Integral Calculus Questions and Answers

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

Answer: B

The coordinate of the centroid is \( \bar{x} = \left(\int x\,dA\right)/\left(\int dA\right) \), a weighted mean in which each element of area is weighted by its distance. Averaging the extreme coordinates would locate the centre of the bounding box, which coincides with the centroid only for a symmetric region

Answer: A

With fixed limits the derivative passes inside the integral, so

\[ \frac{d}{dt}\int_{a}^{b} f(x,t)\,dx \;=\; \int_{a}^{b} \frac{\partial f}{\partial t}\,dx. \]

The boundary terms in the last form appear only when the limits themselves depend on \( t \), and differentiating with respect to \( x \) instead would undo the integration

Answer: D

This is the first form of the fundamental theorem: the integral accumulates the function, and differentiating recovers it,

\[ \frac{d}{dx}\int_{0}^{x} f(t)\,dt \;=\; f(x). \]

It is the statement that integration and differentiation are inverse operations, and it is why the accumulated area is an antiderivative for any continuous integrand