The area of the region bounded by the curve 2 | vedclass

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

Question

(B) Given the curve equation $2y = -x + 8$,we can express $y$ as $y = \frac{-x + 8}{2} = -\frac{1}{2}x + 4$. To find the area bounded by the curve,the

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Given the curve equation $2y = -x + 8$,we can express $y$ as $y = \frac{-x + 8}{2} = -\frac{1}{2}x + 4$. To find the area bounded by the curve,the $X$-axis,and the lines $x = 3$ and $x = 5$,we use the definite integral: $	ext{Area} = \int_{3}^{5} y \, dx = \int_{3}^{5} (-\frac{1}{2}x + 4) \, dx$. Integrating the function: $\int (-\frac{1}{2}x + 4) \, dx = [-\frac{1}{2} \cdot \frac{x^2}{2} + 4x] = [-\frac{x^2}{4} + 4x]$. Now,applying the limits from $3$ to $5$: $	ext{Area} = [-\frac{5^2}{4} + 4(5)] - [-\frac{3^2}{4} + 4(3)]$. $	ext{Area} = [-\frac{25}{4} + 20] - [-\frac{9}{4} + 12]$. $	ext{Area} = [-\frac{25}{4} + \frac{80}{4}] - [-\frac{9}{4} + \frac{48}{4}]$. $	ext{Area} = \frac{55}{4} - \frac{39}{4} = \frac{16}{4} = 4$. Thus,the area is $4$ sq. units. Therefore,the correct option is $B$.

The area of the region bounded by the curve $2y = -x + 8$,the $X$-axis,and the lines $x = 3$ and $x = 5$ is . . . . . . sq. units.

Similar Questions

The straight line through the origin which divides the area formed by the curves $y=2x-x^2$, $y=0$, and $x=1$ into two equal halves is

Area bounded by the curve $y = \sin x$ between $x = 0$ and $x = 2\pi$ is ......... $sq. \text{ unit}$.

Area enclosed by the parabola $ay = 3(a^2 - x^2)$ and $x$-axis is

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:

The area of the region bounded by $y=-\sqrt{16-x^{2}}$ and the $X$-axis is

Vedclass Test Series

Exam Paper Generator

Online Exam Module