Definite Integration Questions - JEE

$\displaystyle\int \frac{x}{(x^2+1)(x^2+4)}\,dx$ equals

The derivative of $x^4 + x^{-5}$ is $-\left(4x^{-5} + 5x^{-6}\right)$. So, $$\int \frac{5x^3 + 4x^5}{\left(x^5 + x + 1\right)^2} dx =$$

If $f(x) = \lim_{n \to \infty} \left[2x + 4x^3 + ... + 2nx^{2n-1}\right]$ $(0 < x < 1)$ then $\int f(x)dx$ is equal to:

Evaluate $I = \displaystyle\int_{e^{\pi/6}}^{e^{\pi/2}} \frac{\sin(\ln(\sin(\ln x)))\cdot\cos(\ln x)}{x\sin(\ln x)}\,dx$. Find $\cos^{-1}(I+1)$.

Let $f: \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = \frac{x^2 - 3x - 6}{x^2 + 2x + 4}$. Then which of the following statements is (are) TRUE ?

$$\int_1 \frac{(2x^3 + 3x^2 - 1)\,\sqrt{4x^6 + 12x^5 + 9x^4 - 6x^3 - 9x^2 + 2}}{x(x+1)}\,dx =$$

For non-negative integers $a$ and $b$, let $I(a,b)=\displaystyle\int_0^{\pi/2}\cos^a x\cos bx\,dx$. Match each entry in List-I to the correct entry in List-II. **List-I:** P) $I(0,5)$; Q) $I(1,4)$; R) $I(2,4)$; S) $I(3,2)$ **List-II:** 1) $\dfrac{1}{5}$; 2) $\dfrac{1}{5}I(2,1)$; 3) $\dfrac{1}{4}[I(1,3)-I(1,5)]$; 4) $\dfrac{1}{5}I(0,3)$; 5) $\dfrac{1}{3}I(2,1)$

If $\displaystyle\int_0^{4\pi}\ln\left|\frac{1}{3}\sin x+\sqrt{3}\cos x\right|\,dx = k\pi\ln 7$, then the value of $k$ is:

The function $f$ is defined for $x>1$ by $f(x)=\displaystyle\int_1^x\frac{t-1}{t+1}\,dt$. Match each entry in List-I to the correct entry in List-II. **List-I:** P) For $x>2$: $\displaystyle\int_2^x\frac{u-2}{u+2}\,du$; Q) For $x>0$: $\displaystyle\int_0^x\frac{u}{u+4}\,du$; R) For $x>5$: $\displaystyle\int_5^x\frac{u-5}{u+1}\,du$; S) $\displaystyle\int_1^2\frac{u^2+2}{u^2+4}\cdot 2u\,du$ **List-II:** 1) $2f\!\left(\tfrac{x+1}{2}\right)$; 2) $2f\!\left(\tfrac{x}{2}\right)$; 3) $f(3)-f(1.5)$; 4) $f(2)-f(1.5)$; 5) $3f\!\left(\tfrac{x-3}{2}\right)$

Let $\displaystyle\sum_{r=1}^\infty\frac{1}{r^2}=a$, $f(x)=\dfrac{1-\ln x}{x}$, $g(x)=\dfrac{1-\ln x}{x^2}$, $I_1=\displaystyle\int_0^1 f(x)\,dx$, $I_2=\displaystyle\int_0^1 g(x)\,dx$. Match each entry in List-I to the correct entry in List-II. **List-I:** P) $I_1$; Q) $I_2$; R) $I_1-I_2$; S) $\displaystyle\lim_{n\to\infty}\sum_{r=1}^n\frac{r(\ln r-\ln n)}{n^2-r^2}$ **List-II:** 1) $a-$?; 2) $-\frac{3}{4}a$; 3) $-a$; 4) $a+$?; 5) $a-2$

If $L=\displaystyle\lim_{n\to\infty}\int_{-\pi/2}^{\pi/2}\frac{x^4\cos x^2}{x}\cdot\frac{1}{\ln\!\left(\sum_{r=0}^n x^{-r^2\cdot 2!}\right)}\,dx$, then $2L$ equals:

If $\displaystyle\int_0^{4\pi}\ln\left|\frac{1}{3}\sin x+\sqrt{3}\cos x\right|\,dx = k\pi\ln 7$, then the value of $k$ is:

If $L=\displaystyle\lim_{n\to\infty}\int_{-\pi/2}^{\pi/2}\frac{x^4\cos x^2}{x}\cdot\frac{1}{\ln\!\left(\sum_{r=0}^n x^{-r^2\cdot 2!}\right)}\,dx$, then $2L$ equals:

Let $\displaystyle\sum_{r=1}^\infty\frac{1}{r^2}=a$, $f(x)=\dfrac{1-\ln x}{x}$, $g(x)=\dfrac{1-\ln x}{x^2}$, $I_1=\displaystyle\int_0^1 f(x)\,dx$, $I_2=\displaystyle\int_0^1 g(x)\,dx$. Match each entry in List-I to the correct entry in List-II. **List-I:** P) $I_1$; Q) $I_2$; R) $I_1-I_2$; S) $\displaystyle\lim_{n\to\infty}\sum_{r=1}^n\frac{r(\ln r-\ln n)}{n^2-r^2}$ **List-II:** 1) $a-$?; 2) $-\frac{3}{4}a$; 3) $-a$; 4) $a+$?; 5) $a-2$

A function $g(\theta)=\displaystyle\int_0^{\sin^2\theta}f(x)\,dx+\int_0^{\cos^2\theta}f(x)\,dx$ is defined on $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ where $f(x)$ is increasing. Then $g(\theta)$ is increasing on:

The function $f$ is defined for $x>1$ by $f(x)=\displaystyle\int_1^x\frac{t-1}{t+1}\,dt$. Match each entry in List-I to the correct entry in List-II. **List-I:** P) For $x>2$: $\displaystyle\int_2^x\frac{u-2}{u+2}\,du$; Q) For $x>0$: $\displaystyle\int_0^x\frac{u}{u+4}\,du$; R) For $x>5$: $\displaystyle\int_5^x\frac{u-5}{u+1}\,du$; S) $\displaystyle\int_1^2\frac{u^2+2}{u^2+4}\cdot 2u\,du$ **List-II:** 1) $2f\!\left(\tfrac{x+1}{2}\right)$; 2) $2f\!\left(\tfrac{x}{2}\right)$; 3) $f(3)-f(1.5)$; 4) $f(2)-f(1.5)$; 5) $3f\!\left(\tfrac{x-3}{2}\right)$

For non-negative integers $a$ and $b$, let $I(a,b)=\displaystyle\int_0^{\pi/2}\cos^a x\cos bx\,dx$. Match each entry in List-I to the correct entry in List-II. **List-I:** P) $I(0,5)$; Q) $I(1,4)$; R) $I(2,4)$; S) $I(3,2)$ **List-II:** 1) $\dfrac{1}{5}$; 2) $\dfrac{1}{5}I(2,1)$; 3) $\dfrac{1}{4}[I(1,3)-I(1,5)]$; 4) $\dfrac{1}{5}I(0,3)$; 5) $\dfrac{1}{3}I(2,1)$

$\int\sqrt{x}\tan\left(2\tan^{-1}\left(\frac{\sqrt{1+x+\sqrt{1}+x}-1}-\sqrt{1+\sqrt{x}-1}}{\sqrt{1+x+\sqrt{1}+x}-1}+\sqrt{1+\sqrt{x}-1}\right)\right)dx$ is equal to $ax^k + K\tan^{-1}\left(\frac{\sqrt{x}}{2}\right) + \frac{a}{\sqrt{1+\sqrt{x}}} + c$ then $a+b$ is equal to (where $a, b, k, a \in \mathbb{R}$)

If the function f: [0, 8] → ℝ is differentiable, then for 0 a, b ∫08 f(t) dt is equal to

If the area of the region bounded by the curve y = x - x2 and the line y = mx equals $\frac{9}{2}$, then the value of m is :

y = f(x) is a function which satisfies : f(0) = 0 f"(x) = f'(x) and f'(0) = 1 then the area bounded by the graph of y = f(x), the lines x = 0, x - 1 = 0 and y + 1 = 0, is :

The value of the integral $\int_\limits{-1}^{1} \log _{e}(\sqrt{1-x}+\sqrt{1+x}) d x$ is equal to:

The value of $\int_{0}^{1}\left(2 x^{3}-3 x^{2}-x+1\right)^{\frac{1}{3}} d x$ is equal to

If $2\displaystyle\int_0^1 \tan^{-1}x\,dx = \int_0^1 \cot^{-1}(1-x+x^2)\,dx$, then $\displaystyle\int_0^1 \tan^{-1}(1-x+x^2)\,dx$ is equal to

$\int \limits_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}} \frac{48}{\sqrt{9-4 x^2}} d x$ is equal to

If $I_1 = \int_0^1 \dfrac{x^{7/2}(1-x)^{9/2}}{30}\,dx$ and $I_2 = \int_0^1 \dfrac{x^{7/2}(1-x)^{9/2}}{(x+5)^{10}}\,dx$ and $\dfrac{I_1}{I_2} = 5a^3\sqrt{a}$, where $a \in N$, then the value of $a$ is:

Which of the following is/are true?

The value of $\dfrac{e^{-\pi/4}+\displaystyle\int_0^{\pi/4}e^{-x}\tan^{50}x\,dx}{\displaystyle\int_0^{\pi/4}e^{-x}(\tan^{49}x+\tan^{51}x)\,dx}$

If $\displaystyle\int\frac{1-5\cos^2x}{\sin^5x\cos^2x}\,dx=f(x)+C$, where $C$ is the constant of integration, then $f\!\left(\dfrac{\pi}{6}\right)-f\!\left(\dfrac{\pi}{4}\right)$ is equal to

Least positive value of $c$ if $c, k, b$ are in A.P. is:

Evaluate $\displaystyle\int_0^1\frac{dx}{x^2+3x+2}$

Let $\alpha>0$. If $\displaystyle\int_0^\alpha\dfrac{x}{\sqrt{x+\alpha}-\sqrt{x}}\,dx=\dfrac{16+20\sqrt{2}}{15}$, then $\alpha$ is equal to:

The value of the integral $\displaystyle\int_{1/2}^{2}\dfrac{\tan^{-1}x}{x}\,dx$ is equal to:

If $I_1 = \int_0^{\sin^2 x} t^2 dt, I_2 = \int_0^{\sin x} \sin t dt$ then, $I_1 : I_2$ is equal to

The value of $\int_0^\pi \sgn\left(\sin^2 x - \sin x + \frac{1}{4}\right)dx$ is equal to (where, $\sgn(x)$ denotes the signum function of $x$)

Evaluate $\displaystyle\int_0^{\pi/2}\ln(\sin x+\cos x)\,dx$ [JEE Main 2021]

If $m, n$ are even integers and $p, q \in \mathbb{R}$, then $\int_{p+ma}^{q+na} g(t)dt$ is equal to:

If $\phi(x)=\dfrac{1}{\sqrt{x}}\displaystyle\int_{\pi/4}^x(4\sqrt{2}\sin t-3\phi'(t))\,dt$, $x>0$, then $\phi'\!\left(\dfrac{\pi}{4}\right)$ is equal to:

Let $f(x)$ be a continuous function such that $f(x) > 0$ for all $x \geq 0$ and $\left(f(x)\right)^{101} = 1 + \int_0^x f(t) dt$, then $(f(10))^{100}$ is equal to ____.

Evaluate $\displaystyle\int_0^1\frac{dx}{\sqrt{x(1-x)}}$

Evaluate $\displaystyle\int_0^{\pi/2}\cos^6 x\,dx$ [JEE Main 2018]

Evaluate $\displaystyle\int_{-1}^{1}[x+[x+[x]]]\,dx$ where $[\cdot]$ is the greatest integer function. [JEE Main 2019]

∫₀⁴ [(y² - 4y + 5)sin(y - 2)]/(2y² - 8y + 11)dy is equal to:

The value of $\displaystyle\int_{\pi/3}^{\pi/2}\dfrac{2+3\sin x}{\sin x(1+\cos x)}\,dx$ is equal to:

Evaluate $\displaystyle\int_0^{\pi/2}\sqrt{1-\sin 2x}\,dx$ [JEE Main 2020]

Let $[x]$ denote the greatest integer $\leq x$. Consider the function $f(x)=\max\{x^2,1+[x]\}$. Then the value of $\displaystyle\int_0^2 f(x)\,dx$ is:

If $y = f(x^2)$, $\forall x \in \mathbb{R}$ is equal to

Let $\alpha$ > - 1 and $\beta$ > - 1, then the value of $\lim _\limits{n \rightarrow \infty} n^{\beta-\alpha}\left(\frac{1^{\alpha}+2^{\alpha}+\ldots+n^{\alpha}}{1^{\beta}+2^{\beta}+\ldots+n^{\beta}}\right)$ is :

Definite Integration Questions (1340)