The region \(R\) in the first quadrant enclosed by the lines \(x = 0\), and \(y = 5\), and the graph of \(y = x^2 + 1\). The volume of the solid generated when \(R\) is revolved about the y-axis is:

Let \(f(x) = \begin{cases} \sin x & \quad \text{if } x < 0 \\ x^2 & \quad \text{if } 0 \leq x < 1 \\ 2 - x & \quad \text{if } 1 \leq x < 2 \\ x - 3 & \quad \text{if } x \geq 2 \end{cases}\). For what values of \(x\) is \(f\) discontinuous?

What is \(\operatorname*{lim}\limits_{x \rightarrow \infty} \dfrac{x^2 - 4}{2 + x - 4 x^2}\)?

If \(r\) is positive and increasing, for what value of \(r\) is the rate of increase of \(r^3\) twelve times that of \(r\)?

The area of the region in the first quadrant between the graph of \(y = x \sqrt{4 - x^2}\) and the x-axis is:

\(\operatorname*{lim}\limits_{t \rightarrow 0} \dfrac{\sin(2 t)}{8 t} =\)

\(\displaystyle\int_{0}^{4} \sqrt{16 - x^2} d x =\)

Evaluate \(\displaystyle\int_{1}^{2} \left(12 + \dfrac{8}{x^3}\right) d x\)

Find the area enclosed by \(y = x - 2\) and \(y = -x^2 + 4 x - 2\).

\(\operatorname*{lim}\limits_{x \rightarrow 0} \dfrac{x^3 + x^2 - 2 x}{x^3 - x} =\)

The region enclosed by the line \(x + y = 1\) and the coordinate axes is rotated about the line \(y = -1\). The volume of the solid is:

\(y = \sin x + \cos x\) is a solution of:
I. \(y + \dfrac{d y}{d x} = 2 \sin x\)
II. \(y + \dfrac{d y}{d x} = 2 \cos x\)
III. \(\dfrac{d y}{d x} - y = -2 \sin x\)

\(f(x) = n + e^{2 x}\) for \(x \geq 0\) and \(f(x) = 4 + m x\) for \(x < 0\) is differentiable at \(x = 0\). \(f(n - m) = ?\)

If \(\dfrac{d y}{d x} = \sin x^3\), \(\dfrac{d^2 y}{d x^2} =\)

Which is an antiderivative of \(3^x\)?

If \(f(x) = e^x\), then \(\ln[f'(2)] =\)

If \(y^2 - 2 x y = 16\), then \(\dfrac{d y}{d x} =\)

A person 2 meters tall walks directly away from a streetlight that is 8 meters above the ground. If the person is walking at a constant rate and the person's shadow is lengthening at a rate of \(\dfrac{4}{9}\) meter per second, at what rate, in meters per second, is the person walking?

Let \(f\) and \(g\) be differentiable functions. If \(g\) is the inverse function of \(f\) and if \(g(-2) = 5\) and \(f'(5) = -\dfrac{1}{2}\), then \(g'(-2) =\)