Pure: Integration Depth
Integration Depth consolidates the standard integrals and integration by parts onto the wider toolkit A Level Maths expects by the end of the course: integration by substitution (including changing the limits of a definite integral), splitting an algebraic fraction into partial fractions before integrating it, rewriting a trig power such as sin^2(x) using a double angle identity so it can be integrated, finding a volume of revolution, and forming and solving a simple first-order differential equation by separating the variables. The examiner is testing whether a student can recognise which technique a given integral needs without being told, and carry a substitution or a partial-fractions split through cleanly to a correct final answer, including the constant of integration where it is required.
Before you start
Make sure you're comfortable with these topics first:
Method
- Decide the technique before integrating. A product where one factor is (or is a multiple of) the derivative of an expression inside the other factor suggests substitution, or inspection (the reverse chain rule) if the inner expression is linear. An algebraic fraction with a factorisable denominator suggests partial fractions first. A product of unrelated function types (polynomial with trig or exponential) suggests integration by parts. An even power of sin or cos suggests a double angle identity.
- For substitution, let u equal the inner expression, find du/dx, and rewrite the ENTIRE integral (including dx) in terms of u and du. For a definite integral, convert the x-limits into u-limits at the same time, so there is no need to substitute back to x at the end.
- For partial fractions, split a proper algebraic fraction over distinct linear factors as A/(x-a) + B/(x-b) (or add a further term C/(x-c)^2 for a repeated factor), find the constants by substituting values of x that make each bracket zero in turn, then integrate each simple term separately to a ln term (or a negative power for a repeated factor).
- For trig powers, use cos(2x) = 1 - 2sin^2(x) rearranged to sin^2(x) = (1 - cos(2x))/2, or cos(2x) = 2cos^2(x) - 1 rearranged to cos^2(x) = (1 + cos(2x))/2, before integrating.
- For a volume of revolution about the x-axis between x = a and x = b, use V = pi x integral of y^2 dx. About the y-axis between y = a and y = b, use V = pi x integral of x^2 dy.
- To form and solve a differential equation, separate the variables so every y-term (with dy) is on one side and every x-term (with dx) is on the other, integrate both sides (only ONE constant of integration is needed, on either side), then use a given initial condition to find the constant and state the particular solution.
Worked example
Using the substitution u = x^2 + 1, evaluate the definite integral of x(x^2+1)^3 with respect to x, from x = 0 to x = 2.
- Let u = x^2 + 1, so du/dx = 2x, which rearranges to x dx = (1/2) du.
- Change the limits: when x = 0, u = 1; when x = 2, u = 5.
- Rewrite the integral entirely in terms of u: the integral from u=1 to u=5 of u^3 x (1/2) du.
- Integrate: (1/2) x [u^4/4] from 1 to 5 = (1/8)[u^4] from 1 to 5.
- Substitute the limits: (1/8)(5^4 - 1^4) = (1/8)(625 - 1) = 624/8.
- Final answer: 78.
Practice questions
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Q1Express 7/((x-1)(x+2)) in partial fractions.Show answer
Answer: (7/3)/(x-1) - (7/3)/(x+2). (Found from 7 = A(x+2) + B(x-1): x=1 gives A=7/3; x=-2 gives B=-7/3.)
Q2Hence find the integral of 7/((x-1)(x+2)) with respect to x.Show answer
Answer: (7/3)ln|x-1| - (7/3)ln|x+2| + c.
Q3Find the integral of sin^2(x) with respect to x.Show answer
Answer: Using sin^2(x) = (1-cos(2x))/2: the integral is x/2 - sin(2x)/4 + c.
Q4Using the substitution u = 2x^3 + 1, find the integral of 6x^2(2x^3+1)^4 with respect to x.Show answer
Answer: du = 6x^2 dx exactly, so the integral becomes the integral of u^4 du = u^5/5 + c = (2x^3+1)^5/5 + c.
Q5The region bounded by the curve y = sqrt(x), the x-axis, and the line x = 4 is rotated 360 degrees about the x-axis. Find the exact volume of the solid formed.Show answer
Answer: V = pi x integral from 0 to 4 of x dx = pi x [x^2/2] from 0 to 4 = pi x 8 = 8*pi.
Q6Given that dy/dx = 6x^2 y, and that y = 4 when x = 0, find y in terms of x.Show answer
Answer: Separating variables: (1/y) dy = 6x^2 dx, so ln|y| = 2x^3 + c. Using y=4 at x=0 gives c = ln(4), so y = 4e^(2x^3).
Q7Explain why the integral of (3x+1)^5 with respect to x can be found by inspection (the reverse chain rule) rather than needing a substitution to be written out in full, and state the result.Show answer
Answer: The expression inside the bracket, 3x+1, is linear, so its derivative is just the constant 3. Integrating as if x alone and then dividing by that constant gives the same result as a full substitution: the integral is (3x+1)^6/18 + c.
Exam-style questions
Written in the style of a A Level Maths exam paper, with a full mark scheme.
Express (5x-4)/((x-2)(x+1)) in partial fractions. Hence find the integral of (5x-4)/((x-2)(x+1)) with respect to x.
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The region R is bounded by the curve y = x^2 + 1, the x-axis, and the lines x = 0 and x = 2. R is rotated 360 degrees about the x-axis to form a solid. Find the exact volume of the solid, giving your answer as a multiple of pi.
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The mass, m grams, of a chemical sample decreases at a rate proportional to the mass present, modelled by the differential equation dm/dt = -0.05m, where t is measured in minutes. The initial mass is 80 grams. (a) Show, by separating the variables, that m = 80e^(-0.05t). (4) (b) Find the mass remaining after 10 minutes, giving your answer to 3 significant figures. (2)
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