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) Given the equation of the curve $x+2y=8$,we can express $y$ in terms of $x$ as: $2y = 8 - x$ $y = \frac{8-x}{2} = 4 - \frac{x}{2}$ To find the are

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(C) Given the equation of the curve $x+2y=8$,we can express $y$ in terms of $x$ as: $2y = 8 - x$ $y = \frac{8-x}{2} = 4 - \frac{x}{2}$ To find the area bounded by the curve,the $X$-axis,and the lines $x=1$ and $x=5$,we use the definite integral: $	ext{Area} = \int_{1}^{5} y \, dx$ $	ext{Area} = \int_{1}^{5} (4 - \frac{x}{2}) \, dx$ Integrating term by term: $	ext{Area} = [4x - \frac{x^2}{4}]_{1}^{5}$ Evaluating at the limits: $	ext{Area} = (4(5) - \frac{5^2}{4}) - (4(1) - \frac{1^2}{4})$ $	ext{Area} = (20 - \frac{25}{4}) - (4 - \frac{1}{4})$ $	ext{Area} = 20 - 6.25 - 4 + 0.25$ $	ext{Area} = 16 - 6 = 10$ Thus,the area is $10$ sq. units. The correct option is $C$.

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.

Similar Questions

The area bounded by the curve $y = x |\sin(\frac{x}{2})|$ and the $X$-axis between the lines $x = 0$ and $x = 4\pi$ (in square units) is: (in $\pi$)

Let the area of the bounded region $\{(x, y): 0 \leq 9x \leq y^2, y \geq 3x-6\}$ be $A$. Then $6A$ is equal to . . . . . .

The area (in sq units) of the region bounded by $x=-1$, $x=2$, $y=x^2+1$ and $y=2x-2$ is

The area (in square units) bounded by the line $y = x$, the $X$-axis and the lines $x = -2$ and $x = 4$ is...

Let $z=x+iy$ be a complex number $(x, y \in R)$. Let $A$ and $B$ be two sets such that $A=\{z:|z| \leq 2\}$ and $B=\{z:(z+2y)+\bar{z} \geq 4\}$. The area of region $A \cap B$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module