Find the area of the region bounded by $y = x^{3}, y = 0, x = 2$ and $x = 4$ .

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Solution

The correct option is B 60
If we draw the graph of $y = x^{3}$ and find out the region whose area is to be found, then it looks as shown below in the figure.

Area $A = \int_{1}^{4} y d x = \int_{2}^{4} (x^{3}) d x$
$= \int_{2}^{4} (x^{3}) d x = {(\frac{x^{4}}{4})}_{2}^{4} = (\frac{4^{4}}{4} - \frac{2^{4}}{4}) = 60$

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y = x^{3}, y = 0, x = 2

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List-I	List-II
1. Area of region bounded by $y = 2 x - x^{2}$ and $x -$ axis	a. $\frac{1}{3}$
2. Area of the region ${(x, y) : x^{2} \leq y \leq \| x \|}$	b. $\frac{1}{2}$
3. Area bounded by $y = x$ and $y = x^{3}$	c. $\frac{2}{3}$
4. Area bounded by $y = x \| x \|$ , $x$ -axis and $x = - 1, x = 1$	d. $\frac{4}{3}$

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Find the area of the region lying in the first quadrant and bounded by $y = 4 x^{2}, x = 0, y = 1$ and y = 4.

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