Q78 / 180 · KCET 2020 · Mathematics
The area of the region bounded by the line $y=2 x+1, X$-axis and the ordinates $x=-1$ and $x=1$ is
- A$\frac{9}{4}$
- B2
- Correct answerC$\frac{5}{2}$
- D5
Solution
Given, equation of line $y=2 x+1...(i)$
Eq. (i) passing through the points $\left(\frac{-1}{2}, 0\right)$ and $(0,1)$.

$\therefore$ Required area of shaded region $\begin{aligned} &=\int_{-1}^{-1 / 2}-(2 x+1) d x+\text { Area of } \triangle A B C \\ &=-\left[2\left(\frac{x^{2}}{2}\right)+x\right]_{-1}^{-1 / 2}+\frac{1}{2} \times A B \times B C \\ &=-\left[x^{2}+x\right]_{-1}^{-1 / 2}+\frac{1}{2} \times \frac{3}{2} \times 3 \\ &=-\left[\left(-\frac{1}{2}\right)^{2}+\left(-\frac{1}{2}\right)-(-1)^{2}-(-1)\right]+\frac{9}{4} \\ &=-\left[\frac{1}{4}-\frac{1}{2}\right]+\frac{9}{4} \\ &=\frac{1}{4}+\frac{9}{4}=\frac{10}{4}=\frac{5}{2} \text { sq units } \end{aligned}$

$\therefore$ Required area of shaded region $\begin{aligned} &=\int_{-1}^{-1 / 2}-(2 x+1) d x+\text { Area of } \triangle A B C \\ &=-\left[2\left(\frac{x^{2}}{2}\right)+x\right]_{-1}^{-1 / 2}+\frac{1}{2} \times A B \times B C \\ &=-\left[x^{2}+x\right]_{-1}^{-1 / 2}+\frac{1}{2} \times \frac{3}{2} \times 3 \\ &=-\left[\left(-\frac{1}{2}\right)^{2}+\left(-\frac{1}{2}\right)-(-1)^{2}-(-1)\right]+\frac{9}{4} \\ &=-\left[\frac{1}{4}-\frac{1}{2}\right]+\frac{9}{4} \\ &=\frac{1}{4}+\frac{9}{4}=\frac{10}{4}=\frac{5}{2} \text { sq units } \end{aligned}$
Hence, the correct answer is option: 'C'.