The area bounded by the curve y = 4x - x^2 an | vedclass

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

Question

(C) Given the curve $y = 4x - x^2$ and the $x$-axis $(y = 0)$.To find the points of intersection with the $x$-axis,set $y = 0$:$4x - x^2 = 0 \implies

Vedclass · Accepted Answer

$\frac{32}{3} 	ext{ sq. unit}$

Vedclass · Answer

(C) Given the curve $y = 4x - x^2$ and the $x$-axis $(y = 0)$.To find the points of intersection with the $x$-axis,set $y = 0$:$4x - x^2 = 0 \implies x(4 - x) = 0 \implies x = 0, 4$.The required area is given by the definite integral:$	ext{Area} = \int_{0}^{4} (4x - x^2) dx$$= \left[ \frac{4x^2}{2} - \frac{x^3}{3} ight]_{0}^{4}$$= \left[ 2x^2 - \frac{x^3}{3} ight]_{0}^{4}$$= \left( 2(4)^2 - \frac{(4)^3}{3} ight) - (0 - 0)$$= 32 - \frac{64}{3}$$= \frac{96 - 64}{3} = \frac{32}{3} 	ext{ sq. unit}$.

The area bounded by the curve $y = 4x - x^2$ and the $x$-axis is:

Similar Questions

Find the area of the region bounded by the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$. (in $\pi$)

The area of the region bounded by the parabola $y = x^2 + 2$,the $X$-axis,and the lines $x = 1$ and $x = 2$ is . . . . . . sq. units.

The area bounded by the curve $y=\sin \left(\frac{x}{3}\right)$,the $x$-axis,and the lines $x=0$ and $x=3 \pi$ is

The area lying between the curves $y^{2}=4x$ and $y=2x$ is

The area bounded by the $x-$ axis,the curve $y = f(x)$ and the lines $x = 1$ and $x = b$ is equal to $\sqrt{b^2 + 1} - \sqrt{2}$ for all $b > 1$. Then $f(x)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module