Integral Calculus Questions (265)

Anti-derivative of $$\frac{x - 1}{(x + 1)\sqrt{x^3 + x^2 + x}}$$ is:

Area bounded by the curves $y = \left[\frac{x^2}{64} + 2\right]$ ([$.$] denotes the greatest integer function), $y = x - 1$ and $x = 0$ above the $x$-axis is:

If $f(x) = x + \int_0^x (y^2 + x^2)f(y) dy$, then:

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

Area bounded by $x^2-y-1=0$, $y=-1$ and $x=0$ (positive side) is

If $\Lim_{n \to \infty} \frac{1}{n^2} \sum_{k=1}^{n-1} k \left[ \int_0^{k/n} \sqrt{(x-k)(k+1-x)} dx \right] = \frac{\pi}{m^n}$, then:

If $A = \int_0^{\sin \theta} \frac{t dt}{1 + t^2}$ and $B = \int_0^{\cos \theta} \frac{dt}{t(1 + t^2)}$, then the value of $e^A e^B \begin{vmatrix} A & A^2 & B \\ 1 & B^2 & -1 \\ 1 & A^2 + B^2 & -1 \end{vmatrix}$ is:

$I = \int \frac{dx}{(\sin x - 2\cos x)(2\cos x + \sin x)}$ is equal to

Let $f(x)$ be a continuous function and 'c' is a constant satisfying $\int_0^x f(t) dt = e^x - ce^{2x} \int_0^x f(t)^2 dt$, then:

$$\int \frac{dx}{\prod_{i=0}^{n}(x+r)} \text{ is equal to:}$$

If $f(x)$ is an even function, then:

Area enclosed by $y=g(x)$, $x=1$ and $x=37$, where $g(x)$ is the inverse of $f(x)=x^3+3x+1$, is

Let $f(x)$ is a quadratic function such that $f(0) = 1 \& f(-1) = 4$. If $\int \frac{f(x)dx}{x^2(x + 1)^2}$ is a rational function, then $f(10) = $

If $f(x)$ is even and periodic with period $T$, $\int_0^a f(x)dx=3$ and $\int_{-T/2}^{3T/2}f(x)dx=18$, then $\int_{-a}^{a+5T}f(x)dx$ is

If $f(x) = \sin x + \displaystyle\int_{-\pi/2}^{\pi/2}(\sin x + t\cos x)f(t)\,dt$, then $f(x)$ may be equal to $\left(-\dfrac{1}{k}\sin x - \dfrac{2}{k}\cos x\right)$, where $k$ is a numerical quantity which equals

The area of the region bounded by $y=x^2$ and $y=\sec^{-1}[-\sin^2 x]$, where $[\cdot]$ is the GIF, is

Value of $\displaystyle\int_0^1 \frac{\sin x}{x}\,dx$ lies in the interval

If $f(x) = \begin{vmatrix}\cos x & e^{x^2} & 2x\cos^2(x/2)\\ x^2 & \sec x & \sin x+x^3\\ 1 & 2 & x+\tan x\end{vmatrix}$ and $\displaystyle\int_{-\pi/2}^{\pi/2}(1+x^4)(f(x)+f''(x))\,dx = 2\lambda+3$, then $\lambda$ is

The value of $\int_0^1 \lim_{n \to \infty} \sum_{k=0}^n \frac{x^{k+2^k}}{k!} dx$ is:

If $a$ is a positive integer, then the number of values of $a$ satisfying $$\int_0^{\pi/2} \left[a^2\left(\frac{\cos 3x}{3} + \cos x\right) + a\sin x - 20\cos x\right] dx \leq -\frac{a^2}{3}$$ is:

Let $f(x)=\begin{cases}|1-2x^2|, & 0\le x<1 \\ [x^2-2x], & 1\le x<2\end{cases}$. If $m,n$ are number of points of discontinuity and non-differentiability of $f(x)$ in $(0,2)$, then

$$\int \frac{x^4 + 1}{x^4 + 1} dx =$$

If $\int \left[\left(\frac{x}{e}\right)^x + \left(\frac{e}{x}\right)^x\right] \ln udx = A\left(\frac{x}{e}\right)^x + B\left(\frac{e}{x}\right)^x + C$, then the value of $A + B$ is

The equation $1012x^{2023}-12138x^{2022}-119x+714=0$ has a root in $(a^{1/2022},b^{1/3})$; $a,b\in\mathbb{N}\geq2$. The value of $4\displaystyle\int_{\sqrt{a}}^{b^{1/3}}\frac{x\cos x^2}{\cos x^2+\cos(263-x^2)}\,dx$ is

If $\int \frac{dx}{\sqrt{x + 7} - \sqrt[3]{x + 7}} = P\sqrt{x + 7} + Q\sqrt[3]{x + 7} + R\ln|x + 7|^{1/4}| + c$. Then find the value of $P + Q + R$.

The function $f:[0,1] \to [0,1]$ is continuous and has the property $f(f(x)) = 1-x$ for all $x \in [0,1]$ and $\alpha = \int_0^1 f(x)dx$, then:

The value of $a$ for which the equation $\int_0^a \sin^2\left(\frac{t}{2}\right)dt = a^2x^2 - \frac{1}{2}(3x-1) + \frac{1}{a^2}$ possesses a solution are:

If $\int (x^9 + x^6 + x^3)(2x^6 + 3x^3 + 6)^{1/3} dx = a(2x^9 + 3x^6 + 6x^3)^{4/3} + c$, then the value of $48a$ must be

Let $P(x)$ be a polynomial of least degree whose graph has three points of inflection $(-1,-1), (1,1)$ and a point with abscissa $0$ at which the curve is inclined to the axis of abscissa at an angle of $60°$. Then $\int_0^1 P(x)dx$ equals to:

If $\int \frac{\cos^2 x + \sin 2x}{(2\cos x - \sin x)^2} dx = \frac{\cos x}{2\cos x - \sin x} + ax + b\ln|2\cos x - \sin x| + c$, then:

If $\int \frac{dx}{\sqrt{9x^2 + 4x + 6}}$ to evaluate $I$, one of the most proper substitution could be:

Let $f: \left[0, \frac{\pi}{2}\right] \to \mathbb{R}$ be such that $f(0) = 3$ and $f'(x) = \frac{1}{1 + \cos x}$. If $a < f\left(\frac{\pi}{2}\right) < b$, then $a$ and $b$ can be

If $I_n = \int \frac{dx}{(x^2 + a^2)^n}$, where $n \in \mathbb{N}$ and $n > 1$. If in and $I_{n-1}$ are related by the relation $PI_n = \frac{x}{(x^2 + a^2)^{n-1}} + QI_{n-1}$. Then $P$ and $Q$ are respectively given by:

The value of the integral $\int_0^\pi \frac{\sin(n+1/2)x}{\sin x/2} dx$ $(n \in \mathbb{N})$ is:

If $I$ is the greatest of the definite integrals $I_1 = \int_0^1 e^{-x}\cos^2 x dx, I_2 = \int_0^1 e^{-x^2}\cos^2 x dx, I_3 = \int_0^1 e^{-x^2} dx, I_4 = \int_0^1 e^{-x^2/2} dx$ then:

The intercepts on the x-axis made by tangents to the curve $y=\displaystyle\int_0^x|t|\,dt$, $x\in\mathbb{R}$, which are parallel to $y=2x$, are equal to

Let $I=\displaystyle\int\frac{\sin^2\theta\cdot\log_e(e\cot\theta)}{\cos^4\theta((\log_e\tan\theta)^4-(\tan\theta)^4)}d\theta=\frac{1}{4}\log_e|f(\theta)|-\frac{1}{2}\tan^{-1}(g(\theta))+c$. The value of $f(\theta)-g(\theta)$ at $\theta=\pi/3$ is

Let $f$ be a function on $(0,2\pi]$ such that $\displaystyle\int_0^x (f'(t)-\sin 2t)\,dt = \int_x^0 f(t)\tan t\,dt$ and $f(0)=1$. If the maximum value of $f(x)$ is $m$, then $8m$ equals

If $x\displaystyle\int_0^x f(t)\,dt = (x+1)\int_0^x tf(t)\,dt$ for $x>0$, and $f(1)=\dfrac{1}{e}$, then $f(-1)$ is

The area bounded by the curve $y = \dfrac{1}{x^2 - 2x + 2}$ and the $x$-axis equals

Let $f$ be continuous and differentiable in $(x_1, x_2)$. If $f(x)f'(x) \geq x\sqrt{1-[f(x)]^4}$ and $\lim_{x\to x_1}(f(x))^2=1$, $\lim_{x\to x_2}(f(x))^2=\frac{1}{2}$. Then minimum value of $\left[x_1^2 - x_2^2\right]$ is ........... (where $[\cdot]$ denotes GIF)

Let $I(x)=\displaystyle\int\frac{x+1}{x(x^2e^{2x}-1)}\,dx=\frac{1}{4}\ln\frac{(xe^x)^\alpha-2(xe^x)^\beta+\gamma}{x^4e^{4x}}+c$, then $\alpha+\beta+\gamma$ equals

Let $I=\displaystyle\int_0^2\!\left[\left|x^2-5x+4\right|+\left[\sin\frac{3\pi}{2}x\right]\right]dx$ (where $[\cdot]$ is GIF). Then $I+\dfrac{2}{3}$ is

If $f(x) = \begin{vmatrix}\cos x & e^{x^2} & 2x\cos^2(x/2)\\ x^2 & \sec x & \sin x+x^3\\ 1 & 2 & x+\tan x\end{vmatrix}$ and $\displaystyle\int_{-\pi/2}^{\pi/2}(1+x^4)(f(x)+f''(x))\,dx = 2\lambda+3$, then $\lambda$ is

The area enclosed by the curves $y=\sin x+\cos x$ and $y=|\cos x-\sin x|$ over the interval $\left[0,\dfrac{\pi}{2}\right]$ is

Let $I=\displaystyle\int\frac{\sin^2\theta\cdot\log_e(e\cot\theta)}{\cos^4\theta((\log_e\tan\theta)^4-(\tan\theta)^4)}d\theta=\frac{1}{4}\log_e|f(\theta)|-\frac{1}{2}\tan^{-1}(g(\theta))+c$. The value of $f(\theta)-g(\theta)$ at $\theta=\pi/3$ is

Let $f$ be a function on $(0,2\pi]$ such that $\displaystyle\int_0^x (f'(t)-\sin 2t)\,dt = \int_x^0 f(t)\tan t\,dt$ and $f(0)=1$. If the maximum value of $f(x)$ is $m$, then $8m$ equals

If $\displaystyle\int\frac{\sec^2 x-2010}{\sin^{2010}x}\,dx=\frac{P(x)}{(\sin x)^{2010}}+C$, then the value of $P\!\left(\dfrac{\pi}{3}\right)$ is

Value of $\displaystyle\int_0^1 x^6(x^3-1)^{2022}\,dx$ is

If $x\displaystyle\int_0^x f(t)\,dt = (x+1)\int_0^x tf(t)\,dt$ for $x>0$, and $f(1)=\dfrac{1}{e}$, then $f(-1)$ is