Definite Integration JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Definite Integration, free to read — no sign-in needed. The full chapter has 52 questions; sign in to attempt the remaining 47 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 2)
   The value of the definite integral $\displaystyle\int_{0}^{2} \dfrac{1}{3^{\frac{x+3}{2}}}\, dx$ is
   1. A.$\dfrac{1}{2}$
   2. B.$\dfrac{1}{3}$
   3. C.$\dfrac{\log_e 3}{3}$
   4. D.$\dfrac{\log_e 3}{2}$
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   Answer: (B)

   Let $I = \int_{0}^{2} \dfrac{1}{3^x + 3} dx$ Substitute $t = 3^x$, which gives $dt = 3^x \ln 3 dx \Rightarrow dx = \dfrac{dt}{t \ln 3}$. When $x = 0$, $t = 1$. When $x = 2$, $t = 9$. The integral becomes: $I = \int_{1}^{9} \dfrac{1}{t + 3} \dfrac{dt}{t \ln 3} = \dfrac{1}{\ln 3} \int_{1}^{9} \dfrac{1}{t(t + 3)} dt$ Using partial fractions: $\dfrac{1}{t(t + 3)} = \dfrac{1}{3} \left( \dfrac{1}{t} - \dfrac{1}{t + 3} \right)$ $I = \dfrac{1}{3 \ln 3} \int_{1}^{9} \left( \dfrac{1}{t} - \dfrac{1}{t + 3} \right) dt$ $I = \dfrac{1}{3 \ln 3} [\ln t - \ln(t + 3)]_{1}^{9}$ $I = \dfrac{1}{3 \ln 3} \left[ \ln\left(\dfrac{t}{t + 3}\right) \right]_{1}^{9}$ $I = \dfrac{1}{3 \ln 3} \left( \ln\left(\dfrac{9}{12}\right) - \ln\left(\dfrac{1}{4}\right) \right)$ $I = \dfrac{1}{3 \ln 3} \left( \ln\left(\dfrac{3}{4}\right) - \ln\left(\dfrac{1}{4}\right) \right)$ $I = \dfrac{1}{3 \ln 3} \ln\left(\dfrac{3/4}{1/4}\right) = \dfrac{1}{3 \ln 3} \ln 3$ $I = \dfrac{1}{3}$
2. Q2JEE Advanced Adv 2025 (Paper 2)
   If $\alpha=\int_{\frac{1}{2}}^2 \frac{\tan ^{-1} x}{2 x^2-3 x+2} d x$ then the value of $\sqrt{7} \tan \left(\frac{2 \alpha \sqrt{7}}{\pi}\right)$ is ____________. (Here, the inverse trigonometric function $\tan ^{-1} x$ assumes values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.)
   Show answer & solution

   Answer: 21

   $\alpha=\int_{\frac{1}{2}}^2 \frac{\tan ^{-1} x}{2 x^2-3 x+2} d x$ ....(i) $\begin{aligned} & \text { Let } x=\frac{1}{t} \\ & \qquad d x=-\frac{1}{t^2} d t \\ & \alpha=\int_2^{\frac{1}{2}} \frac{\tan ^{-1}\left(\frac{1}{t}\right)}{\frac{2}{t^2}-\frac{3}{t}+2}\left(\frac{-1}{t^2}\right) d t\end{aligned}$ $\alpha=\int_{\frac{1}{2}}^2 \frac{\cot ^{-1} \mathrm{t}}{2 \mathrm{t}^2-3 \mathrm{t}+2} \mathrm{dt}$ ....(ii) Now by (i) + (ii) $\begin{aligned} & 2 \alpha=\int_{\frac{1}{2}}^2 \frac{\frac{\pi}{2}}{2 x^2-3 x+2} d x \\ & \alpha=\frac{\pi}{8} \int_{\frac{1}{2}}^2 \frac{d x}{x^2-\frac{3 x}{2}+1} \\ & \alpha=\frac{\pi}{8} \int_{\frac{1}{2}}^2\left(x-\frac{3}{4}\right)^2+\frac{7}{16} \\ & \alpha=\frac{\pi}{8 \times \frac{\sqrt{7}}{4}}\left[\tan ^{-1}\left(\frac{x-\frac{3}{4}}{\frac{\sqrt{7}}{4}}\right)\right]_{\frac{1}{2}}^2 \end{aligned}$ $\begin{aligned} & \alpha=\frac{\pi}{2 \sqrt{7}}\left[\tan ^{-1} \frac{4 x-3}{\sqrt{7}}\right]_{\frac{1}{2}}^2 \\ & \alpha=\frac{\pi}{2 \sqrt{7}}\left[\tan ^{-1} \frac{5}{\sqrt{7}}-\tan ^{-1}\left(-\frac{1}{\sqrt{7}}\right)\right] \\ & \alpha=\frac{\pi}{2 \sqrt{7}} \tan ^{-1} \frac{\left(\frac{5}{\sqrt{7}}+\frac{1}{\sqrt{7}}\right)}{1-\frac{5}{7}} \\ & \alpha=\frac{\pi}{2 \sqrt{7}} \tan ^{-1}(3 \sqrt{7}) \end{aligned}$ Now $\sqrt{7} \tan \left(\frac{2 \sqrt{7} \alpha}{\pi}\right)$ $\begin{aligned} & \sqrt{7} \times \tan \left(\tan ^{-1}(3 \sqrt{7})\right) \\ & \sqrt{7} \times 3 \sqrt{7} \\ & =21 \end{aligned}$
3. Q3JEE Advanced Adv 2024 (Paper 2)
   Let $f:\left[0, \frac{\pi}{2}\right] \rightarrow[0,1]$ be the function defined by $f(x)=\sin ^2 x$ and let $g:\left[0, \frac{\pi}{2}\right] \rightarrow[0, \infty)$ be the function defined by $g(x)=\sqrt{\frac{\pi x}{2}-x^2}$. The value of $\frac{16}{\pi^3} \int_0^{\frac{\pi}{2}} f(x) g(x) d x$ is
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   Answer: 0.25

   Now $I_1=\int_0^{\frac{\pi}{2}} f(x) \cdot g(x) d x=\frac{1}{2} \int_0^{\frac{\pi}{2}} g(x) d x$ (it is explained in previous question solution) i.e. $\frac{1}{2} \int_0^{\frac{\pi}{2}} \sqrt{\left(\frac{\pi}{4}\right)^2-\left(x-\frac{\pi}{4}\right)^2} \mathrm{dx}$ Using $\int \sqrt{a^2-x^2}=\frac{1}{2}\left(x \sqrt{a^2-x^2}+a^2 \sin ^{-1}\left(\frac{x}{a}\right)\right)+C$ $\Rightarrow \frac{1}{2}\left[\frac{\left(x-\frac{\pi}{4}\right)}{2} \sqrt{\frac{\pi x}{2}-x^2}+\frac{\frac{\pi^2}{2}}{2} \sin ^{-1}\left(\frac{x-\frac{\pi}{4}}{\frac{\pi}{4}}\right)\right]_0^{\pi / 2}$ $\begin{aligned} & \Rightarrow \frac{1}{2}\left[\left(0+\frac{\pi^3}{64}\right)-\left(0+\left(\frac{-\pi^3}{64}\right)\right)\right] \\ & \Rightarrow \frac{1}{2} \times \frac{\pi^3}{32}\end{aligned}$ Now $\frac{16}{\pi^3} \times \frac{\pi^3}{64}=\frac{1}{4}=0.25$
4. Q4JEE Advanced Adv 2024 (Paper 2)
   Paragraph: Let $f:\left[0, \frac{\pi}{2}\right] \rightarrow[0,1]$ be the function defined by $f(x)=\sin ^2 x$ and let $g:\left[0, \frac{\pi}{2}\right] \rightarrow[0, \infty)$ be the function defined by $g(x)=\sqrt{\frac{\pi x}{2}-x^2}$. Question: The value of $2 \int_0^{\frac{\pi}{2}} f(x) g(x) d x-\int_0^{\frac{\pi}{2}} g(x) d x$ is
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   Answer: 0

   $I=2 \int_0^{\frac{\pi}{2}} \underbrace{\sin ^2 x \cdot \sqrt{\frac{\pi x}{2}-x^2}}_{I_1}-\int_0^{\frac{\pi}{2}} g(x) d x$ Let $I_1=\int_0^{\frac{\pi}{2}} \sin ^2 x \sqrt{\left(\frac{\pi}{4}\right)^2-\left(x-\frac{\pi}{4}\right)^2}$ (making perfect square) apply kings $I_1=\int_0^{\frac{\pi}{2}} \cos ^2 x \sqrt{\left(\frac{\pi}{4}\right)^2-\left(\frac{\pi}{2}-x\right)^2}$ add both $2 I_1=\int_0^{\frac{\pi}{2}} \sqrt{\left(\frac{\pi}{4}\right)^2-\left(x-\frac{\pi}{4}\right)^2}$ i.e. $2 I_1=\int_0^{\frac{\pi}{2}} g(x)$ Now $I=2 I_1-\int_0^{\frac{\pi}{2}} g(x)=0$
5. Q5JEE Advanced Adv 2023 (Paper 2)
   For $x\in ℝ$, let ${\tan }^{-1}\left(x\right)\in \left(-\dfrac{\pi }{2},\dfrac{\pi }{2}\right)$. Then the minimum value of the function $f:ℝ\rightarrow ℝ$ defined by $f\left(x\right)={\int }_{0}^{x{\tan }^{-1}x}\dfrac{{e}^{\left(t-\cos t\right)}}{1+{t}^{2023}}dt$ is
   Show answer & solution

   Answer: 0

   Given, $f\left(x\right)={\int }_{0}^{x{\tan }^{-1}x}\dfrac{{e}^{\left(t-\cos t\right)}}{1+{t}^{2023}}dt$ Now differentiating both side we get, ${f}^{'}\left(x\right)=\dfrac{{e}^{\left[x{\tan }^{-1}x-\cos \left(x{\tan }^{-1}x\right)\right]}}{1+{\left(x{\tan }^{-1}x\right)}^{2023}}\times \left(\dfrac{x}{1+{x}^{2}}+{\tan }^{-1}x\right)$ $\Rightarrow {f}^{'}\left(x\right)=g\left(x\right)\cdot h\left(x\right)$ $\text{where}g\left(x\right)=\dfrac{{e}^{\left[x{\tan }^{-1}⁡x-\cos ⁡\left(x{\tan }^{-1}⁡x\right)\right]}}{1+{\left(x{\tan }^{-1}⁡x\right)}^{2023}}\gt 0\forall x$ And $h\left(x\right)=\dfrac{x}{1+{x}^{2}}+{\tan }^{-1}x\text{which is}\left\{\begin{matrix}\lt 0\text{for}x\lt 0 \\ =0x=0 \\ \gt 0x\gt 0\end{matrix}\right.$ $∴f\left(x\right)$ has minimum at $x=0$ And $f{\left(x\right)}_{min}=f\left(0\right)=0$