OxfordAQA International AS and A-level Mathematics: Integration by parts, partial fractions and volumes of revolution (9660) – Revision Notes

These notes condense section P2.7 (the second part of Integration) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2: integration by parts, partial fractions and volumes of revolution. The content is part of Unit P2 (Pure Maths) in the International A-level and is not examined at International AS. For full explanations, read the study guide, then test yourself with the practice questions.

Other links: OxfordAQA A-level Mathematics course page, the printable checklist and the free 10-minute diagnostics. Inspection and substitution are taught on a separate page and are not repeated here.

Key results

Topic	Result
Integration by parts	∫ u v′ dx = uv − ∫ u′ v dx, with v′ = dv/dx
Origin	The product rule, (uv)′ = u v′ + u′ v, integrated and rearranged
Linear denominator	k/(ax + b) integrates to (k/a) ln|ax + b| + c
Volume about the x-axis	V = π ∫ y² dx between x-limits
Volume about the y-axis	V = π ∫ x² dy between y-limits

Integration by parts: method

1. Decide which factor is u and which is dv/dx.
2. Differentiate u to get du/dx. Integrate dv/dx to get v. Leave out the constant in v.
3. Write uv − ∫ u′ v dx.
4. Integrate what remains. Add + c at the end.
5. For a definite integral, apply the limits to uv and to the new integral, using the same limits for both.

Choosing u

• x with an exponential: u = x.
• x with a sine or cosine: u = x.
• A power of x with ln x: u = ln x.
• x² with an exponential or trigonometric function: u = x² on the first pass, then u = x on the second.

Worked reminder. ∫ x e^(4x) dx with u = x and v = ¼ e^(4x):

= ¼ x e^(4x) − ¼ ∫ e^(4x) dx = ¼ x e^(4x) − (1/16) e^(4x) + c

Worked reminder. ∫ x ln x dx with u = ln x and v = x²/2:

= (x²/2) ln x − ∫ (x²/2)(1/x) dx = (x²/2) ln x − x²/4 + c

Worked reminder (definite, with limits). ∫ from 0 to 1 of x e^(3x) dx:

u = x, v = (1/3) e^(3x)
= [x e^(3x)/3] from 0 to 1 − (1/3) ∫ from 0 to 1 of e^(3x) dx
= e³/3 − (1/9)(e³ − 1) = (2e³ + 1)/9

Repeated use. If the polynomial factor is x², differentiate it twice. Each pass lowers the power by one, and the last pass leaves a pure exponential or trigonometric integral. Two passes give three terms in the answer.

Check by differentiation. Differentiate your answer. If you get back the original integrand, the result is right. This is quick and catches sign errors.

Partial fractions: method

1. Compare degrees. If the top is not lower than the bottom, divide first.
2. Write the proper fraction as A/(first factor) + B/(second factor).
3. Multiply through by the denominator.
4. Substitute the value of x that makes each factor zero to find A and B.
5. Check with one more value of x.
6. Integrate each term, remembering the factor 1/a for each bracket.
7. Combine logarithms only if the question asks for a single logarithm.

Worked reminder. 7/((x − 2)(x + 5)):

7 = A(x + 5) + B(x − 2)
x = 2: 7 = 7A, A = 1.   x = −5: 7 = −7B, B = −1.
7/((x − 2)(x + 5)) = 1/(x − 2) − 1/(x + 5)
∫ = ln|x − 2| − ln|x + 5| + c = ln|(x − 2)/(x + 5)| + c

Definite version. ∫ from 3 to 6 of 7/((x − 2)(x + 5)) dx:

[ln|x − 2| − ln|x + 5|] from 3 to 6
= (ln 4 − ln 11) − (ln 1 − ln 8) = ln 4 + ln 8 − ln 11 = ln(32/11)

Form of the denominator. Factors such as (3x − 4) are fine. After splitting, 1/(3x − 4) integrates to (1/3) ln|3x − 4|. The integral of a sum of fractions is the sum of the integrals, so treat each term on its own.

Improper fractions. x²/((x − 1)(x + 1)): the denominator is x² − 1, so x² = (x² − 1) + 1 and

x²/((x − 1)(x + 1)) = 1 + 1/((x − 1)(x + 1)) = 1 + (½)/(x − 1) − (½)/(x + 1)

Volumes of revolution: method

1. Identify the axis of rotation. Only the x-axis and the y-axis are used.
2. Write the right expression: y² in terms of x for the x-axis, or x² in terms of y for the y-axis.
3. Take the limits in the same variable as the integration: x-limits for dx, y-limits for dy.
4. Leave π outside the integral until the end.
5. Give the exact answer if the question asks for it. Otherwise round sensibly.

Worked reminder (x-axis). y = 2x from x = 0 to x = 3:

V = π ∫ 4x² dx = π [4x³/3] from 0 to 3 = 36π

This matches a cone of radius 6 and height 3: (1/3)π(6²)(3) = 36π.

Worked reminder (y-axis). y = x² from y = 0 to y = 4, so x² = y:

V = π ∫ y dy = π [y²/2] from 0 to 4 = 8π

Worked reminder (parts inside a volume). y = √(x) e^x, from x = 0 to x = 1, about the x-axis. Then y² = x e^(2x), and

V = π ∫ x e^(2x) dx = π (e² + 1)/4

using the result in question 11 of the self-test.

Worked reminder (y-axis, curve given as y = f(x)). For y = ½x² from y = 2 to y = 8, write x² = 2y. Then V = π ∫ from 2 to 8 of 2y dy = π [y²] from 2 to 8 = 60π.

Must-know distinctions

• dx with x-limits, dy with y-limits. Mixing them is the commonest volume error.
• Square the function, not the integral. π ∫ y² dx is not π (∫ y dx)².
• Parts versus substitution. Use parts for a product of two different types of function, such as x and e^(kx). Use substitution when one factor is the derivative of the inside of the other.
• Proper versus improper. Divide first when the top is not of lower degree than the bottom.
• Logarithm as u. When x^n multiplies ln x, the logarithm must be u, because you cannot yet integrate ln x directly. Differentiating it gives 1/x, which cancels against a power of x and leaves a simple power to integrate.
• Region versus solid. The region is flat. The solid is what you get after a full turn of 360°. The formula sums thin discs, so the radius of each disc is the distance from the curve to the axis.
• ln|…| versus ln(…). The modulus is needed in indefinite integrals, because the bracket may be negative.

Quick self-test

1. Find ∫ x e^(−x) dx.
2. Find ∫ x sin 5x dx.
3. Find ∫ x⁴ ln x dx.
4. Write 7/((x − 2)(x + 5)) in partial fractions.
5. Find ∫ (x + 7)/((x + 1)(x + 3)) dx.
6. Find ∫ 4/(3x − 2) dx.
7. Find the volume formed when y = 2x, 0 ≤ x ≤ 3, is rotated about the x-axis.
8. Find the volume formed when y = x², 0 ≤ y ≤ 4, is rotated about the y-axis (x ≥ 0).
9. Give the exact volume of the solid formed by rotating the region under y = eˣ, for 0 ≤ x ≤ 1, about the x-axis.
10. Split x²/((x − 1)(x + 1)) into a constant and partial fractions.
11. Evaluate ∫ from 0 to 1 of x e^(2x) dx.
12. Differentiate x ln x, and use the result to find ∫ ln x dx.

Answers

1. −(x + 1) e^(−x) + c
2. −(x cos 5x)/5 + (sin 5x)/25 + c
3. (x⁵ ln x)/5 − x⁵/25 + c
4. 1/(x − 2) − 1/(x + 5)
5. Partial fractions 3/(x + 1) − 2/(x + 3), so 3 ln|x + 1| − 2 ln|x + 3| + c
6. (4/3) ln|3x − 2| + c
7. 36π
8. 8π
9. π (e² − 1)/2
10. 1 + (½)/(x − 1) − (½)/(x + 1)
11. (e² + 1)/4
12. ln x + 1. Since ∫ (ln x + 1) dx = x ln x + c, we get ∫ ln x dx = x ln x − x + c.

Where marks are usually lost

• Dropping the minus sign before the new integral, especially when v is negative.
• Differentiating the wrong factor, so the new integral is as hard as the original.
• Forgetting a second pass on an x² integrand, or forgetting to bracket the second result.
• Omitting the 1/a when integrating 1/(ax + b).
• Matching A and B without dividing first on an improper fraction.
• Applying limits to only part of the parts result, instead of the whole uv term and the remaining integral.
• Writing the volume formula with y instead of y², or integrating y² dx with y-limits.
• Not showing the line where you substitute the limits, which makes an arithmetic slip hard to trace.
• Rounding too early when an exact answer is asked for, or leaving ln terms uncombined when the question asks for a single logarithm.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2, published by OxfordAQA, for International AS exams from May/June 2018 and International A-level exams from May/June 2019. See section P2.7 (Integration).