Science:Math Exam Resources/Courses/MATH101/April 2017/Question 03 (a)/Solution 1

Consider the two curves $y=x-8$ and $y=-x^{2}+7x-16$. Note that $y=x-8$ is a line, and $y=-x^{2}+7x-16$ is a parabola whose leading coefficient is negative (so it is shaped like an up-side-down U).

First, we find the intersection points of the two graphs. By solving

$$x-8=-x^{2}+7x-16\iff x^{2}-6x+8=0\iff (x-4)(x-2)=0,$$

the intersection points are obtained at $x=2$ and $x=4$.

Then, the area we are looking for is

$$\int _{2}^{4}|x-8-(-x^{2}+7x-16)|\,dx$$

.

Either drawing the graphs or plugging a number in the interval $(2,4)$, we can see that $-x^{2}+7x-16\geq x-8$ when $2\leq x\leq 4$.

Therefore, we have $|x-8-(-x^{2}+7x-16)|=-x^{2}+7x-16-(x-8)=-x^{2}+6x-8$ on $(2,4)$ and plug this into integral to have

$${\begin{aligned}\int _{2}^{4}|x-8-(-x^{2}+7x-16)|\,dx&=\int _{2}^{4}-x^{2}+6x-8dx=-{\frac {1}{3}}x^{3}+3x^{2}-8x{\bigg |}_{2}^{4}\\&=\left(-{\frac {1}{3}}\cdot 4^{3}+3\cdot 4^{2}-8\cdot 4\right)-\left(-{\frac {1}{3}}\cdot 2^{3}+3\cdot 2^{2}-8\cdot 2\right)=-{\frac {16}{3}}+{\frac {20}{3}}={\frac {4}{3}}\end{aligned}}$$

Here, the Power Rule of differentiation is used. Therefore, the area between the two graphs is ${\frac {4}{3}}$

Answer: $\color {blue}{\frac {4}{3}}$