Let A be the area bounded by the curve y=x|x-3|, the x-axis, and the ordinates x=-1 and x=…

Question

Let $A$ be the area bounded by the curve $y=x|x-3|$, the $x$-axis, and the ordinates $x=-1$ and $x=2$. Then $12A$ is equal to \_\_\_\_\_\_\_.

Answer: 62

Accepted Answer

For $x<3$: $y=x(3-x)=3x-x^2$, negative on $[-1,0)$ and non-negative on $[0,3]$.

$$A=\int_{-1}^0(x^2-3x)\,dx+\int_0^2(3x-x^2)\,dx=\left[\frac{x^3}3-\frac{3x^2}2\right]_{-1}^0+\left[\frac{3x^2}2-\frac{x^3}3\right]_0^2$$
$$=\frac{11}6+\frac{10}3=\frac{31}6.$$

$$12A=12\cdot\frac{31}6=62.$$

Solutions · 1