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 182 questions; sign in to attempt the remaining 177 in the exam simulator.

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.$\dfrac{523}{6}$
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   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 2024 (08 Apr, Shift 1)Area bounded by Miscellaneous Curves
   Let $f(x)$ be a positive function such that the area bounded by $y=f(x), y=0$ from $x=0$ to $x=a\gt 0$ is $e^{-a}+4 a^2+a-1$. Then the differential equation, whose general solution is $y=c_1 f(x)+c_2$, where $c_1$ and $c_2$ are arbitrary constants, is
   1. A.$\left(8 e^x-1\right) \frac{d^2 y}{d x^2}+\frac{d y}{d x}=0$
   2. B.$\left(8 e^x-1\right) \frac{d^2 y}{d x^2}-\frac{d y}{d x}=0$
   3. C.$\left(8 e^x+1\right) \frac{d^2 y}{d x^2}-\frac{d y}{d x}=0$
   4. D.$\left(8 e^x+1\right) \frac{d^2 y}{d x^2}+\frac{d y}{d x}=0$
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   Answer: (D)

   $\begin{aligned} & \int_0^a f(x) d x=e^{-a}+4 a^2+a-1 \\ & f(a)=-e^{-a}+8 a+1 \\ & f(x)=-e^{-x}+8 x+1\end{aligned}$ Now $y=C_1 \mathrm{f}(\mathrm{x})+\mathrm{C}_2$ $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{C}_1 \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{C}_1\left(\mathrm{e}^{-\mathrm{x}}+8\right)$ ...(1) $\frac{d^2 y}{d x^2}=-C_1 e^{-x} \Rightarrow-e^x \frac{d^2 y}{d x^2}$ Put in equation (1) $\begin{aligned} & \frac{d y}{d x}=-e^x \frac{d^2 y}{d x^2}\left(e^{-x}+8\right) \\ & \left(8 e^x+1\right) \frac{d^2 y}{d x^2}+\frac{d y}{d x}=0\end{aligned}$
3. Q3JEE 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
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   Answer: (A)

   $$\begin{aligned} & \text { Required area }(O A B)=\int_{1-e}^0 \ln (x+e) d x \\ & =\left[x \ln (x+e)-\int \frac{1}{x+e} x d x\right]_0^1=1 . \end{aligned}$$
4. Q4JEE 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.$15$
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   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: $9$
5. Q5JEE Main 2026 (05 Apr, Shift 1)Area bounded by two curves
   The area of the region $R = \{(x, y): xy \leq 27, 1 \leq y \leq x^2\}$ is equal to:
   1. A.$78\log_e 3 - \dfrac{52}{3}$
   2. B.$54\log_e 3 - \dfrac{52}{3}$
   3. C.$54\log_e 3 - \dfrac{26}{3}$
   4. D.$54\log_e 3 + \dfrac{26}{3}$
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   Answer: (B)

   The given region is bounded by the curves $y = 1$, $y = x^2$, and $xy = 27$ in the first quadrant. Let us find the points of intersection of these curves: Intersection of $y = 1$ and $y = x^2$ gives $x = 1$. Intersection of $y = x^2$ and $xy = 27$ gives $x(x^2) = 27 \Rightarrow x^3 = 27 \Rightarrow x = 3$, so $y = 9$. Intersection of $y = 1$ and $xy = 27$ gives $x = 27$. We can find the area by integrating with respect to $y$ from $y = 1$ to $y = 9$. For a given $y$, the value of $x$ ranges from the parabola $x = \sqrt{y}$ to the hyperbola $x = \dfrac{27}{y}$. The area $A$ is given by: $A = \int_{1}^{9} \left( \dfrac{27}{y} - \sqrt{y} \right) dy$ Integrating the terms: $A = \left[ 27 \ln y - \dfrac{2}{3} y^{3/2} \right]_{1}^{9}$ Substituting the limits: $A = \left( 27 \ln 9 - \dfrac{2}{3} (9)^{3/2} \right) - \left( 27 \ln 1 - \dfrac{2}{3} (1)^{3/2} \right)$ $A = \left( 27 \ln(3^2) - \dfrac{2}{3} (27) \right) - \left( 0 - \dfrac{2}{3} \right)$ $A = 54 \ln 3 - 18 + \dfrac{2}{3}$ $A = 54 \ln 3 - \dfrac{54}{3} + \dfrac{2}{3}$ $A = 54 \ln 3 - \dfrac{52}{3}$ Answer: $54\log_e 3 - \dfrac{52}{3}$