The area (in sq. units) of the region enclosed between the parabola y2 = 2x and the line x + y = 4 is _______.
Correct Answer: 18
Solution and Explanation
The correct answer is 18
Required Area :
\(= \int_{-4}^{2} \left(4 - y - \frac{y^2}{2}\right) \,dy\)
\(=\left[4y - \frac{y^2}{2} - \frac{y^3}{6}\right]_{2}^{4}\)
= 18 square units.
Top Questions on Area between Two Curves
- Let $A_1$ be the bounded area enclosed by the curves $y=x^2+2$, $x+y=8$ and $y$-axis that lies in the first quadrant. Let $A_2$ be the bounded area enclosed by the curves $y=x^2+2$, $y^2=x$, $x=2$ and $y$-axis that lies in the first quadrant. Then $A_1-A_2$ is equal to
- JEE Main - 2026
- Mathematics
- Area between Two Curves
The eccentricity of the curve represented by $ x = 3 (\cos t + \sin t) $, $ y = 4 (\cos t - \sin t) $ is:
- MHT CET - 2025
- Mathematics
- Area between Two Curves
- If the area of the region $ \{(x, y) : 1 + x^2 \leq y \leq \min(x + 7, 11 - 3x)\} $ is $ A $, then $ 3A $ is equal to:
- JEE Main - 2025
- Mathematics
- Area between Two Curves
If the area of the region $\{ (x, y) : |x - 5| \leq y \leq 4\sqrt{x} \}$ is $A$, then $3A$ is equal to
- JEE Main - 2025
- Mathematics
- Area between Two Curves
- The area of the region bounded by the curve $ y = \max\{|x|, |x-2|\} $, then x-axis and the lines x = -2 and x = 4 is equal to ____.
- JEE Main - 2025
- Mathematics
- Area between Two Curves
Questions Asked in JEE Main exam
Two p-n junction diodes \(D_1\) and \(D_2\) are connected as shown in the figure. \(A\) and \(B\) are input signals and \(C\) is the output. The given circuit will function as a _______.
- JEE Main - 2026
- Electrical Circuits
- If \( g(x) = 3x^2 + 2x - 3 \), \( f(0) = -3 \) and \( 4g(f(x)) = 3x^2 - 32x + 72 \), then \( f(g(2)) \) is equal to:
- JEE Main - 2026
- Functions
- Identify the correct statements: The presence of –NO\(_2\) group in benzene ring A. activates the ring towards electrophilic substitutions. B. deactivates the ring towards electrophilic substitutions. C. activates the ring towards nucleophilic substitutions. D. deactivates the ring towards nucleophilic substitutions. Choose the correct answer from the options given below:
- JEE Main - 2026
- General Chemistry
- Let $\vec{a}=2\hat{i}-\hat{j}-\hat{k}$, $\vec{b}=\hat{i}+3\hat{j}-\hat{k}$ and $\vec{c}=2\hat{i}+\hat{j}+3\hat{k}$. Let $\vec{v}$ be the vector in the plane of $\vec{a}$ and $\vec{b}$, such that the length of its projection on the vector $\vec{c}$ is $\dfrac{1}{\sqrt{14}}$. Then $|\vec{v}|$ is equal to
- JEE Main - 2026
- Vector Algebra
A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.- JEE Main - 2026
- Solutions
Concepts Used:
Area between Two Curves
Integral calculus is the method that can be used to calculate the area between two curves that fall in between two intersecting curves. Similarly, we can use integration to find the area under two curves where we know the equation of two curves and their intersection points. In the given image, we have two functions f(x) and g(x) where we need to find the area between these two curves given in the shaded portion.
Area Between Two Curves With Respect to Y is
If f(y) and g(y) are continuous on [c, d] and g(y) < f(y) for all y in [c, d], then,
