How to find the area bounded by the curve and x-axis [closed]

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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 $y=-x^2+x+4$

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2 Answers

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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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I'll show you how to do the first one. Above is a graph of $y=x^3+1.5x^2$.. As you can see, the region bounded by the curve and x-axis is between $x = -1.5$ and $x = 0$. Therefore you integrate between $-1.5$ and $0$ to get

\begin{equation} \int_{-1.5}^{0} x^3+1.5x^2 dx = [0.25x^4 + 0.5x^3]_{x=-1.5}^0 =2.9531 \end{equation}

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