Q172 / 270 · NDA 2016 (Phase 1) · Mathematics
Consider the function $\mathrm{f}(\mathrm{x})=|\mathrm{x}-1|+\mathrm{x}^2$, where $\mathrm{x} \in \mathrm{R}$.
What is the area of the region bounded by x -axis, the curve $y=f(x)$ and the two ordinates $x=1$ and $x=\frac{3}{2} ?$
- A$\frac{5}{12}$ square unit
- B$\frac{7}{12}$ square unit
- C$\frac{2}{3}$ square unit
- Correct answerD$\frac{11}{12}$ square unit
Solution
Area required for given region is
$\begin{aligned}
A_2 & =\int_1^{3 / 2} f(x) d x \\
& =\int_1^{3 / 2}\left[x^2+x-1\right] d x \\
& =\left[\frac{1}{3} x^3+\frac{1}{2} x^2-x\right]_1^{3 / 2} \\
A_2 & =\left(\frac{1}{3}\left(\frac{27}{8}\right)+\frac{1}{2}\left(\frac{9}{4}\right)-\frac{3}{2}\right)-\left(\frac{1}{3}(1)+\frac{1}{2}(1)-1\right) \\
A_2 & =\frac{11}{12} \text { square unit }
\end{aligned}$
Hence, the correct answer is option: 'D'.