We have R2⇒R2∴R2+R2⇒2R2=R1=∫−23xf(x)dx=∫−23(1−x)f(1−x)dx[Using∫abf(x)dx=∫abf(a+b−x)dx]=∫−23(1−x)f(x)dx=∫−23xf(x)dx+∫−23(1−x)f(x)dx
JEE Main 2013 — Mathematics Calculus
Let f:[−2,3]→[0,∞) be a continuous function such that f(1−x)=f(x) for all x∈[−2,3]. If R1 is the numerical value of the area of the region bounded by y=f(x),x=−2,x=3 and the axis of x and R2=∫−23xf(x)dx, then :
Held on 25 Apr 2013 · Verified 6 Jul 2026.
3R1=2R2
2R1=3R2
R1=R2
R1=2R2
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.