# How to find the area bounded by the curve and x-axis

6,337

I'll tell you how to do the second one.

$$y = -x^3 - 3x^2$$

First of all let's find the extrema of the integration by solving those two equations:

1) Set $x = 0$ we find

$$y = 0$$

2) Set $y = 0$ we find

$$-x^3 - 3x^2 = 0 ~~~~~ \to ~~~~~ -x^2(x + 3) = 0 ~~~~~ \to ~~~~~ x = -3$$

Thence your extrema are $0$ and $-3$.

Finally:

$$A = \int_{-3}^0 -x^3 - 3x^2\ \text{d}x = -\frac{1}{4}x^4 - x^3 \bigg|_{-3}^0 = 0 - \left(-\frac{(-3)^4}{4} - (-3)^3\right) = 0 + \frac{81}{4} - 27 = -6.75$$

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Updated on August 16, 2022

The text says: Determine the area bounded by the curve and $x$-axis. Could somebody solve or at least explain how I should solve these problems.

• $y=x^3+1.5x^2$
• $y=-x^3-3x^2$
• $y=x^2+x+2$ and $y=-x^2+x+4$
• gambler101 about 7 years
Use definite integration.
• amWhy about 7 years
The x-axis is given by $y=0$