Integral Calculus Questions and Answers

Answer: A

Raising the index and dividing by the new index reverses the power rule:

\[ \frac{d}{dx}\left(\frac{x^{n+1}}{n+1}\right) \;=\; \frac{(n+1)x^{n}}{n+1} \;=\; x^{n}. \]

The exclusion \( n = -1 \) is needed because the divisor would vanish; that case integrates to \( \ln|x| \) instead. Lowering the index is differentiation, not integration

Answer: D

Integrate and evaluate at the two limits:

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

The value two belongs to the integral over the full half-period from zero to \( \pi \), which covers twice the area of this quarter-period

Answer: C

The reciprocal integrates to the natural logarithm:

\[ \int_{1}^{e} \frac{dx}{x} \;=\; \left[\ln x\right]_{1}^{e} \;=\; \ln e - \ln 1 \;=\; 1 - 0 \;=\; 1. \]

This is the defining property of \( e \): it is the upper limit at which the area under the reciprocal curve, measured from one, first reaches unity

Answer: B

Substituting \( x = a\tan\theta \) gives \( dx = a\sec^{2}\theta\,d\theta \) and turns the denominator into \( a^{2}\sec^{2}\theta \), leaving

\[ \int \frac{a\sec^{2}\theta\,d\theta}{a^{2}\sec^{2}\theta} \;=\; \frac{\theta}{a} \;=\; \frac{1}{a}\tan^{-1}\frac{x}{a} + C. \]

The inverse sine belongs to \( 1/\sqrt{a^{2}-x^{2}} \), and the logarithm to \( 1/(a^{2}-x^{2}) \)

Answer: A

Integrating \( (uv)' = u'v + uv' \) and rearranging gives \( \int u\,dv = uv - \int v\,du \). So the method undoes a product rule. The chain rule's counterpart is substitution, and the quotient rule has no integral method of its own, quotients being handled by partial fractions or substitution instead

Answer: D

Take \( u = x \) and \( dv = e^{x}dx \), so that differentiating the polynomial factor removes it:

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

Choosing the factors the other way round would raise the power of \( x \) and make the integral harder rather than easier

Answer: C

ILATE ranks inverse trigonometric, logarithmic, algebraic, trigonometric and exponential, and the earlier of the two factors is taken as \( u \). The point is that inverse and logarithmic functions simplify sharply on differentiating but are awkward to integrate, so they belong in the part that gets differentiated

Answer: B

Take \( u = \ln x \) and \( dv = dx \), since the logarithm has no elementary antiderivative of its own:

\[ \int_{1}^{e}\ln x\,dx \;=\; \left[x\ln x\right]_{1}^{e} - \int_{1}^{e} dx \;=\; e - (e - 1) \;=\; 1. \]

The general antiderivative is \( x\ln x - x \), which is worth remembering because the logarithm arises so often as one factor of a product

Answer: A

Write \( 1 = A(x-2) + B(x-1) \). Setting \( x = 1 \) gives \( A = -1 \), and setting \( x = 2 \) gives \( B = 1 \), so

\[ \frac{1}{(x-1)(x-2)} \;=\; \frac{-1}{x-1} + \frac{1}{x-2}. \]

A quick check at \( x = 0 \) gives \( 1/2 \) on both sides, so the signs are the right way round

Answer: D

A factor \( (x-a)^{2} \) needs \( A/(x-a) + B/(x-a)^{2} \), two constants for the two degrees of freedom lost. One term alone cannot represent a general numerator of degree one over the square. A linear numerator over a squared factor is the form reserved for an irreducible *quadratic* factor