The area of the region bounded by the curve y | 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 function $y = \cot x$ from $x = \frac{\pi}{4}$ to $x = \frac{\pi}{2}$.$A = \int_{\frac{\pi}{4}}^{\fra

Vedclass · Accepted Answer

$\frac{1}{2} \log 2$

Vedclass · Answer

(C) The area $A$ is given by the integral of the function $y = \cot x$ from $x = \frac{\pi}{4}$ to $x = \frac{\pi}{2}$.$A = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot x \, dx$We know that $\int \cot x \, dx = \log |\sin x| + C$.Applying the limits:$A = [\log |\sin x|]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$$A = \log |\sin(\frac{\pi}{2})| - \log |\sin(\frac{\pi}{4})|$$A = \log(1) - \log(\frac{1}{\sqrt{2}})$Since $\log(1) = 0$ and $\log(\frac{1}{\sqrt{2}}) = \log(2^{-1/2}) = -\frac{1}{2} \log 2$:$A = 0 - (-\frac{1}{2} \log 2) = \frac{1}{2} \log 2$.Thus,the correct option is $C$.

The area of the region bounded by the curve $y = \cot x$,the lines $x = \frac{\pi}{4}$,$x = \frac{\pi}{2}$,and the $X$-axis is . . . . . . sq. units.

Similar Questions

The area bounded by the parabola ${y^2} = 4ax$ and its latus rectum is:

The area of the region bounded by the curve $xy = 4$,$x = 1$,and $x = 3$ is . . . . . . sq. units.

Find the area of the region bounded by the line $y=3x+2$,the $x$-axis,and the ordinates $x=-1$ and $x=1$.

The area of the region bounded by the curve $y = 2 - x - 3x^2$,the $X$-axis,the $Y$-axis,and the line $x = -2$ is

The area of the region bounded by the curve $y = x|x|$,the $x-$axis,and the ordinates $x = 1$ and $x = -1$ is given by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module