Area Under Curves JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Area Under Curves, free to read — no sign-in needed. The full chapter has 20 questions; sign in to attempt the remaining 15 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 2)
   Passage: Consider the ellipses given by $x^2 + 4y^2 = 1$ and $4x^2 + y^2 = 1$. Question: If $\alpha$ is the area of the common region that lies inside both the given ellipses, then the value of $\cot\alpha$ is ___________.
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   Answer: 0.75

   The given ellipses are $E_1: x^2 + 4y^2 = 1$ and $E_2: 4x^2 + y^2 = 1$. By symmetry, the common region is symmetric about both the coordinate axes and the lines $y = x$ and $y = -x$. The total area $\alpha$ is $8$ times the area of the region in the first quadrant bounded by $\theta = 0$ and $\theta = \dfrac{\pi}{4}$. In polar coordinates, substituting $x = r\cos\theta$ and $y = r\sin\theta$ into the equation of $E_2$ (which is the inner boundary for $0 \le \theta \le \dfrac{\pi}{4}$), we get: $4r^2\cos^2\theta + r^2\sin^2\theta = 1 \Rightarrow r^2 = \dfrac{1}{4\cos^2\theta + \sin^2\theta}$ The area of this sector is given by: $A = \dfrac{1}{2} \int_{0}^{\dfrac{\pi}{4}} r^2 d\theta = \dfrac{1}{2} \int_{0}^{\dfrac{\pi}{4}} \dfrac{1}{4\cos^2\theta + \sin^2\theta} d\theta$ Dividing the numerator and the denominator by $\cos^2\theta$: $A = \dfrac{1}{2} \int_{0}^{\dfrac{\pi}{4}} \dfrac{\sec^2\theta}{4 + \tan^2\theta} d\theta$ Substituting $t = \tan\theta$, we have $dt = \sec^2\theta d\theta$. The limits change from $0$ to $1$: $A = \dfrac{1}{2} \int_{0}^{1} \dfrac{dt}{4 + t^2} = \dfrac{1}{2} \left[ \dfrac{1}{2} \arctan\left(\dfrac{t}{2}\right) \right]_{0}^{1} = \dfrac{1}{4} \arctan\left(\dfrac{1}{2}\right)$ The total area $\alpha$ of the common region is: $\alpha = 8A = 8 \times \dfrac{1}{4} \arctan\left(\dfrac{1}{2}\right) = 2 \arctan\left(\dfrac{1}{2}\right)$ We need to find the value of $\cot\alpha$: $\cot\alpha = \cot\left(2 \arctan\left(\dfrac{1}{2}\right)\right)$ Let $\phi = \arctan\left(\dfrac{1}{2}\right)$, which implies $\tan\phi = \dfrac{1}{2}$. Using the double angle formula for tangent: $\tan(2\phi) = \dfrac{2\tan\phi}{1 - \tan^2\phi} = \dfrac{2 \left(\dfrac{1}{2}\right)}{1 - \left(\dfrac{1}{2}\right)^2} = \dfrac{1}{1 - \dfrac{1}{4}} = \dfrac{4}{3}$ Therefore, $\cot\alpha = \dfrac{1}{\tan(2\phi)} = \dfrac{3}{4}$. Answer: $3/4$
2. Q2JEE Advanced Adv 2026 (Paper 2)
   Passage: Consider the curve $C_1$ given by $y = e^{-x}$ for $x \in [0, 10\pi]$, and the curve $C_2$ given by $y = e^{-x}(\sin x + \cos x)$ for $x \in [0, 10\pi]$. Let $n$ be the total number of points of intersection of the curves $C_1$ and $C_2$. Suppose that $\alpha_1, \alpha_2, \ldots, \alpha_n \in [0, 10\pi]$ are the $x$-coordinates of the points of intersection of the curves $C_1$ and $C_2$ such that $\alpha_1 \lt \alpha_2 \lt \cdots \lt \alpha_n$. Question: Let $\beta$ be the area of the region enclosed between the curves $C_1$, $C_2$, and the lines $x = \alpha_1$ and $x = \alpha_4$. Then the value of $-\dfrac{1}{\pi}\log_e\left(\beta - 2 e^{-\frac{\pi}{2}}\right)$ is ___________.
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   Answer: 2.5

   To find the points of intersection of the curves $C_1$ and $C_2$, we equate their equations: $e^{-x} = e^{-x}(\sin x + \cos x)$ Since $e^{-x} \neq 0$, we have: $\sin x + \cos x = 1$ $\Rightarrow \sqrt{2}\sin\left(x + \dfrac{\pi}{4}\right) = 1 \Rightarrow \sin\left(x + \dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}$ The general solution is $x + \dfrac{\pi}{4} = 2k\pi + \dfrac{\pi}{4}$ or $x + \dfrac{\pi}{4} = 2k\pi + \dfrac{3\pi}{4}$ for any integer $k$. Thus, $x = 2k\pi$ or $x = 2k\pi + \dfrac{\pi}{2}$. For $x \in [0, 10\pi]$, the first four points of intersection are: $\alpha_1 = 0$ $\alpha_2 = \dfrac{\pi}{2}$ $\alpha_3 = 2\pi$ $\alpha_4 = 2\pi + \dfrac{\pi}{2} = \dfrac{5\pi}{2}$ The area $\beta$ enclosed between the curves from $x = \alpha_1$ to $x = \alpha_4$ is given by: $\beta = \int_{0}^{\dfrac{5\pi}{2}} |e^{-x}(\sin x + \cos x - 1)| dx$ We analyze the sign of $\sin x + \cos x - 1$ in the intervals $(0, \dfrac{\pi}{2})$, $(\dfrac{\pi}{2}, 2\pi)$, and $(2\pi, \dfrac{5\pi}{2})$: For $x \in (0, \dfrac{\pi}{2})$, $\sin x + \cos x \gt 1$ For $x \in (\dfrac{\pi}{2}, 2\pi)$, $\sin x + \cos x \lt 1$ For $x \in (2\pi, \dfrac{5\pi}{2})$, $\sin x + \cos x \gt 1$ Let $I = \int e^{-x}(\sin x + \cos x - 1) dx$. Using integration by parts or standard formulas, we get: $\int e^{-x}\sin x dx = -\dfrac{e^{-x}}{2}(\sin x + \cos x)$ $\int e^{-x}\cos x dx = \dfrac{e^{-x}}{2}(\sin x - \cos x)$ $\Rightarrow \int e^{-x}(\sin x + \cos x) dx = -e^{-x}\cos x$ Thus, $I = -e^{-x}\cos x + e^{-x} = e^{-x}(1 - \cos x)$. Now, we evaluate the area $\beta$ by splitting the integral: $\beta = \int_{0}^{\dfrac{\pi}{2}} e^{-x}(\sin x + \cos x - 1) dx - \int_{\dfrac{\pi}{2}}^{2\pi} e^{-x}(\sin x + \cos x - 1) dx + \int_{2\pi}^{\dfrac{5\pi}{2}} e^{-x}(\sin x + \cos x - 1) dx$ $\beta = \left[ e^{-x}(1 - \cos x) \right]_{0}^{\dfrac{\pi}{2}} - \left[ e^{-x}(1 - \cos x) \right]_{\dfrac{\pi}{2}}^{2\pi} + \left[ e^{-x}(1 - \cos x) \right]_{2\pi}^{\dfrac{5\pi}{2}}$ $\beta = (e^{-\dfrac{\pi}{2}} - 0) - (0 - e^{-\dfrac{\pi}{2}}) + (e^{-\dfrac{5\pi}{2}} - 0)$ $\beta = 2e^{-\dfrac{\pi}{2}} + e^{-\dfrac{5\pi}{2}}$ We are required to find the value of $-\dfrac{1}{\pi}\log_e\left(\beta - 2 e^{-\dfrac{\pi}{2}}\right)$: $\beta - 2 e^{-\dfrac{\pi}{2}} = e^{-\dfrac{5\pi}{2}}$ $\Rightarrow -\dfrac{1}{\pi}\log_e\left(e^{-\dfrac{5\pi}{2}}\right) = -\dfrac{1}{\pi} \times \left(-\dfrac{5\pi}{2}\right) = \dfrac{5}{2} = 2.5$ Answer: $2.5$
3. Q3JEE Advanced Adv 2013 (Paper 1)
   The area enclosed by the curves $y=\sin x+\cos x$ and $\text{y}=\left|\text{cos x}-\text{sin x}\right|$ over the interval $\left[0\text{,}\dfrac{\pi }{2}\right]$ is
   1. A.$4\left(\sqrt{2}-1\right)$
   2. B.$2\sqrt{2}\left(\sqrt{2}-1\right)$
   3. C.$2\left(\sqrt{2}+1\right)$
   4. D.$2\sqrt{2}\left(\sqrt{2}+1\right)$
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   Answer: (B)

   The given curves are $y=\sin x+\cos x\&y=\left|\cos x-\sin x\right|$ ${\text{y}}_{1}=\text{sin x}+\text{cos x}=\sqrt{2}\text{sin}\left(\text{x}+\dfrac{\pi }{4}\right)$ ${\text{y}}_{2}=\sqrt{2}|\text{sin}\left(\dfrac{\pi }{4}-\text{x}\right)|$ So, Required Area$=\int _{0}^{\dfrac{\pi }{4}}\left(\left(\text{sin x}+\text{cos x}\right)-\left(\text{cos x}-\text{sin x}\right)\right)\text{dx}+\int _{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}\left(\left(\text{sin x}+\text{cos x}\right)-\left(\text{sin x}-\text{cos x}\right)\right)\text{dx}$ $2{\int }_{0}^{\dfrac{\pi }{4}}\sin xdx+2{\int }_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}\cos xdx$ $=4-2\sqrt{2}=2\sqrt{2}\left(\sqrt{2}-1\right)$.
4. Q4JEE Advanced Adv 2011 (Paper 1)
   Let the straight line $x=b$ divide the area enclosed by $y=(1-x)^2, y=0$ and $x=0$ into two parts $R_1(0 \leq x \leq b)$ and $R_2(b \leq x \leq 1)$ such that $R_1-R_2=\frac{1}{4}$. Then, $b$ equals to
   1. A.$\frac{3}{4}$
   2. B.$\frac{1}{2}$
   3. C.$\frac{1}{3}$
   4. D.$\frac{1}{4}$
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   Answer: (B)

   Here, area between 0 to $b$ is $R_1$ and $b$ to $$\begin{aligned} & 1 \text { is } R_2 \\ & \therefore \int_0^b(1-x)^2 d x-\int_b^1(1-x)^2 d x=\frac{1}{4} \\ & \Rightarrow \quad\left(\frac{(1-x)^3}{-3}\right)_0^b-\left(\frac{(1-x)^3}{-3}\right)_b^1=\frac{1}{4} \\ & \Rightarrow-\frac{1}{3}\left\{(1-b)^3-1\right\}+\frac{1}{3}\left\{0-(1-b)^3\right\} \\ & =\frac{1}{4} \end{aligned}$$ $$\begin{aligned} & \Rightarrow \quad-\frac{2}{3}(1-b)^3=-\frac{1}{3}+\frac{1}{4}=-\frac{1}{12} \\ & \Rightarrow \quad(1-b)^3=\frac{1}{8} \\ & \Rightarrow \quad(1-b)=\frac{1}{2} \Rightarrow b=\frac{1}{2} \\ & \end{aligned}$$
5. Q5JEE Advanced Adv 2008 (Paper 2)
   The area of the region between the curves $y=\sqrt{\frac{1+\sin x}{\cos x}}$ and $y=\sqrt{\frac{1-\sin x}{\cos x}}$ bounded by the lines $x=0$ and $x=\frac{\pi}{4}$ is
   1. A.$\int_0^{\sqrt{2}-1} \frac{t}{\left(1+t^2\right) \sqrt{1-t^2}} d t$
   2. B.$\int_0^{\sqrt{2}-1} \frac{4 t}{\left(1+t^2\right) \sqrt{1-t^2}} d t$
   3. C.$\int_0^{\sqrt{2}+1} \frac{4 t}{\left(1+t^2\right) \sqrt{1-t^2}} d t$
   4. D.$\int_0^{\sqrt{2}+1} \frac{t}{\left(1+t^2\right) \sqrt{1-t^2}} d t$
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   Answer: (B)

   $$\begin{aligned} & \text { Required area }=\int_0^{\pi / 4}\left(\sqrt{\frac{1+\sin x}{\cos x}}-\sqrt{\frac{1-\sin x}{\cos x}}\right) d x \quad\left[\because \frac{1+\sin x}{\cos x}>\frac{1-\sin x}{\cos x}>0\right] \\ & \left.=\int_0^{\pi / 4} \sqrt{\frac{1+\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}}{\frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}}} \sqrt{\frac{1-\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}}{\frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}}}\right) d x \\ & \left.=\int_0^{\pi / 4} \sqrt{\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}}-\sqrt{\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}}\right) d x \\ & \end{aligned}$$ $$=\int_0^{\pi / 4} \frac{1+\tan \frac{x}{2}-1+\tan \frac{x}{2}}{\sqrt{1-\tan ^2 \frac{x}{2}}} d x=\int_0^{\pi / 4} \frac{2 \tan \frac{x}{2}}{\sqrt{1-\tan ^2 \frac{x}{2}}} d x$$ Put $\tan \frac{x}{2}=t$ $$\begin{aligned} \Rightarrow \quad \frac{1}{2} \sec ^2 \frac{x}{2} d x & =d t \\ \therefore \text { Required area } & =\int_0^{\tan \frac{\pi}{8}} \frac{4 t d t}{\left(1+t^2\right) \sqrt{1-t^2}} \\ & =\int_0^{\sqrt{2}-1} \frac{4 t}{\left(1+t^2\right) \sqrt{1-t^2}} d t \quad\left[\text { astan } \frac{\pi}{8}=\sqrt{2}-1\right] \end{aligned}$$