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) Given the curve $y^2 = x^3$, we express $x$ in terms of $y$ as $x = y^{2/3}$.The area $A$ bounded by the curve, the $y$-axis, and the lines $y = 1

Vedclass · Accepted Answer

$\frac{93}{5} 	ext{ sq. units}$

Vedclass · Answer

(B) Given the curve $y^2 = x^3$, we express $x$ in terms of $y$ as $x = y^{2/3}$.The area $A$ bounded by the curve, the $y$-axis, and the lines $y = 1$ and $y = 8$ is given by the integral:$A = \int_{1}^{8} x \, dy = \int_{1}^{8} y^{2/3} \, dy$Integrating $y^{2/3}$ with respect to $y$:$A = \left[ \frac{y^{(2/3) + 1}}{(2/3) + 1} ight]_{1}^{8} = \left[ \frac{y^{5/3}}{5/3} ight]_{1}^{8} = \left[ \frac{3}{5} y^{5/3} ight]_{1}^{8}$Evaluating the definite integral:$A = \frac{3}{5} [8^{5/3} - 1^{5/3}]$Since $8^{5/3} = (2^3)^{5/3} = 2^5 = 32$ and $1^{5/3} = 1$:$A = \frac{3}{5} [32 - 1] = \frac{3}{5} 	imes 31 = \frac{93}{5} 	ext{ sq. units}$.

The area of the region bounded by the curve $y^2 = x^3$, the $y$-axis and the lines $y = 1$ and $y = 8$ is

Similar Questions

The area (in square units) of the region bounded by $x^2=8y$, $x=4$ and the $X$-axis is:

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

The area of the region bounded by the curves $y = |x - 4|$, $x = 3$, $x = 5$, and the $X$-axis is

The area of the region bounded by the parabola $y^2 = 4x$ and its latus rectum is . . . . . . sq. units.

Let the straight line $x=b$ divide the area enclosed by $y=(1-x)^2, y=0$,and $x=0$ into two parts $R_1(0 \leq x \leq b)$ and $R_2(b \leq x \leq 1)$ such that $R_1-R_2=\frac{1}{4}$. Then $b$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module