Area Ques 57
Question
- Find all maxima and minima of the function $y=x(x-1)^{2}, 0 \leq x \leq 2$.
Also, determine the area bounded by the curve $y=x(x-1)^{2}$, the $Y$-axis and the line $x=2$.
$(1989,5 \mathrm{M})$
Show Answer
Answer:
57.$(y _{\max } = \frac{4}{27} , y _{\min }=0 , \frac{10}{3})$ sq units)
Solution:
Formula:
- Given, $y=x(x-1)^{2}$
$\Rightarrow \quad \frac{d y}{d x}=x \cdot 2(x-1)+(x-1)^{2}$
$ \begin{aligned} & =(x-1) \cdot(2 x+ x-1) \\ & =(x-1)(3 x-1) \\ & \quad \quad \frac{+\bullet \quad - \quad \bullet +}{1 / 31 \quad \quad 1} \end{aligned} $
$\therefore \quad$ Maximum at $x=1 / 3$
$ y _{\max }=\frac{1}{3} \quad (\frac{2}{3})^{2}=\frac{4}{9} $
Minimum at $x=1$
$ y _{\min }=0 $
Now, to find the area bounded by the curve $y=x(x-1)^{2}$, the $Y$-axis and line $x=2$.
$\therefore$ Required area $=$ Area of square $O A B C-\int _{0}^{2} y d x$
$=2 \times 2-\int _{0}^{2} x(x-1)^{2} d x$
$=4-[[\frac{x(x-1)^{3}}{3}]^{2} _0-\frac{1}{3} \int _{0}^{2}(x-1)^{3} \cdot 1 d x]$
$=4-[\frac{x}{3}(x-1)^{3}-{\frac{(x-1)^{4}}{12}}]^{2} _0$
$=4-[\frac{2}{3}-\frac{1}{12}+\frac{1}{12}]=\frac{10}{3}$ sq units
BETA
