Required area =2∫023(43x−x2−23x)dx=2∫023(12−(x−23)2−23x)dx=2[2x−2312−(x−23)2+212sin−1(23x−23)−2323x23]023=2{3π−8}=6π−16 sq. units.
JEE Main 2025 — Mathematics Calculus
The area of the region, inside the circle (x−23)2+y2=12 and outside the parabola y2=23x is :
Held on 22 Jan 2025 · Verified 6 Jul 2026.
3π+8
6π−16
3π−8
6π−8
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
Let $[\cdot]$ denote the greatest integer function. Then the value of $\displaystyle\int_0^3 \left(\dfrac{e^x + e^{-x}}{[x]!}\right) dx$ is :
The value of $\sum_{r=1}^{20}\left(\left|\sqrt{\pi\left(\int_{0}^{r} x|\sin \pi x| d x\right)}\right|\right)$ is $\_\_\_\_$
The value of ∫₀¹ x·eˣ dx is:
If the area of the region bounded by $16x^2 - 9y^2 = 144$ and $8x - 3y = 24$ is A, then $3(A + 6 \log_e(3))$ is equal to _______.
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $f\left(\dfrac{x+y}{3}\right) = \dfrac{f(x) + f(y)}{3}$ for all $x, y \in \mathbb{R}$, and $f'(0) = 3$. Then the minimum value of the function $g(x) = 3 + e^x f(x)$, is:
Work through every JEE Main Calculus PYQ, year by year.