Area Under Curves JEE Main previous year questions with solutions

5 solved JEE Main questions on Area Under Curves, free to read — no sign-in needed. The full chapter has 180 questions; sign in to attempt the remaining 175 in the exam simulator.

Answers checked against the official answer keys.

1. Q1JEE Main 2026 (06 Apr, Shift 2)Area bounded by two curves
   The area of the region $\{(x, y) : x^2 - 8x \leq y \leq -x\}$ is :
   1. A.$\dfrac{343}{6}$
   2. B.$\dfrac{637}{6}$
   3. C.$\dfrac{437}{6}$
   4. D.$\dfrac{523}{6}$
   Show answer & solution

   Answer: (A)

   The given region is bounded by the parabola $y = x^2 - 8x$ and the line $y = -x$. To find the points of intersection, equate the two equations: $x^2 - 8x = -x$ $x^2 - 7x = 0$ $x(x - 7) = 0$ The points of intersection are $x = 0$ and $x = 7$. In the interval $[0, 7]$, the line $y = -x$ lies above the parabola $y = x^2 - 8x$. The required area $A$ is given by: $A = \int_{0}^{7} (-x - (x^2 - 8x)) dx$ $A = \int_{0}^{7} (7x - x^2) dx$ Evaluating the integral: $A = \left[ \dfrac{7x^2}{2} - \dfrac{x^3}{3} \right]_{0}^{7}$ $A = \dfrac{7(49)}{2} - \dfrac{343}{3}$ $A = \dfrac{343}{2} - \dfrac{343}{3}$ $A = 343 \left( \dfrac{1}{2} - \dfrac{1}{3} \right)$ $A = \dfrac{343}{6}$ Answer: $\dfrac{343}{6}$
2. Q2JEE Main 2023 (01 Feb, Shift 1)Area bounded by Miscellaneous Curves
   The area enclosed by the closed curve $C$ given by the differential equation $\dfrac{dy}{dx}+\dfrac{x+a}{y-2}=0,y\left(1\right)=0$ is $4\pi$. Let $P$ and $Q$ be the points of intersection of the curve $C$ and the $y$-axis. If normals at $P$ and $Q$ on the curve $C$ intersect $x$-axis at points $R$ and $S$ respectively, then the length of the line segment $RS$ is
   1. A.$2\sqrt{3}$
   2. B.$\dfrac{2\sqrt{3}}{3}$
   3. C.$2$
   4. D.$\dfrac{4\sqrt{3}}{3}$
   Show answer & solution

   Answer: (D)

   Given: $\dfrac{dy}{dx}+\dfrac{x+a}{y-2}=0$ $\Rightarrow \dfrac{dy}{dx}=\dfrac{x+a}{2-y}$ $\Rightarrow \left(2-y\right)dy=\left(x+a\right)dx$ $\Rightarrow 2y-\dfrac{{y}^{2}}{2}=\dfrac{{x}^{2}}{2}+ax+c$ Put $x=1$, then we get $y\left(1\right)=0$. $a+c=-\dfrac{1}{2}$ Now, $4y-{y}^{2}={x}^{2}+2ax+2c$ $\Rightarrow {x}^{2}+{y}^{2}+2ax-4y+2c=0$ $\Rightarrow {x}^{2}+{y}^{2}+2ax-4y-1-2a=0$ This is a circle having radius $r=\sqrt{{a}^{2}+4+1+2a}=\sqrt{{\left(a+1\right)}^{2}+4}$ Now, $\pi {r}^{2}=4\pi$ ${r}^{2}=4$ $\Rightarrow \sqrt{{\left(a+1\right)}^{2}+4}=4$ $\Rightarrow {\left(a+1\right)}^{2}=0$ So, $a=-1$ Hence, circle is ${x}^{2}+{y}^{2}-2x-4y+1=0$ $\Rightarrow {\left(x-1\right)}^{2}+{\left(y-2\right)}^{2}=4$ It cuts $y-axis$ at $x=0$, so $P\equiv \left(0,2+\sqrt{3}\right)\&Q\equiv \left(0,2-\sqrt{3}\right)$ Now, ${x}^{2}+{y}^{2}-2x-4y+1=0$ $\Rightarrow 2x+2y\left(\dfrac{dy}{dx}\right)-2-4\left(\dfrac{dy}{dx}\right)=0$ $\Rightarrow \left(\dfrac{dy}{dx}\right)=\dfrac{1-x}{y-2}$ $\Rightarrow -\left(\dfrac{dx}{dy}\right)=\dfrac{y-2}{x-1}$ Slope of normal at $P$ is ${m}_{1}=\dfrac{2+\sqrt{3}-2}{0-1}=-\sqrt{3}$ Slope of normal at $Q$ is ${m}_{2}=\dfrac{2-\sqrt{3}-2}{0-1}=\sqrt{3}$ Equation of normal at $Q$ is $y-\left(2-\sqrt{3}\right)=\sqrt{3}\left(x-0\right)$ $\Rightarrow y-\left(2-\sqrt{3}\right)=\sqrt{3}x$ Equation of normal at $P$ is $y-2-\sqrt{3}=-\sqrt{3}\left(x-0\right)$ $\Rightarrow y=-\sqrt{3}x+2+\sqrt{3}$ So, $R\equiv \left(1+\dfrac{2}{\sqrt{3}},0\right)$ and $S\equiv \left(1-\dfrac{2}{\sqrt{3}},0\right)$ Hence, $RS=\dfrac{4}{\sqrt{3}}=\dfrac{4\sqrt{3}}{3}$ units
3. Q3JEE Main 2020 (08 Jan, Shift 1)Locus
   For $a\gt 0,$ let the curves ${C}_{1}:{y}^{2}=ax$ and ${C}_{2}:{x}^{2}=ay$ intersect at origin $O$ and a point $P.$ Let the line $x=b\left(0\lt b\lt a\right)$ intersect the chord $OP$ and the $x$ -axis at points $Q$ and $R,$ respectively. If the line $x=b$ bisects the area bounded by the curves, ${C}_{1}$ and ${C}_{2},$ and the area of $∆OQR=\dfrac{1}{2},$ then ‘ $a$ ’ satisfies the equation:
   1. A.${x}^{6}-6{x}^{3}+4=0$
   2. B.${x}^{6}-12{x}^{3}+4=0$
   3. C.${x}^{6}+6{x}^{3}-4=0$
   4. D.${x}^{6}-12{x}^{3}-4=0$
   Show answer & solution

   Answer: (B)

   $\int _{0}^{b}\left(\sqrt{ax}-\dfrac{{x}^{2}}{a}\right)dx=\dfrac{{a}^{2}}{6}$ $\Rightarrow \dfrac{2}{3}\sqrt{a}{b}^{\dfrac{3}{2}}-\dfrac{{b}^{3}}{3a}=\dfrac{{a}^{2}}{6}...\left(1\right)$ Also area of $\Delta OQR=\dfrac{1}{2}$ $\dfrac{1}{2}{b}^{2}=\dfrac{1}{2}\Rightarrow b=1$ Put in $\left(1\right)$ $\Rightarrow 4a\sqrt{a}-2={a}^{3}$ $\Rightarrow {a}^{6}+4{a}^{3}+4=16{a}^{3}$ $\Rightarrow {a}^{6}-12{a}^{3}+4=0$
4. Q4JEE Main 2005Area of Curve along axis
   The area enclosed between the curve $y=\log _e(x+e)$ and the coordinate axes is
   1. A.1
   2. B.2
   3. C.3
   4. D.4
   Show answer & solution

   Answer: (A)

   $$\begin{aligned} & \text { Required area }(O A B)=\int_{1-e}^0 \ln (x+e) d x \\ & =\left[x \ln (x+e)-\int \frac{1}{x+e} x d x\right]_0^1=1 . \end{aligned}$$
5. Q5JEE Main 2026 (06 Apr, Shift 1)Area bounded by two curves
   The area of the region $\{(x, y) : 0 \leq y \leq 6 - x, y^2 \geq 4x - 3, x \geq 0\}$ is:
   1. A.$8$
   2. B.$9$
   3. C.$12$
   4. D.$15$
   Show answer & solution

   Answer: (B)

   The given region is defined by the inequalities: $x \geq 0$ $0 \leq y \leq 6 - x \Rightarrow y \geq 0$ and $x \leq 6 - y$ $y^2 \geq 4x - 3 \Rightarrow x \leq \dfrac{y^2 + 3}{4}$ From these inequalities, for a given $y \geq 0$, the value of $x$ ranges from $0$ to $\min\left(6 - y, \dfrac{y^2 + 3}{4}\right)$. To find the point where the two bounding curves intersect, we equate them: $6 - y = \dfrac{y^2 + 3}{4}$ $24 - 4y = y^2 + 3$ $y^2 + 4y - 21 = 0$ $(y + 7)(y - 3) = 0$ Since $y \geq 0$, the intersection occurs at $y = 3$. For $0 \leq y \leq 3$, the right boundary is the parabola $x = \dfrac{y^2 + 3}{4}$. For $3 \leq y \leq 6$, the right boundary is the line $x = 6 - y$. The total area $A$ can be calculated by integrating with respect to $y$: $A = \int_{0}^{3} \dfrac{y^2 + 3}{4} dy + \int_{3}^{6} (6 - y) dy$ Evaluating the first integral: $\int_{0}^{3} \dfrac{y^2 + 3}{4} dy = \dfrac{1}{4} \left[ \dfrac{y^3}{3} + 3y \right]_{0}^{3} = \dfrac{1}{4} (9 + 9) = \dfrac{18}{4} = \dfrac{9}{2}$ Evaluating the second integral: $\int_{3}^{6} (6 - y) dy = \left[ 6y - \dfrac{y^2}{2} \right]_{3}^{6} = (36 - 18) - \left(18 - \dfrac{9}{2}\right) = 18 - 13.5 = \dfrac{9}{2}$ Total Area = $\dfrac{9}{2} + \dfrac{9}{2} = 9$ Answer: $9$