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 of the region bounded by $y=\cos x$,$x=0$,$x=\pi$,and the $x$-axis is ... sq. units.

The area of the region enclosed by the curve $f(x) = \max \{\sin x, \cos x\}$,$-\pi \leq x \leq \pi$ and the $x$-axis is

Let $f(x) = x^3 - 3x^2 + 3x + 1$ and $g$ be the inverse of $f$. The area bounded by the curve $y = g(x)$ with the $x$-axis between $x = 1$ and $x = 2$ is (in square units):

The area (in square units) of the region bounded by $x^2=8y$, $x=4$ and the $X$-axis is:

The area of the region bounded by the curve $y = x^3$ and the lines $y = 8$ and $x = 0$ is ... square units.

Vedclass Test Series

Exam Paper Generator

Online Exam Module