(B) The area $A$ bounded by the curve $y = \sin x$,the $x$-axis,and the lines $x = 0$ and $x = \pi$ is given by the definite integral:$A = \int_{0}^{\

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(B) The area $A$ bounded by the curve $y = \sin x$,the $x$-axis,and the lines $x = 0$ and $x = \pi$ is given by the definite integral:$A = \int_{0}^{\pi} |\sin x| \, dx$Since $\sin x \geq 0$ for $x \in [0, \pi]$,we have:$A = \int_{0}^{\pi} \sin x \, dx$$A = [-\cos x]_{0}^{\pi}$$A = -(\cos \pi - \cos 0)$$A = -(-1 - 1)$$A = -(-2) = 2$Thus,the area is $2$ square units.

Class 12 Mathematics Application of Integrals Area bounded by region of single curve

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The area bounded by the $x$-axis and the curve $y = \sin x$ between $x = 0$ and $x = \pi$ is:

A
$1$
B
$2$
C
$3$
D
$4$

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Application of Integrals

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Area bounded by region of single curve

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The area bounded by the $x$-axis and the curve $y = \sin x$ between $x = 0$ and $x = \pi$ is:

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The area of the figure bounded by $y = e^x$,$y = e^{-x}$,and the straight line $x = 1$ is

The area of the region bounded by the curve $5y = 5 - x$,the $X$-axis,and the lines $x = 1$ and $x = 4$ is . . . . . . sq. units.

The area of the circle $(x - 2)^2 + (y - 3)^2 = 32$ that lies below the line $y = x + 1$ is:

The area of the region bounded by the curve $y=x|x|$,lines $x=-1$ and $x=1$ is . . . . . . .

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