SATHEE: Area Question 46

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 $x = 2$ .

$(1989, 5 M)$

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Solution:

Given, $y = x (x - 1)^{2}$

$\Rightarrow \frac{d y}{d x} = x \cdot 2 (x - 1) + (x - 1)^{2}$

$\begin{aligned} = (x - 1) \cdot (2 x + x - 1) & = (x - 1) (3 x - 1) & \frac{+ ∙ - +}{1 / 31} \end{aligned}$

$∴$ Maximum at $x = 1 / 3$

$y_{max} = \frac{1}{3} - {\frac{2}{3}}^{2} = \frac{4}{27}$

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 $x = 2$ .

$∴$ 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

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