Science:Math Exam Resources/Courses/MATH101/April 2016/Question 01 (b)/Solution 1

$f(x)=2^{-x}$ 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 ${\color {blue}H=\int _{-1}^{0}((1-x)-2^{-x})dx}$