The slope of the tangent to a curve $y=\mathrm{f}(x)$ at $(x, \mathrm{f}(x))$ is $2 x+1$. If the curve…
The slope of the tangent to a curve $y=\mathrm{f}(x)$ at $(x, \mathrm{f}(x))$ is $2 x+1$. If the curve passes through the point $(1,2)$, then the area (in sq. units), bounded by the curve, the $\mathrm{X}$-axis and the line $x=1$, is
$\frac{3}{2}$
$\frac{4}{3}$
$\frac{5}{6}$
$\frac{1}{12}$
Solution
According to the given condition, $\frac{\mathrm{d} y}{\mathrm{~d} x}=2 x+1$ Integrating w.r.t. $x$, we get $y=x^2+x+\mathrm{c}$ As curve passes through $(1,2)$, we get $2=(1)^2+1+c \Rightarrow c=0$
$\therefore \quad$ The equation of the curve is $y=x^2+x$.
$\therefore \quad$ Required area
$\begin{aligned} & =\int_0^1\left(x^2+x\right) \mathrm{d} x \\ & =\left[\frac{x^3}{3}+\frac{x^2}{2}\right]_0^1 \\ & =\frac{1}{3}+\frac{1}{2} \\ & =\frac{5}{6} \text { sq. units }\end{aligned}$
$\therefore \quad$ Required area
$\begin{aligned} & =\int_0^1\left(x^2+x\right) \mathrm{d} x \\ & =\left[\frac{x^3}{3}+\frac{x^2}{2}\right]_0^1 \\ & =\frac{1}{3}+\frac{1}{2} \\ & =\frac{5}{6} \text { sq. units }\end{aligned}$