Calculus Questions

What is the area bound by the curve $xy=m^2$, $x$ -axis and the ordinates $x=p$ and $x=q$ where $p>q>0$?

${\int }_0^{\pi /2}\log (\tan x)dx$ is equal to:

If at a point on the curve $y=f(x)$ the slope of the line tangent to the curve is equal to $2xy$, then the function $f$ is:

The normal to the curve $y(x-2)(x-3)=x+6$ at the point where the curve intersects the $y$ -axis passes through the point:

If the tangent to the curve $y=x+\sin y$ at a point $(a,b)$ is parallel to the line joining $(0,3/2)$ and $(1/2,2)$, then:

If $f(x)$ is an increasing function and $g(x)$ is a decreasing function, then which of the following is correct for the real values $p$ and $q$? Given that for these values of $x$, both functions are defined.

Find the area bounded by the curves $y\ge x^2$ and $y=|x|$.

The function $f(x)=x^3-3\log x$ has:

Derivative of $x^2$ w.r.t. $x^3$ is:

If $f(x)=a{\tan }^{-1}x+2b\log (1+x)+x+1$ has critical points at $x=0$ and $x=2$, then the values of the constants $a$ and $b$ are respectively:

Let the function $f:[-7,0]\to \mathbb{R}$ be continuous on $[-7,0]$ and differentiable on $(-7,0)$. If $f(-7)=-3$ and $f^{\prime }(x)\le 2$ for all $x\in (-7,0)$, then for all such functions $f$, $f(-1)+f(0)$ lies in the interval:

The function $f(x)=x^2-4x$, $x\in [0,4]$ attains minimum value at:

Find the area between the curve $y=\sin x$ and the $x$ -axis from $x=-\pi /3$ to $x=\pi /3$.

The area under the curve $y=x^4$ and the lines $x=1,x=5$ and the $x$ -axis is:

The value of ${\int }_0^1|5x-3|dx$ is: