\infty }e^{-2x}\,dx=\lim _{b\to \infty }\int _{0}^{b}e^{-2x}\,dx$ Next we integrate: $\displaystyle \lim _{b\to \infty }\int _{0}^{b}e^{-2x}\,dx=\lim _{b\to \infty }{\frac {e^{-2x}}{-2}}{\Bigr |}_{0}^{b}$ Plug in the endpoints: $\displaystyle \lim _{b\to \infty }{\frac {e^{-2x}}{-2}}{\Bigr |}_{0}^{b}=\lim _{b\to \infty }\left({\frac {e^{-2b}}{-2}}-{\frac {e^{-2(0)}}{-2}}\right)$ And we notice that the first limit is 0 (e to a large negative number is 0) and the second is independent of the limit and thus $\displaystyle \lim _{b\to \infty }\left({\frac {e^{-2b}}{-2}}-{\frac {e^{-2(0)}}{-2}}\right)={\frac {1}{2}}$ Thus $\displaystyle \int _{0}^{\infty }e^{-2x}\,dx={\frac {1}{2}}$ completing the question.

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