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

(A) The given curve is $y = |x - 2|$.We need to find the area bounded by $y = |x - 2|$,$x = 1$,$x = 3$,and the $X$-axis.The function $y = |x - 2|$ can

Vedclass · Accepted Answer

$1 	ext{ sq.units}$

Vedclass · Answer

(A) The given curve is $y = |x - 2|$.We need to find the area bounded by $y = |x - 2|$,$x = 1$,$x = 3$,and the $X$-axis.The function $y = |x - 2|$ can be defined as:$y = \begin{cases} -(x - 2) & 	ext{if } x The area $A$ is given by the integral:$A = \int_{1}^{3} |x - 2| \, dx$Splitting the integral at $x = 2$:$A = \int_{1}^{2} -(x - 2) \, dx + \int_{2}^{3} (x - 2) \, dx$$A = \left[ -\frac{(x - 2)^2}{2} ight]_{1}^{2} + \left[ \frac{(x - 2)^2}{2} ight]_{2}^{3}$$A = \left( 0 - (-\frac{(1 - 2)^2}{2}) ight) + \left( \frac{(3 - 2)^2}{2} - 0 ight)$$A = \frac{1}{2} + \frac{1}{2} = 1 	ext{ sq.units}$.

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

Similar Questions

The area bounded by the curve $y = f(x)$,the $x$-axis,and the ordinates $x = 1$ and $x = b$ is $(b - 1) \sin(3b + 4)$. Then $f(x)$ is:

The value of $a$ $(a > 0)$ for which the area bounded by the curves $y = \frac{x}{6} + \frac{1}{x^2}$,$y = 0$,$x = a$,and $x = 2a$ has the least value,is

Find the area of the region bounded by $y^{2}=9x$,$x=2$,$x=4$ and the $x$-axis in the first quadrant.

Area of the region bounded by the curve $y=\cos x$,$x=\frac{\pi}{2}$ and $x=\frac{3 \pi}{2}$ is . . . . . . sq. units.

If the area bounded by the curve $x^2y + y^2x = \alpha xy$ is $2$ units,then the possible value$(s)$ of $\alpha$ is/are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module