The area (in square units) enclosed between t | vedclass

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

Question

(C) The area $A$ is given by the integral of the absolute difference between the two curves over the interval $[\frac{\pi}{4}, \frac{5\pi}{4}]$.$A = \

Vedclass · Accepted Answer

$2 \sqrt{2}$

Vedclass · Answer

(C) The area $A$ is given by the integral of the absolute difference between the two curves over the interval $[\frac{\pi}{4}, \frac{5\pi}{4}]$.$A = \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} |\sin x - \cos x| \, dx$.In the interval $[\frac{\pi}{4}, \frac{5\pi}{4}]$,$\sin x \geq \cos x$.Thus,$A = \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (\sin x - \cos x) \, dx$.$A = [-\cos x - \sin x]_{\frac{\pi}{4}}^{\frac{5\pi}{4}}$.$A = (-(\cos \frac{5\pi}{4} + \sin \frac{5\pi}{4})) - (-(\cos \frac{\pi}{4} + \sin \frac{\pi}{4}))$.$A = (-(-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}})) - (-(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}))$.$A = (\frac{2}{\sqrt{2}}) - (-\frac{2}{\sqrt{2}}) = \sqrt{2} + \sqrt{2} = 2\sqrt{2}$ square units.

The area (in square units) enclosed between the curves $y=\sin x$ and $y=\cos x$ for $\frac{\pi}{4} \leq x \leq \frac{5 \pi}{4}$ is

Similar Questions

The area (in sq. units) of the region described by $\{(x, y) : y^2 \leq 2x \text{ and } y \geq 4x - 1\}$ is

The area (in sq. units) in the first quadrant bounded by the parabola $y = x^2 + 1$,the tangent to it at the point $(2, 5)$,and the coordinate axes is

Find the area common to the curves $y = \sqrt{9 - x^2}$ and $x^2 + y^2 = 6x$.

The area of the region bounded by the curves $x=y^2-2$ and $x=y$ is

The area of the region bounded by $y-x=2$ and $x^{2}=y$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module