Integral Calculus-2 Questions (49)

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

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

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

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:

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:

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:

Which of the following is/are true?

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:

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

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 $k \in \mathbb{R}$.

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

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

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:

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