Science:Math Exam Resources/Courses/MATH101/April 2016/Question 01 (b)

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MATH101 April 2016

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Question 01 (b)
Which integral represents the area between the graphs of $y=2^{-x}$ and $y=1-x$ ? ${\begin{aligned}{{\textbf {A}}{:}}&\,\int _{0}^{1}{\big (}2^{-x}-(1-x){\big )}\,dx&{{\textbf {C}}{:}}&\,\int _{0}^{1}{\big (}(2^{-x})^{2}-(1-x)^{2}{\big )}\,dx&{{\textbf {F}}{:}}&\,\int _{0}^{1}{\big (}(1-x)-2^{-x}{\big )}\,dx\\{{\textbf {B}}{:}}&\,\int _{-1}^{0}{\big (}2^{-x}-(1-x){\big )}\,dx&{{\textbf {D}}{:}}&\,\int _{-1}^{0}{\big (}(1-x)^{2}-(2^{-x})^{2}{\big )}\,dx&{{\textbf {H}}{:}}&\,\int _{-1}^{0}{\big (}(1-x)-2^{-x}{\big )}\,dx\end{aligned}}$

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
Draw the graphs of given functions and find the desired area.

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Solution
$f(x)=2^{-x}$ and $g(x)=1-x$ have intersection points at $x=-1$ and $x=0$ , so the area between the two graphs is in the interval $[-1,0]$ , this means the choices A,C,F cannot be the answer. Next, we need to determine which of the two graphs is above the other, because area is a positive quantity and we must choose the integral with positive integrand. Note that from the graph of exponential functions we know that they are always concave up, and in particular $f(x)=2^{-x}$ must be below the line $g(x)=1-x$ in $[-1,0]$ . We can also determine the concavity of $f(x)=2^{-x}$ by finding the sign of its second derivative: By the formula of derivative of exponential function, we have: $f'(x)=(-\log 2)2^{-x}$ and so $f''(x)=(\log 2)^{2}2^{-x}$ which is always positive, this means that the graph of $f(x)$ is below $g(x)$ in $[-1,0]$ . In other words, $(1-x)-2^{-x}\geq 0$ in $[-1,0]$ . So the solution is ${\color {blue}H=\int _{-1}^{0}((1-x)-2^{-x})dx}$