JEE Challenger
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MathMedAdv 2007 • Paper 1

Match Definite Integrals with Their Evaluated Values

Match the integrals in Column I with the values in Column II and indicate your answer by darkening the appropriate bubbles in the 4×44 \times 4 matrix given in the ORS.

Column IColumn II(A) ∫−11dx1+x2(p) 12log⁡(23)(B) ∫01dx1−x2(q) 2log⁡(23)(C) ∫23dx1−x2(r) π3(D) ∫12dxxx2−1(s) π2\begin{array}{ll} \textbf{Column I} & \textbf{Column II} \\[8pt] \text{(A) } \int_{-1}^1 \frac{dx}{1+x^2} & \text{(p) } \frac{1}{2}\log\left(\frac{2}{3}\right) \\[8pt] \text{(B) } \int_0^1 \frac{dx}{\sqrt{1-x^2}} & \text{(q) } 2\log\left(\frac{2}{3}\right) \\[8pt] \text{(C) } \int_2^3 \frac{dx}{1-x^2} & \text{(r) } \frac{\pi}{3} \\[8pt] \text{(D) } \int_1^2 \frac{dx}{x\sqrt{x^2-1}} & \text{(s) } \frac{\pi}{2} \end{array}

Options

A

(A)→(S),  (B)→(P),  (C)→(S),  (D)→(R)(\text{A}) \rightarrow (\text{S}),\; (\text{B}) \rightarrow (\text{P}),\; (\text{C}) \rightarrow (\text{S}),\; (\text{D}) \rightarrow (\text{R})

B

(A)→(S),  (B)→(S),  (C)→(P),  (D)→(R)(\text{A}) \rightarrow (\text{S}),\; (\text{B}) \rightarrow (\text{S}),\; (\text{C}) \rightarrow (\text{P}),\; (\text{D}) \rightarrow (\text{R})

Correct
C

(A)→(P),  (B)→(S),  (C)→(P),  (D)→(R)(\text{A}) \rightarrow (\text{P}),\; (\text{B}) \rightarrow (\text{S}),\; (\text{C}) \rightarrow (\text{P}),\; (\text{D}) \rightarrow (\text{R})

D

(A)→(S),  (B)→(S),  (C)→(R),  (D)→(P)(\text{A}) \rightarrow (\text{S}),\; (\text{B}) \rightarrow (\text{S}),\; (\text{C}) \rightarrow (\text{R}),\; (\text{D}) \rightarrow (\text{P})

Topics & Concepts

IntegralsDefinite IntegrationProperties of Definite Integrals

Step-by-Step Solution

To match the integrals in Column I with their evaluated values in Column II, we evaluate each integral step-by-step:

(A) Evaluation of ∫−11dx1+x2\int_{-1}^1 \frac{dx}{1+x^2}:
The standard antiderivative of 11+x2\frac{1}{1+x^2} is tan⁡−1x\tan^{-1}x.

∫−11dx1+x2=[tan⁡−1x]−11=tan⁡−1(1)−tan⁡−1(−1)=π4−(−π4)=π2\int_{-1}^1 \frac{dx}{1+x^2} = \left[ \tan^{-1}x \right]_{-1}^1 = \tan^{-1}(1) - \tan^{-1}(-1) = \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{2}

Thus, (A) matches with (s).

(B) Evaluation of ∫01dx1−x2\int_0^1 \frac{dx}{\sqrt{1-x^2}}:
The standard antiderivative of 11−x2\frac{1}{\sqrt{1-x^2}} is sin⁡−1x\sin^{-1}x.

∫01dx1−x2=[sin⁡−1x]01=sin⁡−1(1)−sin⁡−1(0)=π2−0=π2\int_0^1 \frac{dx}{\sqrt{1-x^2}} = \left[ \sin^{-1}x \right]_0^1 = \sin^{-1}(1) - \sin^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}

Thus, (B) matches with (s).

(C) Evaluation of ∫23dx1−x2\int_2^3 \frac{dx}{1-x^2}:
Using the standard formula ∫dxa2−x2=12alog⁡∣a+xa−x∣+C\int \frac{dx}{a^2-x^2} = \frac{1}{2a}\log\left|\frac{a+x}{a-x}\right| + C:

∫23dx1−x2=[12log⁡∣1+x1−x∣]23\int_2^3 \frac{dx}{1-x^2} = \left[ \frac{1}{2}\log\left|\frac{1+x}{1-x}\right| \right]_2^3

Substituting the upper and lower limits:

=12log⁡∣1+31−3∣−12log⁡∣1+21−2∣=12log⁡(2)−12log⁡(3)=12log⁡(23)= \frac{1}{2}\log\left|\frac{1+3}{1-3}\right| - \frac{1}{2}\log\left|\frac{1+2}{1-2}\right| = \frac{1}{2}\log(2) - \frac{1}{2}\log(3) = \frac{1}{2}\log\left(\frac{2}{3}\right)

Thus, (C) matches with (p).

(D) Evaluation of ∫12dxxx2−1\int_1^2 \frac{dx}{x\sqrt{x^2-1}}:
The standard antiderivative of 1xx2−1\frac{1}{x\sqrt{x^2-1}} for x>1x > 1 is sec⁡−1x\sec^{-1}x.

∫12dxxx2−1=[sec⁡−1x]12=sec⁡−1(2)−sec⁡−1(1)=π3−0=π3\int_1^2 \frac{dx}{x\sqrt{x^2-1}} = \left[ \sec^{-1}x \right]_1^2 = \sec^{-1}(2) - \sec^{-1}(1) = \frac{\pi}{3} - 0 = \frac{\pi}{3}

Thus, (D) matches with (r).

Final Matching:

  • (A) →\to (s)
  • (B) →\to (s)
  • (C) →\to (p)
  • (D) →\to (r)
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