Question:
Find the area enclosed by y=x2 and y=x+2
Find the area enclosed by y=x2 and y=x+2
Updated On: May 13, 2024
Hide Solution
Verified By Collegedunia
Solution and Explanation
The correct answer is: \(\frac{5}{6}\).
Was this answer helpful?
0
0
Top Questions on Application of Integrals
- The area of the region, inside the ellipse \(x^2+4y^2=4\) and outside the region bounded by the curves \(y=x-1\) and \(y=1-x\), is:
- JEE Main - 2026
- Mathematics
- Application of Integrals
- If the area of the region \( \{ (x, y) : x^2 + 1 \leq y \leq 3 - x \} \) is divided by the line \( x = -1 \) in the ratio \( m : n \) (where \( m \) and \( n \) are coprime natural numbers), then the value of \( m + n \) is:
- JEE Main - 2026
- Mathematics
- Application of Integrals
- The area bounded by \[ 2 - 4x \leq y \leq x^2 + 4 \quad \text{and} \quad x = \frac{1}{2}, \, x \geq 0, \, y \geq 0 \] (in square units) is:
- JEE Main - 2026
- Mathematics
- Application of Integrals
- Area enclosed by \[ x^2 + 4y^2 \leq 4, \quad |x| \leq 1, \quad y \geq 1 - |x| \text{ is} \]
- JEE Main - 2026
- Mathematics
- Application of Integrals
- \[ \int \frac{x + 5}{(x + 6)^2} e^x \, dx \] is equal to:
- CBSE CLASS XII - 2025
- Mathematics
- Application of Integrals
View More Questions
Questions Asked in VITEEE exam
- Find the value of \( x \) in the following equation: \[ \frac{2}{x} + \frac{3}{x + 1} = 1 \]
- VITEEE - 2025
- Algebra
- How many numbers between 0 and 9 look the same when observed in a mirror?
- VITEEE - 2025
- Odd one Out
- In a code language, 'TIGER' is written as 'JUISF'. How will 'EQUAL' be written in that language?
- VITEEE - 2025
- Odd one Out
- In a code language, 'TIGER' is written as 'JUISF'. How will 'EQUAL' be written in that language?
- VITEEE - 2025
- Data Interpretation
- TUV : VYB :: PRA : ?
- VITEEE - 2025
- Odd one Out
View More Questions