The area bounded by $y = cos x, y = x + 1, y = 0$

\frac{3}{2}

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\frac{2}{3}

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\frac{1}{2}

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\frac{5}{2}

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Solution

The correct option is D $\frac{3}{2}$
Area bounded= $\int y_{1} d x + \int y_{2} d x$
$= \int (x + 1) d x + \int_{0}^{π / 2} cos x d x$
$= {(\frac{x^{2}}{2} + x)}_{- 1}^{0} {(- sin x)}_{0}^{π / 2}$
$= 0 - (\frac{1}{2} - 1) + (1 - 0)$
$= 1 - \frac{1}{2} + 1$
$= 2 - \frac{1}{2}$
$= \frac{3}{2}$

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\sqrt{2} - 1

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List-I	List-II
1. Area of the region bounded by $y = \| 5 sin x \|$ from $x = 0$ to $x = 4 π$ and x-axis	a. 3/2
2. The area bounded by $y =$ cosx in $[0, 2 π]$ and the $X$ -axis	b. $\sqrt{2} - 1$
3. The area bounded by $y = s i n x, y = c o s x$ and the y-axis	c. 4
4. The area bounded by $y = c o s x, y = x + 1, y = 0$	d. 40

\begin{matrix} L i s t I & L i s t I I P . & A r e a b o u n d e d b y y = x | x |, x - a x i s, & 1. & \frac{10}{3} x = 1 a n d x = - 1 Q . & A r e a b o u n d e d b y x - y + 2, & 2. & \frac{64}{3} x = \sqrt{y} a n d x = 0 R . & A r e a e n c l o s e d b y y^{2} = x & 3. & \frac{2}{3} a n d y = | x | S . & A r e a b o u n d e d b y y^{2} = x & 4. & \frac{1}{6} y = 4 a n d x = 0 \end{matrix}

\frac{π}{3}

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