How do you find the area between the curves y=4x-x^2 and y=x?

Feb 20, 2015

First, we need to find where the two functions intersect.

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

${x}^{2} - 4 x + x = 0$

${x}^{2} - 3 x = 0$

$x \left(x - 3\right) = 0$

$x = 0$ and $x = 3$

So our interval of integration is

$0 \le x \le 3$

$y = 4 x - {x}^{2}$ is a parabola that is concave down

$y = x$ is the line that passes through the origin with slope 1

The integral for the area is

$\int 4 x - {x}^{2} - x \mathrm{dx} = \int 3 x - {x}^{2} \mathrm{dx}$

integrating we have

$\frac{3}{2} {x}^{2} - \frac{1}{3} {x}^{3}$

Evaluating at we have

$\left(\frac{3}{2}\right) {3}^{2} - \left(\frac{1}{3}\right) {3}^{3} - 0 = \frac{27}{2} - \frac{27}{3} = \frac{27}{2} - 9$

$\frac{81}{6} - \frac{54}{6} = \frac{27}{6} = \frac{9}{2}$

Feb 20, 2015

Find the points of intersection, $\left(a , {y}_{a}\right)$ and $\left(b , {y}_{b}\right)$
between
$y = 4 x - {x}^{2}$ and $y = x$
then subtract from the integral of the first (between $a$ and $b$)
the integral of the second (again, between $a$ and $b$)

Part 1:
Points of intersection occurs when
$4 x - {x}^{2} = x$
This occurs when either $x = 0$ or $x = 3$
(we could, but don't actually need to calculate ${y}_{a}$ and ${y}_{b}$)

Part 2:
${\int}_{a}^{b} \left(4 x - {x}^{2}\right) \mathrm{dx} - {\int}_{a}^{b} x \mathrm{dx}$

$= \left(\left(2 {x}^{2} - {x}^{3} / 3\right) {|}_{a}^{b}\right) - \left(\left(\frac{{x}^{2}}{2}\right) {|}_{a}^{b}\right)$

with $a = 0$ and $b = 3$
this becomes
$\left(\left(2 \cdot 9 - \frac{9}{3}\right) - \left(0\right)\right) - \left(\left(\frac{9}{2}\right) - \left(0\right)\right)$
$= 10.5$