Calculus Questions (384)

If $\int \frac{dx}{x^2\left(x^7 - 6\right)} = A\left[\ln\left(p^9 + 9p^2 - 2p^3 - 18p\right)\right] + c$, then:

If the anti-derivative of $\frac{x^3}{\sqrt{4+2x^2}}$ which passes through $(1, 2)$ is $\frac{1}{m}\left(1+2x^2\right)^{1/2}\left(x^2-1\right)+c$. Then:

If $I = \int \frac{\sin x + \sin^3 x}{\cos 2x} dx = P\cos x + Q\ln |f(x)| + R$, then:

Let $f$ is a differentiable function such that $f(x) = x^2 + \int_0^x e^{-t}f(x-t)dt$, then:

Let $f$ is a differentiable function such that $f'(x) = f(x) + \int_0^2 f(x)dx, f(0) = \frac{4 - e^2}{3}$, then:

If $p, q, r, s$ are in arithmetic progression and $f(x) = \begin{vmatrix} p + \sin x & q + \sin x & p - r + \sin x \\ q + \sin x & r + \sin x & -1 + \sin x \\ r + \sin x & s + \sin x & s - q + \sin x \end{vmatrix}$ such that $\int_0^2 f(x) dx = -4$, then the common difference of the progression is:

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:

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:

The value of $\int_1^a \frac{x^a - 1}{\log x} dx$ is:

The value of $\int_0^{\pi/2} \log(\sin^2 \theta + k^2 \cos^2 \theta) d\theta$, where $k \geq 0$, is:

The value of $\frac{dI}{da}$ when $I = \int_0^{\pi/2} \log\left(\frac{1 + a \sin x}{1 - a \sin x}\right) \frac{dx}{\sin x}$ (where $|a| < 1$) is:

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

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

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 $G(x,t) = \begin{cases} x(t-1), & \text{when } x \leq t \\ t(x-1), & \text{when } t < x \end{cases}$ and if $f$ is continuous function of $x$ in $[0,1]$. Let $g(x) = \int_0^1 f(t)G(x,t)dt$. Then which is incorrect:

If $\int_0^1 \frac{\sin t}{1+t} dt = a$, then the value of $\int_{4\pi-2}^{4\pi} \frac{\sin t}{4\pi + 2 - t} dt$ is:

The value of $\int_1^8 x\sin[x^2 - \pi] dx$, where $[.]$ denotes the greatest integer function is:

If $a \leq \int_0^1 \frac{dx}{\sqrt{4-x^2-x^3}} \leq b$, then $(a,b) =$

Let $I = \int_{\pi/4}^{\pi/3} \frac{\sin x}{x} dx$, then $I$ belongs to:

The sum of the series as $n \to \infty$ $\frac{\sqrt{n}}{(3+4\sqrt{n})^2} + \frac{\sqrt{n}}{\sqrt{2}(3\sqrt{2}+4\sqrt{n})^2} + \frac{\sqrt{n}}{\sqrt{3}(3\sqrt{3}+4\sqrt{n})^2} + \ldots + \frac{1}{49n}$ is:

If $P = \int_0^\infty \frac{x^2}{1+x^4} dx$; $Q = \int_0^\infty \frac{xdx}{1+x^4}$ and $R = \int_0^\infty \frac{dx}{1+x^4}$, then:

Let $u = \int_0^{\pi/4} \left(\frac{\cos x}{\sin x + \cos x}\right)^2 dx$ and $v = \int_0^{\pi/4} \left(\frac{\sin x + \cos x}{\cos x}\right)^2 dx$, then:

If $\int_0^2 \frac{\ln(1+2x)}{1+x^2} dx = (\tan^{-1}a)(\ln\sqrt{b})$ where $a,b \in \mathbb{N}$, then:

Comment upon the nature of roots of the quadratic equation $x^2 + 2x + k = \int_0^k |1+k| dr$ depending on the value of $k \in \mathbb{R}$.

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:

Let $g(x) = x^c e^{2x}$ & let $f(x) = \int_0^x e^{2t}(3t^2+1)^{1/2} dt$. For a certain value of 'c', the limit of $\frac{f'(x)}{g'(x)}$ as $x \to \infty$ is finite and non-zero, then:

Which of the following is/are true?

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:

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

Let $f$ be a function defined by $f(x)=\begin{cases}(x-r)^2, & r-1\leq x<r+1\\ 1, & r+1\leq x\leq r+2\end{cases}$ where $r=3k$, $k\in I$. Find $\displaystyle\sqrt{\int_0^{45}f(x)\,dx}$

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:

If $I = \int_3^4 \frac{1}{\sqrt[3]{\ln x}} dx$, then:

$$\int \frac{x^4 - 2}{x^2\sqrt{x^4 + x^2 + 2}} dx =$$

The sum of the series as $n \to \infty$ $\frac{\sqrt{n}}{(3+4\sqrt{n})^2} + \frac{\sqrt{n}}{\sqrt{2}(3\sqrt{2}+4\sqrt{n})^2} + \frac{\sqrt{n}}{\sqrt{3}(3\sqrt{3}+4\sqrt{n})^2} + \ldots + \frac{1}{49n}$ is:

Comment upon the nature of roots of the quadratic equation $x^2 + 2x + k = \int_0^k |1+k| dr$ depending on the value of $k \in \mathbb{R}$.

In which of the following cases the given equations has atleast one root in the indicated interval?

Let $f: \mathbb{R} \to \mathbb{R}$ be a function satisfying $f(x+2y) = f(x)e^{2y} + f(2y)e^x + x^2\left(1-e^{2y}\right) + 4y^2\left(1-e^x\right) + 4xyy$ for all $x, y \in \mathbb{R}$ and $f'(0) = 1$, then:

Let $I = \int_{\pi/4}^{\pi/3} \frac{\sin x}{x} dx$, then $I$ belongs to:

If $I_{m,n} = \int \cos^m x \sin nx dx$, then $7I_{4,3} - 4I_{3,2} =$$

$\lim f(x)$ does not exist when (where $[x]$ denotes the greatest integer less than or equal to $x$)

A function $f(x)$ continuous on $\mathbb{R}$ and periodic with $2\pi$ satisfies $f(x) + (\sin x) f(x + \pi) = \sin^2 x$ then,

Assume that $\lim_{\theta \to 1} f(0)$ exists and $\frac{\theta^2 + 0 - 2}{\theta + 3} \leq \frac{f(0)}{\theta^2} \leq \frac{\theta^2 + 20 - 1}{\theta + 3}$ holds for certain interval containing the point $\theta = -1$ then $\lim_{\theta \to 1} f(0)$ and $\lim_{\theta \to 1} \frac{f(0)}{\theta^2}$ is :

Let $f: R \to R$ be defined by $f(x) = \begin{cases} x + 2x^2 \sin \frac{1}{x} & \text{for } x \neq 0 \\ 0 & \text{for } x = 0 \end{cases}$ then

The value of $\int_1^8 x\sin[x^2 - \pi] dx$, where $[.]$ denotes the greatest integer function is:

If $\int \frac{dx}{x^2\left(x^7 - 6\right)} = A\left[\ln\left(p^9 + 9p^2 - 2p^3 - 18p\right)\right] + c$, then:

Let a function $f$ be defined as $f(x) = \begin{cases} \frac{|x-1|}{x^2+1} & \text{if } x > -1 \\ x^2 & \text{if } x \leq -1 \end{cases}$. Then the number of critical point(s) on the graph of this function is/are:

If $I = \int \frac{\sin x + \sin^3 x}{\cos 2x} dx = P\cos x + Q\ln |f(x)| + R$, then:

If $\int \sqrt{\cos ecx + 1}\,dx = kfog(x) + c$, where $k$ is a real constant, then :

A function is defined as $f(x) = [\tan x] + \sqrt{\tan x - [\tan x]}$ $0 \leq x 2 \\ 5x - 7 & \text{if } x \leq 2 \end{cases}$ then:

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