Integral Calculus Questions and Answers

Answer: D

The fundamental theorem guarantees that \( \int_{a}^{x} f(t)\,dt \) is an antiderivative of a continuous \( f \), so one always exists. Whether it can be written with the elementary functions is a separate question, and \( e^{-x^{2}} \) is the standard case where it cannot

Answer: C

Although the rule interpolates with a parabola, its error term involves the fourth derivative, which vanishes for a cubic. So the rule integrates cubics exactly, gaining a degree over what the fitting polynomial suggests. The multiple-of-three condition belongs to the three-eighths rule

Answer: B

Use \( \sin^{2}x = \tfrac{1}{2}(1 - \cos 2x) \), whose second term integrates to zero over the quarter period:

\[ \int_{0}^{\pi/2}\sin^{2}x\,dx \;=\; \frac{1}{2}\left[x - \frac{\sin 2x}{2}\right]_{0}^{\pi/2} \;=\; \frac{1}{2}\cdot\frac{\pi}{2} \;=\; \frac{\pi}{4}. \]

By symmetry \( \cos^{2}x \) gives the same value, and the two together sum to \( \pi/2 \)

Answer: A

Split off one factor and substitute \( u = \cos x \), so \( du = -\sin x\,dx \):

\[ \int_{0}^{\pi/2}\sin^{3}x\,dx \;=\; \int_{0}^{1}\left(1 - u^{2}\right)du \;=\; \left[u - \frac{u^{3}}{3}\right]_{0}^{1} \;=\; \frac{2}{3}. \]

Odd powers of the sine give rational values this way, while even powers bring in \( \pi \) because their average over the quarter period is not rational

Answer: D

A reduction formula such as \( I_{n} = \tfrac{n-1}{n} I_{n-2} \) steps the index down until a base case is reached, so a whole family is handled by one relation plus one or two starting values. It is a recursion, not a closed form, and nothing about it is numerical or a differentiation

Answer: C

With \( u = x^{2} \) we get \( du = 2x\,dx \), so \( x\,dx = \tfrac{1}{2}du \), and the limits \( x = 0, 1 \) become \( u = 0, 1 \):

\[ \int_{0}^{1} x e^{x^{2}}dx \;=\; \frac{1}{2}\int_{0}^{1} e^{u}du \;=\; \frac{e-1}{2}. \]

The lone factor of \( x \) is exactly what the substitution absorbs, which is why it works here and not for \( \int e^{x^{2}}dx \) alone

Answer: B

Write the tangent as a quotient and substitute \( u = \cos x \), with \( du = -\sin x\,dx \):

\[ \int \frac{\sin x}{\cos x}dx \;=\; -\int \frac{du}{u} \;=\; -\ln|u| + C \;=\; -\ln|\cos x| + C. \]

The same antiderivative is often written \( \ln|\sec x| + C \), the two differing only by the sign taken inside the logarithm

Answer: A

An element of the region is a thin sector of radius \( r \) and angle \( d\theta \), whose area is \( \tfrac{1}{2}r^{2}d\theta \), so

\[ A \;=\; \frac{1}{2}\int_{\alpha}^{\beta} r^{2}\,d\theta. \]

Setting \( r = a \) and integrating over a full turn returns \( \pi a^{2} \), which is the check worth doing to fix the half in memory

Answer: D

Apply the polar formula over a complete revolution:

\[ A \;=\; \frac{1}{2}\int_{0}^{2\pi} 2^{2}\,d\theta \;=\; \frac{1}{2}(4)(2\pi) \;=\; 4\pi. \]

This agrees with \( \pi r^{2} = \pi(4) \). Dropping the half would give \( 8\pi \), and omitting the square would give \( 2\pi \), which is the circumference instead

Answer: C

A band of the surface has circumference \( 2\pi y \) and width equal to the arc element \( ds \), not \( dx \), so

\[ S \;=\; 2\pi\int y\,ds \;=\; 2\pi\int y\sqrt{1 + y'^{2}}\,dx. \]

Using \( dx \) in place of \( ds \) is the standard error: it measures the band along the axis rather than along the slanting surface, and understates the area