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 area $A$ of the region bounded by the curve $y = f(x)$,the $x$-axis,and the lines $x = a$ and $x = b$ is given by the formula $A = \int_{a}^{b

Vedclass · Accepted Answer

$\frac{4}{3}$

Vedclass · Answer

(A) The area $A$ of the region bounded by the curve $y = f(x)$,the $x$-axis,and the lines $x = a$ and $x = b$ is given by the formula $A = \int_{a}^{b} |f(x)| \, dx$. Given the curve $y = -2\sqrt{x}$,the lines $x = 0$ and $x = 1$,and the $x$-axis $(y = 0)$,the area is: $A = \int_{0}^{1} |-2\sqrt{x}| \, dx$ $A = \int_{0}^{1} 2\sqrt{x} \, dx$ $A = 2 \int_{0}^{1} x^{1/2} \, dx$ $A = 2 \left[ \frac{x^{3/2}}{3/2} ight]_{0}^{1}$ $A = 2 	imes \frac{2}{3} [x^{3/2}]_{0}^{1}$ $A = \frac{4}{3} (1^{3/2} - 0^{3/2})$ $A = \frac{4}{3}$ sq. units. Thus,the correct option is $A$.

The area of the region bounded by the curve $y = -2\sqrt{x}$ and the lines $x = 0$,$x = 1$,and $y = 0$ is . . . . . . sq. units.

Similar Questions

The area of the region bounded by the curves $y=3x+1$,$y=4x+1$ and the line $x=2$ is

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.

Area enclosed by the curve $y = f(x)$ that is defined parametrically as $x = \frac{1 - t^2}{1 + t^2}, y = \frac{2t}{1 + t^2}$ (where $t \in R$) is equal to

Find the area of the region bounded by the curve $y=x^{2}$ and the line $y=4$. (in $/3$)

The area of the curve $xy^2 = a^2(a - x)$ bounded by the $y$-axis is

Vedclass Test Series

Exam Paper Generator

Online Exam Module