Find the area bounded by the curve y=sin x be | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The graph of $y=\sin x$ is shown in the figure. The area bounded by the curve between $x=0$ and $x=2 \pi$ is given by the sum of the areas of the

Vedclass · Accepted Answer

$4 	ext{ sq. units}$

Vedclass · Answer

(B) The graph of $y=\sin x$ is shown in the figure. The area bounded by the curve between $x=0$ and $x=2 \pi$ is given by the sum of the areas of the two loops.$	ext{Required Area} = \int_{0}^{\pi} \sin x \, dx + \left| \int_{\pi}^{2 \pi} \sin x \, dx ight|$$= [-\cos x]_{0}^{\pi} + \left| [-\cos x]_{\pi}^{2 \pi} ight|$$= (-\cos \pi - (-\cos 0)) + |(-\cos 2 \pi - (-\cos \pi))|$$= (-(-1) + 1) + |(-1 - (1))|$$= (1 + 1) + |-2|$$= 2 + 2 = 4 	ext{ sq. units}$.

Find the area bounded by the curve $y=\sin x$ between $x=0$ and $x=2 \pi$.

Similar Questions

The area of the region bounded by the curve $y=|x-2|$ between $x=1, x=3$ and the $X$-axis is ......

The area lying in the first quadrant and bounded by the circle $x^{2}+y^{2}=4$ and the lines $x=0$ and $x=2$ is

Area enclosed by the ellipse $9x^2 + 4y^2 = 1$ in the first quadrant is . . . . . . .

The area of the region $\{(x, y): x^{2}+y^{2} \leq 1 \leq x+y\}$ is

The area bounded by the curve $y = x^2 + 4x + 5$,the coordinate axes,and the minimum ordinate is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module