Integral Calculus Questions and Answers

Answer: C

Substitution undoes a chain rule, so it wants the inner function's derivative as a factor. Parts undoes a product rule and suits a product of functions of different kinds. Partial fractions splits a rational function into pieces each of which integrates to a logarithm or an inverse tangent

Answer: B

The trapezoidal rule joins two ordinates by a chord and is exact for a straight line. Simpson's one-third fits a parabola through three ordinates but is exact up to a cubic, the cubic error terms cancelling. The three-eighths rule uses four ordinates and a cubic fit, so it too is exact to that degree

Answer: A

The factors must be found before the shape of the decomposition is known, so factorisation comes first. The sum is then written with unknown numerators, whose values follow from comparing coefficients or substituting convenient values. Only with numbers in place is each piece integrated

Answer: D

The first two are standard properties. The third fails: \( \int_{0}^{1} dx/\sqrt{x} = 2 \) converges although the integrand is unbounded at the origin. Convergence turns on how fast the integrand grows, not on whether it stays finite

Answer: C

The first two hold. The third has the recursion inverted: the correct relation is \( \Gamma(n+1) = n\,\Gamma(n) \), so the gamma of the *larger* argument carries the factor. Written the other way it would make the function decrease along the integers instead of growing like the factorials

Answer: B

For the tail at infinity, convergence needs the exponent above one, so only \( x^{-3} \) qualifies; \( x^{-1} \) and \( x^{-1/2} \) both diverge. The integral of \( 1/x \) from zero to one diverges as well, since \( \ln x \to -\infty \) at the lower end, so it fails at the other kind of singularity

Answer: A

Parts rewrites \( \int u\,dv \) as \( uv - \int v\,du \), which is only an improvement if the new integral is simpler; a poor choice of factors makes it worse, which is why ILATE exists. That it comes from the product rule, that the choice matters, and that repetition is often needed are all correct

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 \( \pi/2 \)