Q231 / 270 · NDA 2018 (Phase 1) · Mathematics
What is $\int_1^{\mathrm{e}} \mathrm{x} \ln \mathrm{x} \mathrm{dx}$ equal to?
- A$\frac{\mathrm{e}+1}{4}$
- Correct answerB$\frac{\mathrm{e}^2+1}{4}$
- C$\frac{\mathrm{e}-1}{4}$
- D$\frac{\mathrm{e}^2-1}{4}$
Solution
$\begin{aligned} & \text { } \int_1^{\mathrm{e}} \mathrm{x} \cdot \ln \mathrm{x} \cdot \mathrm{dx}=\left[\ln \mathrm{x} \cdot \frac{\mathrm{x}^2}{2}\right]_1^{\mathrm{e}}-\int_1^{\mathrm{e}} \frac{1}{\mathrm{x}} \cdot \mathrm{x}^2 \cdot \mathrm{dx} \\ & =\left(\frac{\mathrm{e}^2}{2}-0\right)-\frac{1}{2}\left(\frac{1}{2}\right)\left(\mathrm{x}^2\right)_1^{\mathrm{e}}=\frac{\mathrm{e}^2}{2}-\frac{1}{4}\left(\mathrm{e}^2-1\right)=\frac{2 \mathrm{e}^2-\mathrm{e}^2+1}{4} \\ & =\frac{\mathrm{e}^2+1}{4}\end{aligned}$
Hence, the correct answer is option: 'B'.