Question

Find the area between the following curves. y=x^3-x^2+x+8 ; y=5x^2-7x+8

Answer 1

Find where the two curves meet.

$x^3 - x^2 + x + 8 = 5x^2 - 7x + 8 \\\\ \implies x^3 - 6x^2 + 8x = 0 \\\\ \implies x (x-2) (x - 4)= 0 \implies x=0, x=2, x=4$

The area between the curves is

$\displaystyle \int_0^4 \left|\left(x^3-x^2+x+8\right) - \left(5x^2 - 7x + 8\right)\right| \, dx = \int_0^4 \left|x(x-2)(x-4)\right| \, dx$

When $x$ is between 0 and 2, $x(x-2)(x-4)$ is positive; when $x$ is between 2 and 4, $x(x-2)(x-4)$ is negative. So we split the integral at $x=2$ to get

$\displaystyle \int_0^2 x(x-2)(x-4) \, dx - \int_2^4 x(x-2)(x-4)\,dx$

In the second integral, substitute $y=x-2$ to get

$\displaystyle \int_0^2 x(x-2)(x-4) \, dx - \int_0^2 (y+2)y(y-2)\,dy$

$\displaystyle \int_0^2 x(x-2) \bigg((x-4) - (x+2)\bigg) \, dx$

$\displaystyle -6 \int_0^2 x(x-2) \, dx$

$\displaystyle 6 \int_0^2 \left(2x - x^2\right) \, dx$

$\displaystyle 6 \left(x^2 - \frac13x^3\right)\bigg|_0^2 = \boxed{8}$