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 $12 A$ is equal to _____

Answer 1

Correct Answer: 62

To find the area $A$ bounded by the curve $y=x|x-3|$, the $x$-axis, and the ordinates $x=-1$ and $x=2$, we need to analyze the behavior of the function $y=x|x-3|$. For $x\lt 3$, $|x-3|=3-x$, so the function becomes $y=x(3-x)=3x-x^2$.

Consider the interval $x\in[-1,2]$ where $y=3x-x^2$ is valid:

1. Compute the definite integral of $3x-x^2$ from $-1$ to $2$: \(\int_{-1}^{2}(3x-x^2)\,dx\).

2. Antiderivative calculation: \(\int(3x-x^2)dx=\frac{3}{2}x^2-\frac{1}{3}x^3+C\).

3. Evaluate the definite integral: \(\left[\frac{3}{2}x^2-\frac{1}{3}x^3\right]_{-1}^{2}\).

4. Calculate at $x=2$: \(\frac{3}{2}(2)^2-\frac{1}{3}(2)^3=6-\frac{8}{3}=\frac{18}{3}-\frac{8}{3}=\frac{10}{3}\).

5. Calculate at $x=-1$: \(\frac{3}{2}(-1)^2-\frac{1}{3}(-1)^3=\frac{3}{2}+\frac{1}{3}=\frac{9}{6}+\frac{2}{6}=\frac{11}{6}\).

6. Area $A$ computation: \(\frac{10}{3}-\frac{11}{6}=\frac{20}{6}-\frac{11}{6}=\frac{9}{6}=\frac{3}{2}\).

7. Compute \(12A\): \(12\times\frac{3}{2}=18\).