Q113 / 160 · WBJEE 2012 · Mathematics
The value of the integral $\int_{1}^{5}[|x-3|+|1-x|] d x$ is equal to
- A4
- B8
- Correct answerC12
- D16
Solution
$\int_{1}^{5}[|x-3|+|1-x|] d x$
$=\int_{1}^{5}|x-3| d x+\int_{1}^{5}|1-x| d x$
$=\int_{1}^{3}|x-3| d x+\int_{3}^{5}|x-3| d x+\int_{1}^{5}|1-x| d x$
$=\int_{1}^{3}-(x-3) d x+\int_{3}^{5}(x-3) d x$
$+\int_{1}^{5}-(1-x) d x$
$=\int_{1}^{3}(3-x) d x+\int_{3}^{5}(x-3) d x+\int_{1}^{5}(x-1) d x$
$=\left[3 x-\frac{x^{2}}{2}\right]^{3}+\left[\frac{x^{2}}{2}-3 x\right]_{3}^{5}+\left[\frac{x^{2}}{2}-x\right]_{0}^{5}$
$=\left(3 \times 3-\frac{9}{2}\right)-\left(3 \times 1-\frac{1}{2}\right)+\left(\frac{5 \times 5}{2}-3 \times 5\right)$
$\begin{aligned}-\left(\frac{3 \times 3}{2}-3 \times 3\right)+\left(\frac{5 \times 5}{2}-5\right)-\left(\frac{1}{2}-1\right) \\=\left(9-\frac{9}{2}\right)-\left(3-\frac{1}{2}\right)+\left(\frac{25}{2}-15\right)-\left(\frac{9}{2}-9\right) \\ &+\left(\frac{25}{2}-5\right)-\left(-\frac{1}{2}\right) \\=& \frac{9}{2}-\frac{5}{2}-\frac{5}{2}+\frac{9}{2}+\frac{15}{2}+\frac{1}{2} \\=& \frac{9-5-5+9+15+1}{2}=\frac{24}{2}=12 \end{aligned}$
Hence, the correct answer is option: 'C'.