The curve $y=a x^2+b x$ passes through the point $(1,2)$ and lies above the $X$-axis for $0 \leq x \leq 8$. If the area enclosed by this curve, the $X$-axis and the line $x=6$ is 108 square units, then $2 b-a=$

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Aicharya · Accepted Answer

Given, curve $y=a x^2+b x$ passes through point $(1,2)$. $ \therefore \quad 2=a(1)^2+b(1) $

$\begin{aligned} & \text { Given, area under curve }=\int_0^6\left(a x^2+b x\right) d x=108 \\ & \Rightarrow \quad\left[\frac{a x^3}{3}+\frac{b x^2}{2}\right]_0^6=108 \\ & \Rightarrow \quad 72 a+18 b=108\end{aligned}$

By solving Eqs. (i) and (ii), we get $ \begin{aligned} & a=\frac{4}{3} \text { and } b=\frac{2}{3} \\ \therefore \quad & 2 b-a=2 \times \frac{2}{3}-\frac{4}{3}=\frac{4}{3}-\frac{4}{3}=0 \end{aligned} $

The curve $y=a x^2+b x$ passes through the point $(1,2)$ and lies above the $X$-axis for $0 \leq x \leq 8$.…

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