Integral Calculus Questions and Answers

Answer: D

Integrating \( \Gamma(n+1) = \int_{0}^{\infty} x^{n}e^{-x}dx \) by parts gives

\[ \Gamma(n+1) \;=\; \left[-x^{n}e^{-x}\right]_{0}^{\infty} + n\int_{0}^{\infty} x^{n-1}e^{-x}dx \;=\; n\,\Gamma(n). \]

With \( \Gamma(1) = 1 \) this recursion gives \( \Gamma(n+1) = n! \) for positive integers, so the function interpolates the factorials

Answer: C

Substituting \( x = t^{2} \) turns the defining integral into a Gaussian:

\[ \Gamma\!\left(\tfrac{1}{2}\right) \;=\; \int_{0}^{\infty} x^{-1/2}e^{-x}dx \;=\; 2\int_{0}^{\infty} e^{-t^{2}}dt \;=\; 2 \cdot \frac{\sqrt{\pi}}{2} \;=\; \sqrt{\pi}. \]

It is the one non-integer argument whose value is met constantly, appearing wherever a Gaussian is integrated

Answer: B

The relation is

\[ B(m,n) \;=\; \int_{0}^{1} x^{m-1}(1-x)^{n-1}dx \;=\; \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}, \]

which is what makes the beta integral computable in closed form. The reciprocal arrangement is the binomial coefficient's shape, not the beta function's

Answer: A

An element of arc satisfies \( ds^{2} = dx^{2} + dy^{2} \), so

\[ ds \;=\; \sqrt{1 + \left(\frac{dy}{dx}\right)^{2}}\,dx, \]

and integrating gives the length. The slope, not the function value or the second derivative, is what enters, since it is the slope that fixes how much longer the curve is than its projection

Answer: D

On this interval the line lies above the parabola, so integrate the difference:

\[ \int_{0}^{1}\left(x - x^{2}\right)dx \;=\; \left[\frac{x^{2}}{2} - \frac{x^{3}}{3}\right]_{0}^{1} \;=\; \frac{1}{2} - \frac{1}{3} \;=\; \frac{1}{6}. \]

The separate areas are \( 1/2 \) and \( 1/3 \), and taking either alone is the usual slip

Answer: C

A thin slice perpendicular to the axis is a disc of radius \( y \) and thickness \( dx \), so its volume is \( \pi y^{2}dx \) and

\[ V \;=\; \pi\int_{a}^{b} y^{2}\,dx. \]

The form \( 2\pi\int y\,dx \) is the surface of revolution's shape rather than the volume's, carrying one power of \( y \) because a surface has one dimension fewer

Answer: B

Apply the disc formula with \( y = x \):

\[ V \;=\; \pi\int_{0}^{3} x^{2}\,dx \;=\; \pi\left[\frac{x^{3}}{3}\right]_{0}^{3} \;=\; 9\pi. \]

The solid is a cone of radius three and height three, and the elementary formula \( \tfrac{1}{3}\pi r^{2}h = \tfrac{1}{3}\pi(9)(3) = 9\pi \) agrees

Answer: A

The mean is the total divided by the length of the interval, so that a constant function returns its own value:

\[ \bar{f} \;=\; \frac{1}{b-a}\int_{a}^{b} f(x)\,dx. \]

Averaging the endpoint values ignores everything between them, and the integral of \( f^{2} \) leads to the root-mean-square instead, which needs a square root as well

Answer: D

Square, average over the period and take the root, using \( \sin^{2}\theta = \tfrac{1}{2}(1 - \cos 2\theta) \) whose mean is \( 1/2 \):

\[ \text{r.m.s.} \;=\; \sqrt{\frac{1}{2}} \;=\; \frac{1}{\sqrt{2}} \;\approx\; 0.707. \]

The value \( 2/\pi \approx 0.637 \) is the mean of the *rectified* sine over a half cycle, a different average often confused with this one

Answer: C

The rule fits a parabola through each group of three consecutive ordinates, consuming two subintervals at a time, so the total must be even. The three-eighths rule fits a cubic through four ordinates and needs a multiple of three subintervals. Only the trapezoidal rule accepts any number