The area of the region bounded by the line y= | vedclass

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

Question

(D) 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 $A = \int_{a}^{b} f(x) dx$. Here,$

Vedclass · Accepted Answer

$10$ sq units

Vedclass · Answer

(D) 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 $A = \int_{a}^{b} f(x) dx$. Here,$f(x) = x+1$,$a=3$,and $b=5$. $	herefore$ Required area,$A = \int_{3}^{5} (x+1) dx$ $= \left[ \frac{x^2}{2} + x ight]_{3}^{5}$ $= \left( \frac{5^2}{2} + 5 ight) - \left( \frac{3^2}{2} + 3 ight)$ $= \left( \frac{25}{2} + 5 ight) - \left( \frac{9}{2} + 3 ight)$ $= \left( \frac{25+10}{2} ight) - \left( \frac{9+6}{2} ight)$ $= \frac{35}{2} - \frac{15}{2}$ $= \frac{20}{2} = 10 	ext{ sq units}$.

The area of the region bounded by the line $y=x+1$ and the lines $x=3$ and $x=5$ is

Similar Questions

The area of the smaller part between the circle $x^2+y^2=4$ and the line $x=1$ is . . . . . . sq. units.

The area of the region bounded by the lines $2x - y + 1 = 0$, $y = -1$, $y = 3$ and the y-axis is

The area of the region bounded by the curves $y = \sin x$, $y = \cos x$ and the lines $x = 0$ and $x = \frac{\pi}{4}$ is

Area lying in the first quadrant and bounded by the ellipse $4x^2 + 9y^2 = 144$ is . . . . . . . (in $\pi$)

The area of the region bounded by the curve $y = \cos x$,$x = 0$,and $x = 3\pi$ is . . . . . . sq. units.

Vedclass Test Series

Exam Paper Generator

Online Exam Module