Question

Let $I = \int_{1}^{3}|(x-1)(x-2)(x-3)|dx$. The value of $I^{-1}$, is equal to

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

2

Answer 2

To evaluate $I = \int_{1}^{3}|(x-1)(x-2)(x-3)|dx$, we can make a substitution. Let $u = x-2$. Then $du = dx$. When $x=1$, $u=1-2=-1$. When $x=3$, $u=3-2=1$. The expression $(x-1)(x-2)(x-3)$ becomes $(u+2-1)(u)(u+2-3) = (u+1)(u)(u-1) = u(u^2-1) = u^3-u$. So, $I = \int_{-1}^{1}|u^3-u|du$.

The function $g(u) = u^3-u$ has roots at $u=-1, 0, 1$. For $u \in (-1, 0)$, $u^3-u > 0$, so $|u^3-u| = u^3-u$. For $u \in (0, 1)$, $u^3-u < 0$, so $|u^3-u| = -(u^3-u) = u-u^3$.

The integral can be written as: $I = \int_{-1}^{0}(u^3-u)du + \int_{0}^{1}(u-u^3)du$.

Alternatively, notice that $u^3-u$ is an odd function, so $|u^3-u|$ is an even function. For an even function $f(u)$, $\int_{-a}^{a}f(u)du = 2\int_{0}^{a}f(u)du$. Thus, $I = 2\int_{0}^{1}|u^3-u|du$. In the interval $[0, 1]$, $|u^3-u| = u-u^3$. $I = 2\int_{0}^{1}(u-u^3)du = 2 \left[\frac{u^2}{2} - \frac{u^4}{4}\right]_{0}^{1} = 2 \left(\left(\frac{1}{2} - \frac{1}{4}\right) - (0-0)\right) = 2 \left(\frac{1}{4}\right) = \frac{1}{2}$.

The question asks for $I^{-1}$. $I^{-1} = \left(\frac{1}{2}\right)^{-1} = 2$.

The value of $I^{-1}$ is 2.