SATHEE: Area Question 5

Area Question 5

Question 5

The area (in $sq$ units) of the region $Missing or unrecognized delimiter for \left$ is (2019 Main, 9 April I) (a) $\frac{13}{6}$ (b) $\frac{9}{2}$ (c) $\frac{31}{6}$ (d) $\frac{10}{3}$

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Solution:

Given region is $Missing or unrecognized delimiter for \left$

Now, the region is shown in the following graph

For intersecting points $A$ and $B$

$\begin{array}{ll} Taking, & x^{2} = x + 2 \Rightarrow x^{2} - x - 2 = 0 \Rightarrow & x^{2} - 2 x + x - 2 = 0 \Rightarrow & x (x - 2) + 1 (x - 2) = 0 \Rightarrow & x = - 1, 2 \Rightarrow y = 1, 4 \end{array}$

So, $A (- 1, 1)$ and $B (2, 4)$ .

Now, shaded area $= \int_{- 1}^{2} [(x + 2) - x^{2}] d x$

$= \frac{x^{2}}{2} + 2 x - {\frac{x^{3}}{3}}^{2} = \frac{4}{2} + 4 - \frac{8}{3} - \frac{1}{2} - 2 + \frac{1}{3}$

$= 8 - \frac{1}{2} - \frac{9}{3} = 8 - \frac{1}{2} - 3 = 5 - \frac{1}{2} = \frac{9}{2}$ sq units

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