Integral Calculus Questions (265)

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

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

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

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

Let $f: [0, \infty) \to \mathbb{R}$ be a continuous strictly increasing function such that $f'(x) = \int_{0}^{x} tf^2(t)dt$ for every $x \geq 0$, then value of $f(6)$ is _____.

$$\int \frac{x^2 - 1}{\left(x^4 + 3x^2 + 1\right)\tan^{-1}\left(x + \frac{1}{x}\right)} dx =$$

If $I = \int \frac{dx}{(x-2)\left(1+\sqrt{7x-10-x^2}\right)} = f(t) + c$ (Where $t = \sqrt{\frac{5-x}{x-2}}$ and $f(0) = k\ln\frac{3-\sqrt{5}}{3+\sqrt{5}}$ $(k > 0)$ then $k^2$ is equal to ____.

If $\int \frac{\sqrt{2-x-x^2}}{x^2} dx = \frac{A\sqrt{2-x-x^2}}{x} + \frac{B}{4\sqrt{2}} \ln\left|\frac{4-x+4\sqrt{2-x-x^2}}{x}\right| - \sin^{-1}\left(\frac{2x+1}{3}\right) + c$ then $|A+B|$ is equal to ____.

Let $k(x) = \int \frac{(x^2 + 1)dx}{\sqrt[4]{x^4 + 3x + 6}}$ and $k(-1) = \frac{1}{3\sqrt{2}}$, then the value of $k(-2)$ is

Suppose $\begin{vmatrix} f'(x) & f(x) \\ f''(x) & f'(x) \end{vmatrix} = 0$ where $f(x)$ is differentiable function with $f'(x) \neq 0$ & satisfies $f(0) = 1, f'(0) = 2$. If $f(x) = e^{\lambda x} + \mu$ then $\lambda + \mu$ is ____.

Let $P(x)$ be a quadratic polynomial with $P(1)=-1$. If $\displaystyle\int\frac{P(x)\,dx}{(2x-3)^2(3x-2)^2}=\frac{1}{5}\ln|f(x)|+C$ where $f(x)$ is rational and $\lim_{x\to\infty}f(x)=\frac{4}{3}$, then $|f(1)|$ is

If $\int \frac{\sqrt{1+x^{2n}}\ln(1+x^{2n}) - 2n\ln x}{x^{2n+1}} dx = \frac{\alpha p}{\beta n}[1-3\ln p] + c$ where $p = \sqrt{1+\frac{1}{x^{2n}}}, \alpha,\beta \in \mathbb{N}$ then $\alpha + \beta = $

If $\int \frac{2\cos x - \sin x + \lambda}{\cos x + \sin x + 2} dx = A\ln|\cos x + \sin x - 2| + Bx + C$. Then the value of $A + B + |\lambda|$ is

Consider a real valued continuous function $f$ such that $f(x) = \sin x + \int_{-\pi/2}^{\pi/2} (\sin x + tf(t))dt$. If $M$ and $m$ are maximum and minimum value of the function $f$, then the value of $M/m$ is _____.

Let $g(x) = \int \frac{1 + 2\cos x}{(\cos x + 2)^2} dx$ and $g(0) = 0$, then the value of $8g(\pi/2)$ is

If $\int\left[\left(\frac{x^{-6}-64}{4+2x^{-1}+x^{-2}}\right)\left(\frac{x^2}{4-4x^{-1}+x^{-2}}\right) - \frac{4x^2(2x+1)}{(1-2x)}\right] dx$ is equal to $f(x)$ where $f(1) = 2$ then $f(3) = $

To evaluate $\int \frac{dx}{(x-1)\sqrt{-x^2 + 3x - 2}}$, one of the most suitable substitution could be:

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

$\int \frac{(x + \sqrt{1 + x^2})^{15}}{\sqrt{1 + x^2}} dx$ is equal to:

If $\int(\sin(2020x))(\sin^{2018} x) dx$ is equal to $\frac{(\sin(a)).(\sin x)^b}{c} + k$ (where $k$ is integration constant) then $\frac{a+b+c}{3} = $

If $f(x) = \int_0^x [f(t)]^{-1} dt$ and $\int_0^1 [f(x)]^{-1} dx = \sqrt{2}$, then:

If the primitive of the function $f(x) = \frac{x^{2009}}{(1 + x^2)^{1006}}$ w.r.t $x$ is equal to $\frac{1}{n}\left(\frac{x^2}{1 + x^2}\right)^m + C$ then find the value of $(m + n)$ (where $m,n \in \mathbb{N}$)

If $\displaystyle\int e^{\frac{1}{2}\left(x^2+\frac{1}{x^2}\right)}\cdot\frac{x^4+x^2-1}{x^2}\,dx = f(x)+c$, then $\bigl(f(\sqrt{2})\bigr)^4$ is

$\int\dfrac{x(x\tan^{-1}x+(\ln x)(\ln(\ln x)))+\tan^{-1}x}{(x^3+x)\ln x}=f(x)+c$ where $f(e)=0$. Then $\left[\lim_{x\to 1^+}\dfrac{f(x)}{\tan\left(\frac{\pi x}{2}\right)}+\frac{11}{10}\right]=$

$\displaystyle\int\dfrac{\cos 2x-1}{\cos 2x+1}dx$

Area bounded by $y=\lfloor e^x\rfloor$, $y=\lfloor e^{-x}\rfloor$, and $|x|=1$ is

If $\displaystyle\int e^{\frac{1}{2}\left(x^2+\frac{1}{x^2}\right)}\cdot\frac{x^4+x^2-1}{x^2}\,dx = f(x)+c$, then $\bigl(f(\sqrt{2})\bigr)^4$ is

Area enclosed by the curve $y=\dfrac{1}{1+x^2}$ and the $x$-axis from $x=-\infty$ to $x=+\infty$ is

$$\int \frac{(ax^2 - b)dx}{\sqrt{c^2x^2 - (ax^2 + b)^2}} =$$

Consider curves $C_1: y^2-x=0$; $C_2: y-x^2=0$; $0\leq x\leq\frac{\sqrt{3}}{2}$ and $C_3: y=f(x)$; $f(x)<0$ $\forall x\in\left(0,\frac{\sqrt{3}}{2}\right)$. From any point $P$ on $C_2$, lines parallel to coordinate axes intersect $C_1$ at $Q$ and $C_3$ at $R$. If area of region $OPRO$ is twice the area of region $OPQO$ (O = origin), then $\left|32f\!\left(\frac{1}{2}\right)\right|$ is

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

If $\int \frac{dx}{\sqrt[3]{(x - 1)^3(x + 2)^2}} = k\sqrt[3]{\frac{x - 1}{x + 2}} + c$, then $3k$ is equal to

If $I=\int(x^2+1)\left((x+1)e^x\right)^2dx=A(f(x))^2+C$, where $C$ is constant and $f(-1)=\dfrac{2}{e}$, then $2A+f(0)$ is

Let $f(x) = x^3 - \frac{3x^2}{2} + x + \frac{1}{4}$. Then the value of $\left[\int_{1/4}^{3/4} f(f(x))dx\right]^{-1}$ is _____.

Which must be true for: I) For $x\in\mathbb{R}$, let $\tan^{-1}x\in(-\pi/2,\pi/2)$. Then minimum value of $f(x)=\int_0^{x\tan^{-1}x}\frac{e^{t-\cos t}}{1+t^{2023}}dt$ is zero. II) If $f(x)$ is periodic with period $T$, then $\int_{a+nT}^{b+nT}f(x)dx=\int_a^b f(x)dx$ for $n\in\mathbb{Z}$. III) If $\psi(x)\le\phi(x)$ for $a\le x\le b$, then $\int_a^b|\psi(x)|dx\le\int_a^b|\phi(x)|dx$.

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 $f(x) = x + \sin x$. Suppose $g$ denotes the inverse function of $f$. If the value of $g'\left(\frac{\pi}{4} + \frac{1}{\sqrt{2}}\right)$ is $l$ then $2l = $

Let $e$ be the eccentricity of a hyperbola and $f(e)$ be the eccentricity of its conjugate hyperbola. Then $\int(f(e)+f(f(e)))de=g(e)$ and $g(\sqrt{2})=2$, then $g(e)=$

Let $I_1=\int_0^x e^{tx}\cdot e^{-t^2}\,dt$ and $I_2=\int_0^x e^{-t^2/4}\,dt$ where $x>0$. Then the value of $\dfrac{I_1}{I_2}$ is

$\displaystyle\int_1^{81}\dfrac{dx}{\sqrt{x}+\sqrt[4]{x}}$

Consider: I) If $f(x)$ is bounded for $x\in[a,b]$, then area bounded by $y=f(x)$, $x$-axis, $x=a$ and $x=b$ is $\int_a^b f(x)dx$. II) If $f(x)$ is bounded and differentiable on $[a,b]$, then the area between $y=f(x)$ and $y=f^{-1}(x)$ is equal to double the value of $\int_a^b|f(x)-x|dx$.

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

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

$\int_{-5}^{5}\dfrac{x^{10}}{x^{10}+(5-x)^{10}}\cdot\dfrac{dx}{x^2+16}$

Let $y=f(x)$ be a differentiable function satisfying $\int_2^x f(t)\,dt+2=\dfrac{x^2}{2}+\int_x^2 t^2 f(t)\,dt$. Then $\int_{-\pi/4}^{\pi/4}\dfrac{f(x)+x^9-x^3+x+1}{\cos^2 x}\,dx=$

$\displaystyle\int_0^{10}[x^2]dx$ (where $[.]$ is GIF) equals

$\displaystyle\int\sqrt{\dfrac{x}{1-x^3}}dx=$

The area enclosed between the curves $y=ax^2$ and $x=ay^2$ ($a>0$) is 1 sq. unit. Then the value of $a$ is

$\lim_{n\to\infty}\left(\dfrac{n^2}{n^2+1}\cdot\dfrac{n^2}{n^2+4}\cdots\dfrac{n^2}{n^2+n^2}\right)^{1/n}$

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