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Let $y$ be the function which passes through $(1,2)$ having slope $(2 x+1)$. The area bounded between the curve and $x$-axis is
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$1 / 6$ sq. unit
$\begin{aligned} \frac{d y}{d x}=2 x+1 \Rightarrow & y=x^2+x+c \\ \Rightarrow y=x^2+x, & {[\boxtimes \quad c=0 \text { by putting } x=1, y=2) } \\ & \quad\left(x+\frac{1}{2}\right)^2=y+\frac{1}{4}, \text { which is a equation of parabola, whose vertices is, } V\left(\frac{-1}{2}, \frac{-1}{4}\right)\end{aligned}$
$\begin{aligned}& =\left|\int_{-1}^0\left(x^2+x\right) d x\right|=\left(\frac{x^3}{3}+\frac{x^2}{2}\right)_{-1}^0 \\
& =\left|\frac{-1}{3}+\frac{1}{2}\right|=\frac{1}{6}\end{aligned}$ sq. unit.
$\begin{aligned}& =\left|\int_{-1}^0\left(x^2+x\right) d x\right|=\left(\frac{x^3}{3}+\frac{x^2}{2}\right)_{-1}^0 \\
& =\left|\frac{-1}{3}+\frac{1}{2}\right|=\frac{1}{6}\end{aligned}$ sq. unit.
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