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

(D) The area $A$ is given by the integral of the absolute value of the function: $A = \int_{0}^{3\pi} |\cos x| \, dx$ Since the function $\cos x$ chan

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(D) The area $A$ is given by the integral of the absolute value of the function: $A = \int_{0}^{3\pi} |\cos x| \, dx$ Since the function $\cos x$ changes sign at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$ and $x = \frac{5\pi}{2}$,we split the integral: $A = \int_{0}^{\pi/2} \cos x \, dx - \int_{\pi/2}^{3\pi/2} \cos x \, dx + \int_{3\pi/2}^{5\pi/2} \cos x \, dx - \int_{5\pi/2}^{3\pi} \cos x \, dx$ Evaluating each part: $|\sin x|_{0}^{\pi/2} = |1 - 0| = 1$ $|\sin x|_{\pi/2}^{3\pi/2} = |-1 - 1| = |-2| = 2$ $|\sin x|_{3\pi/2}^{5\pi/2} = |1 - (-1)| = |2| = 2$ $|\sin x|_{5\pi/2}^{3\pi} = |0 - 1| = |-1| = 1$ Summing these values: $A = 1 + 2 + 2 + 1 = 6$ sq. units. Thus,the correct option is $D$.

The area of the region bounded by the curve $y = \cos x$,$x = 0$,and $x = 3\pi$ is . . . . . . sq. units.

Similar Questions

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 bounded by the curve $y = \ln(x)$ and the lines $y = 0$,$y = \ln(3)$,and $x = 0$ is equal to

The area bounded by the curves $y = \cos x$ and $y = \sin x$ and the ordinates $x = 0$ and $x = \frac{\pi}{4}$ is:

The odd natural number $a$ such that the area of the region bounded by $y = 1, y = 3, x = 0,$ and $x = y^a$ is $\frac{364}{3}$ is equal to:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module