# How do you find the area bounded by y=x^2+2x-4 and y=-2x^2+4x-3?

Jul 30, 2017

$A = \frac{32}{27}$

#### Explanation:

Consider the function:

$f \left(x\right) = \left({x}^{2} + 2 x - 4\right) - \left(- 2 {x}^{2} + 4 x - 3\right)$

$f \left(x\right) = 3 {x}^{2} - 2 x - 1$

The values of $x$ for which the two curves intersect are the solutions of the equation:

$f \left(x\right) = 0$

$3 {x}^{2} - 2 x - 1 = 0$

$x = \frac{1 \pm \sqrt{1 + 3}}{3}$

${x}_{1} = - \frac{1}{3}$, ${x}_{2} = 1$

Note now that as $f \left(x\right)$ is a second degree polynomial with leading positive coefficient, its value is negative in the interval between the roots.

The area bounded by the curves is then:

$A = {\int}_{- \frac{1}{3}}^{1} \left\mid f \left(x\right) \right\mid \mathrm{dx} = {\int}_{- \frac{1}{3}}^{1} \left(- 3 {x}^{2} + 2 x + 1\right) \mathrm{dx}$

$A = {\left[- {x}^{3} + {x}^{2} + x\right]}_{- \frac{1}{3}}^{1}$

$A = - 1 + 1 + 1 - \frac{1}{27} - \frac{1}{9} + \frac{1}{3} = \frac{27 - 1 - 3 + 9}{27} = \frac{32}{27}$

Jul 30, 2017

Use the integral of the difference between the two functions:

"Area" = int_a^bf(x)-g(x)dx; f(x)>g(x), (a,b)

#### Explanation:

Here is a graph of

$y = {x}^{2} + 2 x - 4 \text{ [1]}$ (in red)

and

$y = - 2 {x}^{2} + 4 x - 3 \text{ [2]}$ (in blue)

Pleases observe that equation [2] is greater than equation [1] in the enclosed region; this means that the integral is of the form:

${\int}_{a}^{b} - 2 {x}^{2} + 4 x - 3 - \left({x}^{2} + 2 x - 4\right) \mathrm{dx}$

Simplify the integrand:

${\int}_{a}^{b} - 3 {x}^{2} + 2 x + 1 \mathrm{dx}$

We need to find the values of "a" and "b" by finding the two x coordinates where the two parabolas intersect. We can do this by setting the integrand equal to 0 and then solving for the two values of x:

$- 3 {x}^{2} + 2 x + 1 = 0$

Multiply by -1:

$3 {x}^{2} - 2 x - 1 = 0$

This factors into:

$\left(x - 1\right) \left(3 x + 1\right)$

$x = 1 \mathmr{and} x = - \frac{1}{3}$

The above are the values of "a" and "b":

${\int}_{- \frac{1}{3}}^{1} - 3 {x}^{2} + 2 x + 1 \mathrm{dx} =$

${\left[- {x}^{3} + {x}^{2} + x\right]}_{- \frac{1}{3}}^{1} =$

$- {\left(1\right)}^{3} + {\left(1\right)}^{2} + 1 + {\left(\frac{1}{3}\right)}^{3} - {\left(\frac{1}{3}\right)}^{2} - \frac{1}{3} = \frac{32}{27}$