# Question #16bc5

Apr 8, 2017

#### Explanation:

The curves intersect where $f \left(x\right) = g \left(x\right)$, so we need to solve

${x}^{2} - 4 x + 3 = - {x}^{2} + 2 x + 3$

$2 {x}^{2} - 6 x = 0$ at $0$ and $3$.

We need to integrate from $0$ to $3$.

To make sure we get the correct sign, we should determine which function is greater on the interval $\left(0 , 3\right)$.
One way to do this is to simply choose a test number in the interval, say $x = 1$. Since $f \left(1\right) = 0$ and $g \left(1\right) = 4$, the graph of $g$ is above the graph of $f$. (Also see Note below.)

The desired area is

${\int}_{0}^{3} \left(g \left(x\right) - f \left(x\right)\right) \mathrm{dx} = {\int}_{0}^{3} \left(- 2 {x}^{2} + 6 x\right) \mathrm{dx}$

$= {\left[- \frac{2 {x}^{3}}{3} + 3 {x}^{2}\right]}_{0}^{3}$

$= \left[\left(- 18 + 27\right) - \left(- 0 + 0\right)\right] = 9$

Note

Another way to determine which graph is on top is to sketch the graphs.

It is also possible to get the correct answer by choosing an order of subtraction and the making the final answer positive. That is, we can find the absolute value of either integral.
Be careful! Some graders do not approve of this method. They want you to put the subtraction in the order that will give a positive answer. If possible, check with the person who will be evaluating your work.

$\left\mid {\int}_{0}^{3} \left(g \left(x\right) - f \left(x\right)\right) \mathrm{dx} \right\mid = \left\mid {\int}_{0}^{3} \left(f \left(x\right) - g \left(x\right)\right) \mathrm{dx} \right\mid$