Question

The value of integral $\int_{0}^{1}\int_{x}^{1}\frac{1}{1+y^{2}}\cdot dydx$ is equal to

• A. $1-\ln(2)$
• B. $\frac{1}{2}\ln(2)$
• C. $\frac{\pi}{4}$
• D. $\frac{\pi}{2}-\ln 2$

Hint:

In double integrals, if the inner integral seems "unfriendly," check if changing the order produces a $y dy$ or $x dx$ term. This often allows for a simple logarithmic substitution.

Answer 1

The Correct Option is B

Evaluating the inner integral results in $\tan^{-1}(y)$, which is hard to integrate against $x$. We will change the order of integration.
Step 1: The limits for $y$ are $x \le y \le 1$ and for $x$ are $0 \le x \le 1$. This represents a triangle in the $xy$-plane bounded by the line $y=x$, the vertical line $x=0$ (y-axis), and the horizontal line $y=1$.
Step 2: In the new order, we look at horizontal strips. For a fixed $y$, $x$ starts at 0 and goes up to the line $y=x$, meaning $x$ ends at $y$. The total range for $y$ is from 0 to 1. The integral becomes: $\int_{0}^{1} \int_{0}^{y} \frac{1}{1+y^2} dx dy$.
Step 3: Integrate with respect to $x$ while treating $y$ as a constant: $\int_{0}^{y} \frac{1}{1+y^2} dx = \left[ \frac{x}{1+y^2} \right]_{0}^{y} = \frac{y}{1+y^2} - 0 = \frac{y}{1+y^2}$.
Step 4: Now solve the resulting single integral: $\int_{0}^{1} \frac{y}{1+y^2} dy$. Let $u = 1 + y^2$, then $du = 2y dy \implies y dy = \frac{1}{2} du$. Limits: $y=0 \to u=1$, $y=1 \to u=2$. $\frac{1}{2} \int_{1}^{2} \frac{1}{u} du = \frac{1}{2} [\ln u]_1^2 = \frac{1}{2}(\ln 2 - \ln 1) = \frac{1}{2} \ln 2$. The result is $\frac{1}{2} \ln 2$, matching Option (2).