The area (in sq. units) in the first quadrant bounded by the curve $y=x^2+2$ and the lines $y=x+1$,$x=0$,and $x=2$ is:
- A$\frac{1}{3}$
- B$\frac{2}{3}$
- C$\frac{5}{3}$
- D$\frac{8}{3}$
Explore More
Similar Questions
The area of the region bounded by the curve $y = \cos x$,$x = 0$,and $x = 3\pi$ is . . . . . . sq. units.
The area of the region bounded by the parabola $y^{2}=8x$ and its latus rectum is
The area bounded by the curve $y = 4x - x^2$ and the $X$-axis in square units is:
The area of the region bounded by the curve $y = 2 \sqrt{1 - x^2}$ and the $X$-axis is . . . . . . sq. units.
The part of the straight line $y = x + 1$ between $x = 2$ and $x = 3$ is revolved about the $x$-axis. The curved surface area of the solid thus generated is
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo