Question

Find the area bounded by the curve y=4x(x-1)(x-2) and the x axis.

Answer 1

The equation of the curve is $y=4x(x-1)(x-2)$

$\{\left(0\right),\left(0\right)\},\{1,\left(0\right)\}$ and $\{2,\left(0\right)\}$ from $\left(1\right)$ we see that $y$ is positive in $\left(0\right)<x<1$ and negative in $1<x<2$.

A rough sketch of the graph of functions $\left(1\right)$ is shown in figure.

The shaded region is the required area bounded by the curve $\left(1\right)$ and the $x$ axis and this area is the sum of area $\left(\right)CA$ and area $ADB$.

Now Area $OCA={\int }_0^{1}ydx$

$={\int }_0^1{x}^3-{3x}^2+2xydx$

$=4{[\frac{x^4}{4}-x^3+x^2]}_0^1=4[(\frac{1}{4}-1+1)-\left(0\right)]1$ sq unit

and area $ADB={\int }_1^{2}ydx=4{[\frac{x^4}{4}-x^3+x^2]}_1^2$

$=4\{(\frac{16}{4}-8+4)-(\frac{1}{4}-1+1)\}$

$=4(0-\frac{1}{4})=-1=1$ sq unit

Hence required area $=$ area $OCD+$ area $ADB$

$=1$ sq unit $+$ $1$ sq.unit $=2$ sq units.