
Required Area
=∫01(1+x)dx+∫12(3−x)dx−∫024x2dx
=[x+32x23]01+[3x−2x2]12−[12x3]02
=(1+32)+(4−25)−32
=25sq.units
JEE Main 2017 — Mathematics Calculus
The area (in sq. units) of the region (x,y):x≥0,x+y≤3,x2≤4yandy≤1+x is
Held on 2 Apr 2017 · Verified 6 Jul 2026.
1259sq.units
23sq.units
37sq.units
25sq.units
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
The value of $\int_{-\pi / 6}^{\pi / 6}\left(\frac{\pi+4 x^{11}}{1-\sin (|x|+\pi / 6)}\right) d x$ is equal to:
The product of all possible values of $\alpha$, for which $\displaystyle\lim_{x \to 0}\left(\dfrac{1 - \cos(\alpha x)\cos((\alpha+1)x)\cos((\alpha+2)x)}{\sin^2((\alpha+1)x)}\right) = 2$, is:
The value of the integral $\displaystyle\int_0^\infty \dfrac{\log_e(x)}{x^2 + 4}\,dx$ is:
Let $f$ be a differentiable function satisfying $f(x)=1-2 x+\int_{0}^{x} \mathrm{e}^{(x-t)} f(t) \mathrm{dt}, x \in \mathbf{R}$ and let $\mathrm{g}(x)=\int_{0}^{x}(f(\mathrm{t})+2)^{15}(\mathrm{t}-4)^{6}(\mathrm{t}+12)^{17} \mathrm{dt}, x \in \mathbf{R}$. If p and q are respectively the points of local minima and local maxima of g, then the value of $|\mathrm{p}+\mathrm{q}|$ is equal to $\_\_\_\_$.
Let the area of the region bounded by the curve $y=\max \{\sin x, \cos x\}$, lines $x=0, x=\frac{3 \pi}{2}$, and the $x$-axis be A. Then, $\mathrm{A}+\mathrm{A}^{2}$ is equal to $\_\_\_\_$.
Work through every JEE Main Calculus PYQ, year by year.