The area bounded by $y = x^{2} + 2, X$ -axis, $x = 1$ and $x = 2$ is:

\frac{16}{3}

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\frac{17}{3}

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\frac{13}{3}

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\frac{20}{3}

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Solution

The correct option is A $\frac{13}{3}$
$y = x^{2} + 2$
area of the shaded region.
= $2 \int 1 y d x$
$= 2 \int 1 (x^{2} + 2) d x = \frac{x^{3}}{3} |_{1}^{2} + 2 x |_{1}^{2}$

$= \frac{7}{3} + 2 (1)$

$= \frac{13}{3} s q u n i t s$ .

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Similar questions

y = x^{2} + 2

x = 2

x = 3

The area bounded by the curve, x-axis and the ordinates x = –1 and x = 1 is given by

[Hint: y = x² if x > 0 and y = –x² if x < 0]

A. 0

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. $\frac{4}{3}$