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

Task: Evaluate four definite integrals and match them to their corresponding values.
Methodology: Employ integration techniques including substitution, properties of odd/even functions, and partial fractions.
Integral Evaluations:
(A) $\int_{0}^{1} \frac{2x}{1+x^2} dx$:
Let $u = 1 + x^2$, so $du = 2x dx$.
Limits change: $x=0 \implies u=1$; $x=1 \implies u=2$.
Integral becomes $\int_{1}^{2} \frac{1}{u} du = [\ln|u|]_{1}^{2} = \ln(2) - \ln(1) = \ln(2)$.
Matches value (III).
(B) $\int_{-1}^{1} \sin^3x \cos^4x dx$:
Let $f(x) = \sin^3x \cos^4x$.
Test for odd/even: $f(-x) = \sin^3(-x) \cos^4(-x) = (-\sin x)^3 (\cos x)^4 = -\sin^3x \cos^4x = -f(x)$.
The function is odd. The integral of an odd function over $[-a, a]$ is 0.
Matches value (IV).
(C) $\int_{0}^{\pi} \sin x dx$:
Antiderivative of $\sin x$ is $-\cos x$.
$\int_{0}^{\pi} \sin x dx = [-\cos x]_{0}^{\pi} = (-\cos(\pi)) - (-\cos(0)) = (-(-1)) - (-1) = 1 + 1 = 2$.
Matches value (I).
(D) $\int_{2}^{3} \frac{2}{x^2 - 1} dx$:
Partial fraction decomposition: $\frac{2}{x^2 - 1} = \frac{2}{(x-1)(x+1)} = \frac{1}{x-1} - \frac{1}{x+1}$.
Integral: $\int_{2}^{3} (\frac{1}{x-1} - \frac{1}{x+1}) dx = [\ln|x-1| - \ln|x+1|]_{2}^{3} = [\ln|\frac{x-1}{x+1}|]_{2}^{3}$.
$= \ln|\frac{3-1}{3+1}| - \ln|\frac{2-1}{2+1}| = \ln|\frac{2}{4}| - \ln|\frac{1}{3}| = \ln(\frac{1}{2}) - \ln(\frac{1}{3}) = \ln(\frac{1/2}{1/3}) = \ln(\frac{3}{2})$.
Matches value (II).
Conclusion:
The correct matches are (A) - (III), (B) - (IV), (C) - (I), (D) - (II). This corresponds to option (4).