Further Mathematics

1. 1.

   Find integral ln x dx for x > 0 using integration by parts.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Treat ln x as a product with 1, which can be integrated easily.
   2. Differentiate x ln x - x to check: ln x + 1 - 1 = ln x.

   Marking points

   • Choose u = ln x, dv = dx, so du = dx/x and v = x.
   • The integral becomes x ln x - integral 1 dx.
   • Answer: x ln x - x + C.

   Examiner tip: Do not apply a power rule to ln x; treat it as ln x times 1 for integration by parts.

2. 2.

   Evaluate integral from 0 to 1 of x/(1 + x^2) dx exactly.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The numerator is half the derivative of the denominator, indicating a logarithm.
   2. Evaluate ln u/2 between 1 and 2; ln 1 contributes zero.

   Marking points

   • Set u = 1 + x^2; du = 2x dx.
   • New limits are 1 and 2; integrate (1/2)/u.
   • The answer is ln(2)/2.

   Examiner tip: Keep the factor 1/2 when replacing x dx.

3. 3.

   Determine for which real p the improper integral from 1 to infinity of x^(-p) dx converges, and give its value when it does.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Infinity is a limiting process, not a number substituted into an antiderivative.
   2. Even though x^(-p) tends to zero for p > 0, its accumulated area need not be finite.

   Marking points

   • For p != 1, integrate to x^(1 - p)/(1 - p) and take an upper-limit R.
   • The R term tends to zero only if p > 1, giving 1/(p - 1).
   • At p = 1 the integral is ln R and diverges; p < 1 also diverges.

   Examiner tip: Handle p = 1 separately because the power antiderivative has a zero denominator.

4. 4.

   Evaluate integral from 0 to 1 of x^2 e^x dx exactly.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Differentiate the polynomial and integrate the exponential on each pass to lower the degree.
   2. The endpoint values are e and 2; the nonzero lower value must be subtracted.

   Marking points

   • First integration by parts gives x^2 e^x - 2 integral x e^x dx.
   • A second gives antiderivative e^x(x^2 - 2x + 2).
   • The definite integral is e - 2.

   Examiner tip: A lower limit of zero does not make an exponential antiderivative zero.

5. 5.

   Let I_n = integral from 0 to pi/2 of sin^n x dx. For n >= 2 derive I_n = (n - 1)I_(n - 2)/n, and find I_4 exactly.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The differentiated power contributes (n - 1)sin^(n - 2)x cos x; multiplying by the integrated -cos x leaves cos^2 x after the parts subtraction.
   2. Step down twice: I_4 = (3/4)I_2 and I_2 = (1/2)I_0.

   Marking points

   • Integrate sin^(n - 1)x times sin x by parts; the boundary term -sin^(n - 1)x cos x vanishes.
   • I_n = (n - 1)(I_(n - 2) - I_n), using cos^2 x = 1 - sin^2 x.
   • Rearrange to the recurrence; I_0 = pi/2, so I_4 = 3pi/16.

   Examiner tip: Move the I_n term to the left before dividing by n.

6. 6.

   A learner claims integral from -1 to 1 of 1/x^2 dx equals -2 by using -1/x at the endpoints. Explain the error and determine convergence.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Split at zero and require both one-sided integrals to be finite independently.
   2. The integrand is positive on both intervals; a negative area is already a warning that a theorem's conditions failed.

   Marking points

   • There is an interior singularity at x = 0, so the ordinary endpoint rule is invalid.
   • For epsilon > 0 the integral from epsilon to 1 is 1/epsilon - 1, which tends to positive infinity.
   • The left side also diverges positively; the improper integral does not converge.

   Examiner tip: Never integrate across a pole without defining one-sided limits.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.