The area bounded by the curves y-1=cos x,y=si | vedclass

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

Question

(D) The curves are $y = \cos x + 1$ and $y = \sin x$. We need to find the area bounded by these curves and the $X$-axis between $x=0$ and $x=\pi$.From

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(D) The curves are $y = \cos x + 1$ and $y = \sin x$. We need to find the area bounded by these curves and the $X$-axis between $x=0$ and $x=\pi$.From the graph,the intersection point $A$ occurs where $\cos x + 1 = \sin x$. However,looking at the region bounded by the $X$-axis,the area is split at $x = \pi/2$.For $0 \le x \le \pi/2$,the region is bounded above by $y = \sin x$ and below by the $X$-axis.For $\pi/2 \le x \le \pi$,the region is bounded above by $y = \cos x + 1$ and below by the $X$-axis.Area $= \int_0^{\pi/2} \sin x \, dx + \int_{\pi/2}^{\pi} (\cos x + 1) \, dx$$= [-\cos x]_0^{\pi/2} + [\sin x + x]_{\pi/2}^{\pi}$$= (-\cos(\pi/2) - (-\cos 0)) + ((\sin \pi + \pi) - (\sin(\pi/2) + \pi/2))$$= (0 + 1) + (0 + \pi - 1 - \pi/2)$$= 1 + \pi - 1 - \pi/2 = \pi/2$.

The area bounded by the curves $y-1=\cos x$,$y=\sin x$ and the $X$-axis between $x=0$ and $x=\pi$ is

Similar Questions

The area enclosed between the parabolas ${y^2 = 4x}$ and ${x^2 = 4y}$ is

The area of the region bounded by the curves $x + 3y^2 = 0$ and $x + 4y^2 = 1$ is equal to: (in $/3$)

The area between the parabolas $y^2 = 4ax$ and $x^2 = 8ay$ is

The area (in sq. units) of the region $A = \{(x, y) / \frac{y^2}{2} \leq x \leq y+4\}$ is:

Let $S = \{(x, y) \in R \times R : x \geq 0, y \geq 0, y^2 \leq 4x, y^2 \leq 12 - 2x \text{ and } 3y + \sqrt{8}x \leq 5\sqrt{8}\}$. If the area of the region $S$ is $\alpha \sqrt{2}$,then $\alpha$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module