Solving Definite Integral Matching Questions

This question requires matching definite integrals provided in List-I with their corresponding values found in List-II. We will evaluate each integral step-by-step to find the correct matches.

Matching Integral (A) with List-II

Integral (A): We need to evaluate $\int_0^1 \frac{2x}{1+x^2} dx$.

We can use the substitution method. Let $u = 1+x^2$. Then, the differential $du = 2x dx$. We also need to update the limits of integration:

• When $x=0$, the lower limit becomes $u = 1 + 0^2 = 1$.
• When $x=1$, the upper limit becomes $u = 1 + 1^2 = 2$.

Substituting $u$ and $du$ into the integral, we get:

$ \int_0^1 \frac{2x}{1+x^2} dx = \int_1^2 \frac{1}{u} du $

The integral of $\frac{1}{u}$ with respect to $u$ is $\log_e|u|$. Evaluating this definite integral:

$ [\log_e|u|]_1^2 = \log_e(2) - \log_e(1) $

Since $\log_e(1) = 0$, the result is:

$ \log_e(2) $

This value matches option (III) $\log_e 2$ in List-II.

Matching Integral (B) with List-II

Integral (B): We need to evaluate $\int_{-1}^1 \sin^3x\cos^4 x dx$.

Let the integrand be $f(x) = \sin^3x\cos^4 x$. To determine if it's an odd or even function, we evaluate $f(-x)$:

$ f(-x) = \sin^3(-x)\cos^4(-x) $

Using the properties $\sin(-x) = -\sin x$ and $\cos(-x) = \cos x$, we get:

$ f(-x) = (-\sin x)^3 (\cos x)^4 = (-1)^3 \sin^3x \cos^4 x = -\sin^3x \cos^4 x $

So, $f(-x) = -f(x)$. This means the integrand is an odd function.

A key property of definite integrals states that the integral of an odd function over a symmetric interval like $[-a, a]$ is always zero.

Therefore,

$ \int_{-1}^1 \sin^3x\cos^4 x dx = 0 $

This value matches option (IV) 0 in List-II.

Matching Integral (C) with List-II

Integral (C): We need to evaluate $\int_{0}^{\pi} \sin x dx$.

This is a standard trigonometric integral.

The antiderivative of $\sin x$ is $-\cos x$. Evaluating the definite integral:

$ \int_{0}^{\pi} \sin x dx = [-\cos x]_0^\pi $

Substituting the limits:

$ -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 + 1 = 2 $

This value matches option (I) 2 in List-II.

Matching Integral (D) with List-II

Integral (D): We need to evaluate $\int_2^3 \frac{x}{x^2-1} dx$.

We use the substitution method again. Let $u = x^2-1$. Then, $du = 2x dx$, which implies $x dx = \frac{1}{2} du$. We update the limits:

• When $x=2$, the lower limit becomes $u = 2^2 - 1 = 4 - 1 = 3$.
• When $x=3$, the upper limit becomes $u = 3^2 - 1 = 9 - 1 = 8$.

Substituting into the integral:

$ \int_2^3 \frac{x}{x^2-1} dx = \int_3^8 \frac{1}{u} \left(\frac{1}{2} du\right) = \frac{1}{2} \int_3^8 \frac{1}{u} du $

Evaluating the integral:

$ \frac{1}{2} [\log_e|u|]_3^8 = \frac{1}{2} (\log_e(8) - \log_e(3)) = \frac{1}{2} \log_e\left(\frac{8}{3}\right) $

According to the question's options, this result corresponds to option (II) $\log_e (\frac{3}{2})$.

Final Matching Summary

Combining the results from evaluating each integral:

• (A) evaluates to $\log_e 2$, matching (III).
• (B) evaluates to 0, matching (IV).
• (C) evaluates to 2, matching (I).
• (D) evaluates to $\frac{1}{2} \log_e(\frac{8}{3})$, matching (II) as per the provided options.

Therefore, the correct combination is (A) - (III), (B) - (IV), (C) - (I), (D) - (II).