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

(B) The area $A$ is given by the integral of the absolute value of the function: $A = \int_{-\pi}^{\frac{3\pi}{2}} |\sin x| dx$ We split the integral

Vedclass · Accepted Answer

$5 	ext{ (unit)}^2$

Vedclass · Answer

(B) The area $A$ is given by the integral of the absolute value of the function: $A = \int_{-\pi}^{\frac{3\pi}{2}} |\sin x| dx$ We split the integral based on the sign of $\sin x$: $A = \int_{-\pi}^{0} |\sin x| dx + \int_{0}^{\pi} |\sin x| dx + \int_{\pi}^{\frac{3\pi}{2}} |\sin x| dx$ Since $\sin x \le 0$ for $x \in [-\pi, 0]$ and $x \in [\pi, \frac{3\pi}{2}]$,and $\sin x \ge 0$ for $x \in [0, \pi]$: $A = \int_{-\pi}^{0} (-\sin x) dx + \int_{0}^{\pi} (\sin x) dx + \int_{\pi}^{\frac{3\pi}{2}} (-\sin x) dx$ $A = [\cos x]_{-\pi}^{0} + [-\cos x]_{0}^{\pi} + [\cos x]_{\pi}^{\frac{3\pi}{2}}$ $A = (\cos 0 - \cos(-\pi)) + (-\cos \pi + \cos 0) + (\cos(\frac{3\pi}{2}) - \cos \pi)$ $A = (1 - (-1)) + (-(-1) + 1) + (0 - (-1))$ $A = (1 + 1) + (1 + 1) + (0 + 1) = 2 + 2 + 1 = 5 	ext{ (unit)}^2$

The area of the region bounded by the curve $y=\sin x$ between $x=-\pi$ and $x=\frac{3\pi}{2}$ is

Similar Questions

The volume of the solid formed by rotating the area enclosed between the curve $y = x^2$ and the line $y = 1$ about the $y$-axis is (in cubic units):

The area of the region bounded by the curves $y = |x - 2|$,$x = 1$,$x = 3$ and the $x$-axis is

The area bounded by the curve $x = 2 - y - y^2$ and the $Y$-axis is

The area bounded by the curve $y=x|x|$,the $x$-axis,and the ordinates $x=-1$ and $x=1$ is given by [Hint: $y=x^{2}$ if $x>0$ and $y=-x^{2}$ if $x < 0$].

The area of the region bounded by the line $y = 3 - x$,the $X$-axis,and the lines $x = 2$ and $x = 5$ is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module