Area of the region bounded by the curve y = c | 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 value of the function $y = \cos x$ from $x = -\frac{\pi}{2}$ to $x = \pi$. $A = \int_{-\frac

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) The area $A$ is given by the integral of the absolute value of the function $y = \cos x$ from $x = -\frac{\pi}{2}$ to $x = \pi$. $A = \int_{-\frac{\pi}{2}}^{\pi} |\cos x| \, dx$. Since $\cos x \ge 0$ for $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ and $\cos x \le 0$ for $x \in [\frac{\pi}{2}, \pi]$,we split the integral: $A = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x \, dx + \int_{\frac{\pi}{2}}^{\pi} (-\cos x) \, dx$. Evaluating the first part: $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x \, dx = [\sin x]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \sin(\frac{\pi}{2}) - \sin(-\frac{\pi}{2}) = 1 - (-1) = 2$. Evaluating the second part: $\int_{\frac{\pi}{2}}^{\pi} -\cos x \, dx = [-\sin x]_{\frac{\pi}{2}}^{\pi} = -(\sin \pi - \sin \frac{\pi}{2}) = -(0 - 1) = 1$. Total area $A = 2 + 1 = 3$ sq. units.

Area of the region bounded by the curve $y = \cos x$,$x = -\frac{\pi}{2}$ and $x = \pi$ is . . . . . . sq. units.

Similar Questions

The area between the curve $y = 4 + 3x - x^2$ and the $x$-axis is:

Find the area of the smaller region bounded by the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ and the line $\frac{x}{a}+\frac{y}{b}=1$.

If the area bounded by the curve $y = x^3 + ax$ (where $a > 0$), the $x$-axis, and the lines $x = -2$ and $x = 1$ is $\frac{37}{4}$ square units, find the value of $a$.

The area of the region bounded by the curve $x+2y=8$,the $X$-axis,and the lines $x=1$ and $x=5$ using integration is . . . . . . sq. units.

The area bounded by the curve $y = -x^2$,the $x$-axis,$x = 1$,and $x = 4$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module