Calculus Questions (384)

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

Water is filled at rate $\pi$ cm$^3$/s in right circular conical vessel (vertex up) of height 5 cm and diameter 8 cm. When water height is 3 cm, rate of increase of wet conical surface area is (cm$^2$/s)

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

Which of the following functions is differentiable at $x=0$?

For $x>0$, $y>0$ with $x^2y^3=6$, find $\min(3x+4y)$.

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

If $f(x)=\displaystyle\int x^{1/3}(x^{2/3}+x^{1/3}+1)(2x^{2/3}+3x^{1/3}+6)^3\,dx$, $f(0)=0$, then $\dfrac{16}{11}f(1)$ is equal to

For positive integer $n$, let $I_n=\displaystyle\int_{-\pi}^{\pi}\!\left(\frac{\pi}{2}-|x|\right)\cos nx\,dx$. Find $[I_1+I_2+I_3+I_4]$ (GIF).

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 $f(x)=\displaystyle\int x^{1/3}(x^{2/3}+x^{1/3}+1)(2x^{2/3}+3x^{1/3}+6)^3\,dx$, $f(0)=0$, then $\dfrac{16}{11}f(1)$ is equal to

Let $f(x)=\displaystyle\int e^x(x-1)(x-2)\,dx$. Then $f(x)$ decreases in the interval

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

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

$\displaystyle\int_0^{\pi/2}\dfrac{dx}{a^2\cos^2x+b^2\sin^2x}=$

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

$f(x)=\min(x-[x],1-x+[x])$ for $x\in\mathbb{R}$. Then $\displaystyle 2\int_0^2 f(x)dx=$

$\displaystyle\int_{\pi/4}^{3\pi/4}\dfrac{x}{1+\sin x}dx$

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

If $\displaystyle\int (x^2+1)(x+1)^2 e^x\,dx = A(f(x))^2 + C$ and $f(-1) = \frac{2}{e}$, then $2A + f(0)$ equals

The area bounded by the $x$-axis, part of the curve $y=1+x^{-2}$ and the ordinates $x=1$, $x=2$ is divided into equal parts by the ordinate at $x=a$ such that $a=\dfrac{3+\sqrt{p}}{8}$. The value of $p$ is

Let $f:(0,\infty)\to\mathbb{R}$ be a differentiable function such that $f'(x)=2-\dfrac{f(x)}{x}$ for all $x\in(0,\infty)$ and $f(1)\ne 1$. 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 $\displaystyle\sqrt{\int_0^{45}f(x)\,dx}$