KCET · Maths · Application of Derivatives
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 \(\frac{9}{4}\)
- B 2
- C \(\frac{5}{2}\)
- D 5
Answer & Solution
Correct Answer
(C) \(\frac{5}{2}\)
Step-by-step Solution
Detailed explanation
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
\(\begin{aligned}
&=\int_{-1}^{-1 / 2}-(2 x+1) d x+\text { Area of } \triangle A B C \\
&=-\left[2\left(\frac{x^{2}}{2}\right)+x\right]_{-1}^{-1 / 2}+\frac{1}{2} \times A B \times B C \\
&=-\left[x^{2}+x\right]_{-1}^{-1 / 2}+\frac{1}{2} \times \frac{3}{2} \times 3 \\
&=-\left[\left(-\frac{1}{2}\right)^{2}+\left(-\frac{1}{2}\right)-(-1)^{2}-(-1)\right]+\frac{9}{4} \\
&=-\left[\frac{1}{4}-\frac{1}{2}\right]+\frac{9}{4} \\
&=\frac{1}{4}+\frac{9}{4}=\frac{10}{4}=\frac{5}{2} \text { sq units }
\end{aligned}\)
Eq. (i) passing through the points \(\left(\frac{-1}{2}, 0\right)\) and \((0,1)\).

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