Area Question 46
Question 46
- 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 $x=2$.
$(1989,5 \mathrm{M})$
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Solution:
- Given, $y=x(x-1)^{2}$
$\Rightarrow \quad \frac{d y}{d x}=x \cdot 2(x-1)+(x-1)^{2}$
$$ \begin{aligned} & =(x-1) \cdot(2 x+x-1) \ & =(x-1)(3 x-1) \ & \frac{+\bullet-\quad+}{1 / 31} \end{aligned} $$
$\therefore \quad$ Maximum at $x=1 / 3$
$$ y_{\max }=\frac{1}{3} \quad-\frac{2}{3}^{2}=\frac{4}{27} $$
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 $O A B C-\int_{0}^{2} y d x$
$=2 \times 2-\int_{0}^{2} x(x-1)^{2} d x$
$=4-\frac{x(x-1)^{3}}{3}-\frac{1}{3} \int_{0}^{2}(x-1)^{3} \cdot 1 d x$
$=4-\frac{x}{3}(x-1)^{3}-{\frac{(x-1)^{4}}{12}}^{2}$
$=4-\frac{2}{3}-\frac{1}{12}+\frac{1}{12}=\frac{10}{3}$ sq units
