Integral Calculus Questions and Answers

Answer: C

The integrand is an odd function, \( (-x)^{3} = -x^{3} \), and the interval is symmetric about the origin, so the two halves cancel:

\[ \int_{-a}^{a} x^{3}dx \;=\; \left[\frac{x^{4}}{4}\right]_{-a}^{a} \;=\; \frac{a^{4}}{4} - \frac{a^{4}}{4} \;=\; 0. \]

For an even integrand the halves would add instead, giving twice the integral over half the range

Answer: B

An even function satisfies \( f(-x) = f(x) \), so the graph is a mirror image about the vertical axis and the two halves of the interval contribute equally:

\[ \int_{-a}^{a} f(x)\,dx \;=\; 2\int_{0}^{a} f(x)\,dx. \]

Vanishing is the odd case, and shifting the range to \( [0, 2a] \) samples different values of the function altogether

Answer: A

By definition \( \int_{b}^{a} f = -\int_{a}^{b} f \), which is what makes the additivity rule \( \int_{a}^{b} + \int_{b}^{c} = \int_{a}^{c} \) hold for any arrangement of the three points. The convention is not a separate rule but a consequence of writing the increment of the antiderivative as \( F(b) - F(a) \)

Answer: D

The Riemann definition partitions the interval, forms \( \sum f(x_{i})\,\Delta x_{i} \), and takes the limit as the largest width tends to zero. That the result equals \( F(b) - F(a) \) is the content of the fundamental theorem, a proved connection rather than the definition itself

Answer: C

Evaluate to a finite upper limit and then let it grow:

\[ \int_{1}^{t} \frac{dx}{x^{2}} \;=\; \left[-\frac{1}{x}\right]_{1}^{t} \;=\; 1 - \frac{1}{t} \;\longrightarrow\; 1 \quad (t \to \infty). \]

The tail dies fast enough for the total area to be finite, whereas the same integral with \( 1/x \) in place of \( 1/x^{2} \) diverges logarithmically

Answer: B

For \( p \neq 1 \) the antiderivative is \( x^{1-p}/(1-p) \), which stays finite as \( x \to \infty \) only when \( 1 - p < 0 \). At \( p = 1 \) the antiderivative is \( \ln x \), which diverges. So the boundary case itself fails, and the condition is strict: \( p > 1 \)

Answer: A

The first kind has an infinite limit of integration; the second has a finite interval but an integrand that blows up at an endpoint or an interior point, as \( \int_{0}^{1} dx/\sqrt{x} \) does at the origin. A change of sign raises no difficulty at all, and a removable discontinuity does not affect the value

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