The area enclosed between the curves y = sin | vedclass

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

Question

(B) The curves $y = \sin x$ and $y = \cos x$ intersect at $x = \frac{\pi}{4}$.For $0 \le x \le \frac{\pi}{4}$,the area is bounded by $y = \sin x$ and

Vedclass · Accepted Answer

$2 - \sqrt{2}$

Vedclass · Answer

(B) The curves $y = \sin x$ and $y = \cos x$ intersect at $x = \frac{\pi}{4}$.For $0 \le x \le \frac{\pi}{4}$,the area is bounded by $y = \sin x$ and the $x$-axis.For $\frac{\pi}{4} \le x \le \frac{\pi}{2}$,the area is bounded by $y = \cos x$ and the $x$-axis.The total area $A$ is given by:$A = \int_{0}^{\frac{\pi}{4}} \sin x \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos x \, dx$$A = [-\cos x]_{0}^{\frac{\pi}{4}} + [\sin x]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$$A = \left(-\cos \frac{\pi}{4} - (-\cos 0)\right) + \left(\sin \frac{\pi}{2} - \sin \frac{\pi}{4}\right)$$A = \left(-\frac{1}{\sqrt{2}} + 1\right) + \left(1 - \frac{1}{\sqrt{2}}\right)$$A = 2 - \frac{2}{\sqrt{2}} = 2 - \sqrt{2}$

The area enclosed between the curves $y = \sin x$,$y = \cos x$ and the $x$-axis for $0 \le x \le \frac{\pi}{2}$ is:

Similar Questions

Let $A_1$ be the bounded area enclosed by the curves $y=x^2+2$, $x+y=8$ and the y-axis that lies in the first quadrant. Let $A_2$ be the bounded area enclosed by the curves $y=x^2+2$, $y^2=x$, $x=2$, and the y-axis that lies in the first quadrant. Then $A_1-A_2$ is equal to

The area (in sq. units) of the smaller portion enclosed between the curves,$x^2 + y^2 = 4$ and $y^2 = 3x$,is

The area in the first quadrant between $x^2 + y^2 = \pi^2$ and $y = \sin x$ is

Using the method of integration,find the area of the region bounded by the lines: $2x + y = 4$,$3x - 2y = 6$,and $x - 3y + 5 = 0$.

The area (in sq. units) enclosed by the curves $y=2x-x^2$ and $y=x^2-2x-6$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module