Match List-I with List-II (x > 0; the constant of integration is left out in List-II)
| List-I (Integral) | List-II (Value) | ||
| (A) | \(\displaystyle\int x\,e^x\,dx\) | (I) | \(x\,e^x\) |
| (B) | \(\displaystyle\int \log x\,dx\) | (II) | \(e^x(x - 1)\) |
| (C) | \(\displaystyle\int x\log x\,dx\) | (III) | \(\displaystyle\frac{x^2}{2}\log x - \frac{x^2}{4}\) |
| (D) | \(\displaystyle\int e^x(x + 1)\,dx\) | (IV) | \(x\log x - x\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(II), (B)-(IV), (C)-(III), (D)-(I)
Each integral is found by integration by parts, \(\int u\,dv = uv - \int v\,du\).
(A) \(\int x\,e^x\,dx\)
This matches (II) \(e^x(x - 1)\). With u = x and v = ex:
\( \displaystyle \int x\,e^x\,dx = x e^x - \int e^x\,dx = x e^x - e^x \)
(B) \(\int \log x\,dx\)
This matches (IV) \(x\log x - x\). With u = log x and v = x:
\( \displaystyle \int \log x\,dx = x\log x - \int x \times \frac{1}{x}\,dx = x\log x - x \)
(C) \(\int x\log x\,dx\)
This matches (III) \(\frac{x^2}{2}\log x - \frac{x^2}{4}\). With u = log x and \(v = \frac{x^2}{2}\):
\( \displaystyle \int x\log x\,dx = \frac{x^2}{2}\log x - \int \frac{x^2}{2} \times \frac{1}{x}\,dx \)
\( \displaystyle = \frac{x^2}{2}\log x - \frac{x^2}{4} \)
(D) \(\int e^x(x + 1)\,dx\)
This matches (I) \(x\,e^x\). It has the form \(e^x[f(x) + f'(x)]\) with f(x) = x, so the integral is \(e^x f(x) = x e^x\).