Pure: Integration
Integration is the reverse process of differentiation, used to find the area under a curve, the equation of a curve from its gradient function, and volumes of revolution. A Level Maths covers integrating polynomials, trig, exponential and 1/x functions, integration by substitution and by parts, and using definite integrals to find areas between curves and lines.
Before you start
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Method
- Reverse the differentiation rule for polynomials: integral of ax^n dx = a*x^(n+1)/(n+1) + c (for n not equal to -1), remembering the constant of integration c for indefinite integrals.
- Learn the standard integrals: integral of cos(x) dx = sin(x) + c, integral of sin(x) dx = -cos(x) + c, integral of e^x dx = e^x + c, and integral of 1/x dx = ln|x| + c.
- Spot integration by inspection (reverse chain rule) when the integrand is a function of a linear expression, such as (ax+b)^n or e^(ax+b): integrate as if x alone, then divide by the coefficient of x inside the bracket.
- For products that do not simplify by inspection, use integration by parts: integral of u(dv/dx) dx = uv - integral of v(du/dx) dx, choosing u to be the term that gets simpler when differentiated.
- For a definite integral, integrate normally then substitute the upper limit and subtract the result of substituting the lower limit; take the absolute value of any negative result when finding an actual area.
- To find the area between a curve and a line (or two curves), subtract one equation from the other before integrating, then apply the limits given by their points of intersection.
Worked example
Find the integral of (6x + 1)^4 with respect to x, and hence evaluate the definite integral from x = 0 to x = 1.
- Recognise this is integration by inspection: raise the power by 1 to get (6x+1)^5.
- Divide by the new power (5) and by the coefficient of x inside the bracket (6): integral = (6x+1)^5/30 + c.
- For the definite integral, substitute x = 1: (6(1)+1)^5/30 = 7^5/30 = 16807/30.
- Substitute x = 0: (6(0)+1)^5/30 = 1^5/30 = 1/30.
- Subtract: 16807/30 - 1/30 = 16806/30.
- Final answer: the definite integral equals 16806/30, which simplifies to 2801/5, or 560.2.
Practice questions
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Q1Find the integral of (4x^3 - 2x + 5) dx.Show answer
Answer: x^4 - x^2 + 5x + c.
Q2Find the integral of 3e^(2x) dx.Show answer
Answer: (3/2)e^(2x) + c.
Q3Find the integral of sin(4x) dx.Show answer
Answer: -(1/4)cos(4x) + c.
Q4Evaluate the definite integral from x = 1 to x = 3 of (2x - 1) dx.Show answer
Answer: 6 (antiderivative x^2 - x, evaluated at 3 minus at 1).
Q5Find the integral of 5/(2x+3) dx.Show answer
Answer: (5/2)ln|2x + 3| + c.
Q6Use integration by parts to find the integral of x*cos(3x) dx.Show answer
Answer: (x/3)sin(3x) + (1/9)cos(3x) + c.
Exam-style questions
Written in the style of a A Level Maths exam paper, with a full mark scheme.
Find the integral of (2x - 5)^3 dx, and hence evaluate the definite integral from x = 2 to x = 3.
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The curve C has equation y = x^2 - 2x, and the line l has equation y = 3x - 4. Find the coordinates of the points of intersection of C and l, and find the area of the region enclosed between C and l.
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A curve has parametric equations x = t^2, y = 2t + 1, for 0 <= t <= 2. Find dx/dt, and use the formula Area = integral of y*(dx/dt) dt to find the area under the curve between t = 0 and t = 2.
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