Sample 2023-24 Class 12 Maths Q24 Solution: Evaluate ∫ from −1 to 1 of log((2 − x)/(2 + x)) dx

The question

Evaluate : $\displaystyle\int_{-1}^{1}\log\left(\frac{2 - x}{2 + x}\right)dx$ ⁠.

Exam answer

Let $f(x) = \log\left(\dfrac{2 - x}{2 + x}\right)$ ⁠, $x \in [-1, 1]$ ⁠.

Why?

The limits $-1$ and $1$ are of the form $-a$ and $a$ ⁠. With such symmetric limits, first test whether the integrand is odd or even. Integrating this log directly would be long, and the property below makes it unnecessary:

\begin{aligned} &\int_{-a}^{a} f(x)\,dx = 0 \\ &\text{if } f(-x) = -f(x) \end{aligned}

For $x$ in $[-1, 1]$ ⁠, both $2 - x$ and $2 + x$ are positive, so $f$ is defined there.

\begin{aligned} f(-x) &= \log\left(\frac{2 + x}{2 - x}\right) \\ &= -\log\left(\frac{2 - x}{2 + x}\right) \\ &= -f(x) \end{aligned}

Why?

Replace every $x$ by $-x$ ⁠: $2 - (-x) = 2 + x$ and $2 + (-x) = 2 - x$ ⁠. So the fraction turns upside down.

Turning the fraction inside a log upside down changes the sign of the log:

\log\frac{b}{a} = -\log\frac{a}{b}

This holds because $\log\frac{b}{a} = \log b - \log a$ ⁠.

\begin{aligned} &\log\left(\frac{2 + x}{2 - x}\right) \\ &= \log(2 + x) - \log(2 - x) \\ &= -\big(\log(2 - x) - \log(2 + x)\big) \\ &= -\log\left(\frac{2 - x}{2 + x}\right) \end{aligned}

So, $f(x)$ is an odd function, and $\int_{-a}^{a} f(x)\,dx = 0$ for an odd function.

Why?

The graph of an odd function on the left of 0 is the graph on the right turned upside down. The area below the axis on one side cancels the area above it on the other side.

Watch out: show the $f(-x)$ working. Just writing "odd, so 0" without showing why may not get full credit.

$\therefore\ \displaystyle\int_{-1}^{1}\log\left(\dfrac{2 - x}{2 + x}\right)dx = 0$ ⁠.

Where this comes from

Chapter 7, Integrals: Definite integrals: odd function property.

Question type: Limits from $-a$ to $a$

How to spot it: click to see

Type 11. Limits from $-a$ to $a$ ⁠: check odd or even

Spot it: the two limits are the same number with opposite signs.

Here: The limits are $-1$ and $1$ ⁠. Putting $-x$ turns $\frac{2 - x}{2 + x}$ upside down, which changes the sign of the log, so the integrand is odd and the integral is 0.

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Q24: Evaluate $\int_{-1}^{1}\log\left(\frac{2 - x}{2 + x}\right)dx$