Area integrals calculus

Question

Determine the area bounded by the following curve: y = x^4 − 2x^3 − 3x^2 and the x-axis

Answer 1

The graph of y = x⁴ − 2x³ − 3x² looks like this:

so I can only assume you mean _-1∫³(x⁴ − 2x³ − 3x²) dx

_-1∫³(x⁴ − 2x³ − 3x²) dx = [(1/5x⁵ - (1/2)x⁴ - x³] evaluated between -1 and 3 =

[(1/5(3)⁵ - (1/2)(3)⁴ - (3)³] - [(1/5(-1)⁵ - (1/2)(-1)⁴ - (-1)³] =

243/5 - 81/2 - 27 - [-1/5 - 1/2 + 1] =

486/10 - 405/10 - 270/10 - (-2/10 - 5/10 + 10/10) =

-189/10 - 3/10 = -192/10 = -19.2

Answer 2

For this problem, we are asked about the area between a curve and the x-axis, suggesting that we're going to need perform an integral across one or more intervals.

I would highly suggest using a calculator or some graphing software to graph this function to get a sense for the integrals that we are about to do. (example here: https://www.wolframalpha.com/input/?i=graph+of+x%5E4+-+2x%5E3+-+3x%5E2+on+the+interval+%5B-1.1%2C3.1%5D)

The first step is going to be to find the x - intercepts of the function so we know the bounds for our integrals.

To find the x-intercepts, we are solving the polynomial

x⁴ - 2x³ - 3x² = 0

factoring out an x^2, we get that

x² (x² - 2x - 3) = 0

factoring further,

x²(x - 3)(x + 1) = 0

So we know our x intercepts are going to be x = -1, x = 0, x = 3,

Now to find the total area bounded by the curve and the x axis, we just need to find the absolute values of the integrals of the function from -1 to 0, then from 0 to 3 and add them together. (Absolute values because we only care about the total area, not whether that area is above or below the x axis).

The antiderivative of our function y is x⁵/5 - (1/2)x⁴ - x³.

Thus the integral, from -1 to 0 is -0.3 (you can check for yourself). That means that the area bounded by the function and the curve from -1 to 0 is positive 0.3.

Given that antiderivative, we also know that the integral from 0 to 3 is -18.9, meaning that the area bounded by the function and the curve along that interval is 18.9.

To get the total area bounded by the curve and the x axis, we just add those two values together, and get 19.2.