The area (in sq. units) of the region bounded by the $X$-axis and the curve $y=1-x-6 x^2$ is

1. $\frac{125}{216}$
2. $\frac{125}{512}$
3. $\frac{25}{216}$
4. $\frac{25}{512}$

Solution

We have, $ \begin{aligned} & y=1-x-6 x^2 \\ & \Rightarrow \quad y=-\left[6 x^2+x-1\right] \\ & \Rightarrow \quad y=-6\left[x^2+\frac{x}{6}-\frac{1}{6}\right] \\ & \Rightarrow \quad y=-6\left[\left(x+\frac{1}{12}\right)^2-\frac{1}{144}-\frac{1}{6}\right] \\ & \Rightarrow \quad y=-6\left(x+\frac{1}{12}\right)^2+\frac{1}{24}+1 \\ & \Rightarrow \quad y=-6\left(x+\frac{1}{12}\right)^2+\frac{25}{24} \\ & \Rightarrow 6\left(x+\frac{1}{12}\right)^2=-\left(y-\frac{25}{24}\right) \\ & \Rightarrow\left(x+\frac{1}{12}\right)^2=-\frac{1}{6}\left(y-\frac{25}{24}\right) \\ & \end{aligned} $
$\begin{aligned} & \therefore \text { Required area }=\int_{-1 / 2}^{1 / 3} y d x \\ & =\int_{-1 / 2}^{1 / 3}\left(1-x-6 x^2\right) d x=\left[x-\frac{x^2}{2}-2 x^3\right]_{-1 / 2}^{1 / 3} \\ & =\left(\frac{1}{3}-\frac{1}{18}-\frac{2}{27}\right)-\left(-\frac{1}{2}-\frac{1}{8}+\frac{1}{4}\right) \\ & =\left(\frac{18-3-4}{54}\right)-\left(\frac{-4-1+2}{8}\right) \\ & =\frac{11}{54}+\frac{3}{8}=\frac{44+81}{216}=\frac{125}{216} \text { sq unit }\end{aligned}$

Asked in: AP EAMCET 2018 (22 Apr Shift 1)

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