The area (in sq. units) of the region bounded | vedclass

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

Question

(A) The area $A$ is given by the integral of the absolute difference between the two functions from $x=0$ to $x=\frac{\pi}{2}$.$A = \int_{0}^{\frac{\p

Vedclass · Accepted Answer

$2(\sqrt{2}-1)$

Vedclass · Answer

(A) The area $A$ is given by the integral of the absolute difference between the two functions from $x=0$ to $x=\frac{\pi}{2}$.$A = \int_{0}^{\frac{\pi}{2}} |\sin x - \cos x| \, dx$.The curves $\sin x$ and $\cos x$ intersect at $x=\frac{\pi}{4}$ in the interval $[0, \frac{\pi}{2}]$.For $0 \le x \le \frac{\pi}{4}$,$\cos x \ge \sin x$. For $\frac{\pi}{4} \le x \le \frac{\pi}{2}$,$\sin x \ge \cos x$.Thus,$A = \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\sin x - \cos x) \, dx$.Evaluating the first integral: $[\sin x + \cos x]_{0}^{\frac{\pi}{4}} = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1) = \sqrt{2} - 1$.Evaluating the second integral: $[-\cos x - \sin x]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = (-0 - 1) - (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) = -1 + \sqrt{2}$.Total area $A = (\sqrt{2} - 1) + (\sqrt{2} - 1) = 2(\sqrt{2} - 1)$ sq. units.

The area (in sq. units) of the region bounded by the lines $x=0, x=\frac{\pi}{2}$ and $f(x)=\sin x, g(x)=\cos x$ is

Similar Questions

One of the points of intersection of the curves $y=1+3x-2x^2$ and $y=\frac{1}{x}$ is $\left(\frac{1}{2}, 2\right)$. Let the area of the region enclosed by these curves be $\frac{1}{24}(\ell \sqrt{5}+m)-n \log_{e}(1+\sqrt{5})$,where $\ell, m, n \in N$. Then $\ell+m+n$ is equal to

The area in the first quadrant between the ellipses $x^{2} + 2y^{2} = a^{2}$ and $2x^{2} + y^{2} = a^{2}$ is:

Find the area of the region enclosed between the two circles $:$ $x^{2}+y^{2}=4$ and $(x-2)^{2}+y^{2}=4$.

The area of the region enclosed by the curves $y^2=4(x+1)$ and $y^2=5(x-4)$ is

The area bounded by the curves $y = \log_e x$ and $y = (\log_e x)^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module