Science:Math Exam Resources/Courses/MATH101/April 2016/Question 02 (b)

MATH101 April 2016

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Question 02 (b)

Which integral equals $\displaystyle \lim _{n\rightarrow \infty }{\bigg (}\sum _{i=1}^{n}e^{-1-2i/n}\cdot {\frac {2}{n}}{\bigg )}$? ${\begin{aligned}{{\textbf {F}}\!:}&\,\int _{0}^{1}e^{-1-2x}\ dx&{{\textbf {J}}\!:}&\,\int _{0}^{2}e^{-1-2x}\ dx\\{{\textbf {G}}\!:}&\ 2\int _{0}^{2}e^{-1-2x}\ dx&{{\textbf {K}}\!:}&\,\int _{1}^{3}e^{-x}\ dx\\{{\textbf {H}}\!:}&\,\int _{0}^{2}e^{-x}\ dx&\end{aligned}}$

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
Recall $\lim _{n\to \infty }\sum _{i=1}^{n}f(x_{i})\Delta x=\int _{a}^{b}f(x)dx$, where $\Delta x={\dfrac {b-a}{n}},x_{i}=a+i\Delta x$ . Compare the given summation with the above formula to find $\Delta x$ and then $a,b$.

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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. By the definition of the integral as a summation we have $\lim _{n\to \infty }\sum _{i=1}^{n}f(x_{i})\Delta x=\int _{a}^{b}f(x)dx$ Now comparing this with the question we see that $\Delta x={\frac {2}{n}}={\frac {b-a}{n}}$ so we must have $b-a=2$ So far we know that the interval of the integration has length 2. Also note that $x_{i}=a+i\Delta x$ Now let's see what $f(x_{i})$ is: Again compare the formula with the question $f(x_{i})=f(a+i\Delta x)=e^{-(1+i{\frac {2}{n}})}=e^{-(1+i\Delta x)}=e^{-x_{i}}$ we see that $f(x)=e^{-x}$ and $a=1$ since $b-a=2$, so we have $b=3$ now we have all the elements of the integral therefore $\lim _{n\to \infty }\sum _{i=1}^{n}e^{-(1+i{\frac {2}{n}})}{\frac {2}{n}}=\int _{1}^{3}e^{-x}dx$ ${\displaystyle }$