Question:

Consider the curves $y = x^3 + 2x^2 - 5x$ and $y = 2x^2 - 4x$ on the interval $-1 \leq x \leq 2$. Find the area of the region enclosed by these two curves.

Answer:

To find the area between the curves, we need to set up an integral and find the difference between the upper and lower curves.

First, we'll find the points of intersection of the two curves by setting them equal to each other:

$x^3 + 2x^2 - 5x = 2x^2 - 4x$

$x^3 - 3x = 0$

Factoring out an x gives us:

$x(x^2 - 3) = 0$

So, $x = 0$ and $x = \sqrt{3}$ are the points of intersection.

Next, we'll set up the integral to find the area between the curves:

$A = \int_{-1}^{0} [(2x^2 - 4x) - (x^3 + 2x^2 - 5x)] dx + \int_{0}^{\sqrt{3}} [(2x^2 - 4x) - (x^3 + 2x^2 - 5x)] dx$

Simplify the integrands:

$A = \int_{-1}^{0} (-x^3) dx + \int_{-1}^{0} (-4x) dx + \int_{0}^{\sqrt{3}} (-x^3 - 2x^2) dx + \int_{0}^{\sqrt{3}} (x) dx$

Now, integrate each term:

$A = [(-\frac{1}{4}x^4)_{-1}^{0}] + [(-2x^2)_{-1}^{0}] + [(-\frac{1}{4}x^4 - \frac{2}{3}x^3)_{0}^{\sqrt{3}}] + [(\frac{1}{2}x^2)_{0}^{\sqrt{3}}]$

Evaluate the definite integrals:

$A = [-0 - (-\frac{1}{4}(-1)^4)] + [0 - (-2(-1)^2)] + [(-\frac{1}{4}(\sqrt{3})^4 - \frac{2}{3}(\sqrt{3})^3) - (-\frac{1}{4}(0)^4 - \frac{2}{3}(0)^3)] + [(\frac{1}{2}(\sqrt{3})^2 - \frac{1}{2}(0)^2)]$

$A = [\frac{1}{4}] + [2] + [\frac{9\sqrt{3}}{4} - \frac{2\sqrt{3}}{3}] + [\frac{3}{2}]$

So, the area between the curves is:

$A = \frac{17}{4} + \frac{9\sqrt{3}}{4} - \frac{2\sqrt{3}}{3} \approx 10.805$

Thus, the area between the curves is approximately 10.805 square units.