The area of the region bounded by the curve x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The given equation of the curve is $x + 2y + 8 = 0$,which can be rewritten as $x = -2y - 8$.The area $A$ of the region bounded by the curve $x = f

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(C) The given equation of the curve is $x + 2y + 8 = 0$,which can be rewritten as $x = -2y - 8$.The area $A$ of the region bounded by the curve $x = f(y)$ and the lines $y = c$ and $y = d$ is given by the integral $A = \int_{c}^{d} |x| \, dy$.Here,$c = -3$ and $d = -1$.So,$A = \int_{-3}^{-1} |-2y - 8| \, dy$.Since for $y \in [-3, -1]$,the value of $-2y - 8$ is negative (e.g.,at $y = -2$,$-2(-2) - 8 = 4 - 8 = -4$),we have $|-2y - 8| = -(-2y - 8) = 2y + 8$.Thus,$A = \int_{-3}^{-1} (2y + 8) \, dy$.Integrating,we get $A = [y^2 + 8y]_{-3}^{-1}$.Evaluating at the limits: $A = ((-1)^2 + 8(-1)) - ((-3)^2 + 8(-3))$.$A = (1 - 8) - (9 - 24) = -7 - (-15) = -7 + 15 = 8$.Therefore,the area is $8$ sq. units.

The area of the region bounded by the curve $x + 2y + 8 = 0$ and the lines $y = -3$ and $y = -1$ is . . . . . . sq. units.

Similar Questions

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

The area between the curve $y = \cos x$ and the $x$-axis for $0 \le x \le 2\pi$ is:

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

The area of the region bounded by $x^2=4y$,$y=1$,$y=4$ and the y-axis lying in the first quadrant is $ . . . . . . $ square units.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module