Integral Calculus Questions and Answers

Answer: B

Each strip contributes the area of a trapezium, \( \tfrac{h}{2}(y_{i} + y_{i+1}) \). Summing, every interior ordinate is counted twice and the two end ordinates once:

\[ I \;\approx\; \frac{h}{2}\left[y_{0} + y_{n} + 2\sum_{i=1}^{n-1} y_{i}\right]. \]

The divisor three and the weight four belong to Simpson's rule

Answer: A

With \( h = 1 \) the ordinates are \( y_{0} = 0 \), \( y_{1} = 1 \), \( y_{2} = 4 \), so

\[ I \;\approx\; \frac{1}{2}\left[0 + 4 + 2(1)\right] \;=\; 3. \]

The exact value is \( 8/3 \approx 2.667 \), so the rule overestimates by a third. It always does for a convex integrand, the chords lying above the curve

Answer: D

With the same ordinates and \( h = 1 \),

\[ I \;\approx\; \frac{1}{3}\left[0 + 4(1) + 4\right] \;=\; \frac{8}{3}. \]

Simpson's rule fits a parabola, and the integrand *is* a parabola, so the fit is exact. The rule is in fact exact for any cubic as well, which is why its error term involves the fourth derivative

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