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

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MATH101 April 2016

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Question 02 (b)
Which integral equals $\displaystyle \lim _{n\rightarrow \infty }{\bigg (}\sum _{i=1}^{n}e^{-1-2i/n}\cdot {\frac {2}{n}}{\bigg )}$ ? ${\begin{aligned}{{\textbf {F}}\!:}&\,\int _{0}^{1}e^{-1-2x}\ dx&{{\textbf {J}}\!:}&\,\int _{0}^{2}e^{-1-2x}\ dx\\{{\textbf {G}}\!:}&\ 2\int _{0}^{2}e^{-1-2x}\ dx&{{\textbf {K}}\!:}&\,\int _{1}^{3}e^{-x}\ dx\\{{\textbf {H}}\!:}&\,\int _{0}^{2}e^{-x}\ dx&\end{aligned}}$

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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
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
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 $\lim _{n\to \infty }\sum _{i=1}^{n}e^{-(1+i{\frac {2}{n}})}{\frac {2}{n}}=\int _{1}^{3}e^{-x}dx$