Question

Match List-I with List-II

List-I (Definite integral)	List-II (Value)
(A) \( \int_{0}^{1} \frac{2x}{1+x^2}\, dx \)	(I) 2
(B) \( \int_{-1}^{1} \sin^3x \cos^4x\, dx \)	(II) \(\log_e\!\left(\tfrac{3}{2}\right)\)
(C) \( \int_{0}^{\pi} \sin x\, dx \)	(III) \(\log_e 2\)
(D) \( \int_{2}^{3} \frac{2}{x^2 - 1}\, dx \)	(IV) 0

Choose the correct answer from the options given below:

• (A) - (I), (B) - (II), (C) - (III), (D) - (IV)
• (A) - (III), (B) - (II), (C) - (I), (D) - (IV)
• (A) - (III), (B) - (I), (C) - (IV), (D) - (II)
• (A) - (III), (B) - (IV), (C) - (I), (D) - (II)

Hint:

Before starting a definite integral, always check for simplifying properties. Is the interval symmetric (like [-a, a])? If so, check if the function is odd (integral is 0) or even. Is it of the form that allows using King's property? These checks can save a lot of calculation time.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We need to evaluate four different definite integrals and match them with their correct values from List-II. This involves using various integration techniques such as substitution, properties of odd/even functions, and partial fractions.
Step 3: Detailed Explanation:


(A) $\int_{0^{1} \frac{2x}{1+x^2} dx$:} Use substitution. Let $u = 1 + x^2$. Then $du = 2x dx$.
When $x=0$, $u=1$. When $x=1$, $u=2$.
The integral becomes $\int_{1}^{2} \frac{1}{u} du = [\ln|u|]_{1}^{2} = \ln(2) - \ln(1) = \ln(2)$. This matches (III).

(B) $\int_{-1^{1} \sin^3x \cos^4x dx$:} Let $f(x) = \sin^3x \cos^4x$. Check if it's an odd or even function.
$f(-x) = \sin^3(-x) \cos^4(-x) = (-\sin x)^3 (\cos x)^4 = -\sin^3x \cos^4x = -f(x)$.
Since $f(-x) = -f(x)$, the function is odd. The integral of an odd function over a symmetric interval $[-a, a]$ is 0. This matches (IV).

(C) $\int_{0^{\pi} \sin x dx$:} The antiderivative of $\sin x$ is $-\cos x$.
$\int_{0}^{\pi} \sin x dx = [-\cos x]_{0}^{\pi} = (-\cos(\pi)) - (-\cos(0)) = (-(-1)) - (-1) = 1 + 1 = 2$. This matches (I).

(D) $\int_{2^{3} \frac{2}{x^2 - 1} dx$:} Use partial fraction decomposition: $\frac{2}{x^2 - 1} = \frac{2}{(x-1)(x+1)} = \frac{1}{x-1} - \frac{1}{x+1}$.
The integral is $\int_{2}^{3} (\frac{1}{x-1} - \frac{1}{x+1}) dx = [\ln|x-1| - \ln|x+1|]_{2}^{3} = [\ln|\frac{x-1}{x+1}|]_{2}^{3}$.
$= \ln|\frac{3-1}{3+1}| - \ln|\frac{2-1}{2+1}| = \ln|\frac{2}{4}| - \ln|\frac{1}{3}| = \ln(\frac{1}{2}) - \ln(\frac{1}{3}) = \ln(\frac{1/2}{1/3}) = \ln(\frac{3}{2})$. This matches (II).

Step 4: Final Answer:
The correct matching is (A) - (III), (B) - (IV), (C) - (I), (D) - (II). This corresponds to option (4).