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Integration and area under a curve

Further Mathematics · WAEC and JAMB · SS2 and SS3

Integration reverses differentiation and is examined both as straight calculation and as area. Candidates who are careful with limits and with the constant of integration score heavily, because the arithmetic itself is light.

What you need to know

  • Integration is the reverse of differentiation, so the integral of ax^n is a x^(n+1)/(n + 1) + c. Add one to the power, then divide by the new power. This is exactly the opposite of what you do when differentiating.
  • The rule fails when n = -1, because dividing by zero is impossible. Instead, the integral of 1/x is ln x + c, and this special case is examined often enough that forgetting it is expensive.
  • Every indefinite integral must carry the arbitrary constant c. If the question gives a boundary condition, such as the curve passing through a named point, substitute it to find c and then write the particular equation.
  • The standard integrals are: integral of sin x dx = -cos x + c, integral of cos x dx = sin x + c, integral of sec^2 x dx = tan x + c and integral of e^x dx = e^x + c. Note the minus sign on the cosine, which is the reverse of the differentiation pattern.
  • For a definite integral, find the antiderivative, write it in square brackets with the limits, substitute the upper limit, then subtract the value at the lower limit. There is no constant of integration in a definite integral, since it cancels in the subtraction.
  • Integration by substitution handles a function of a linear function: the integral of (ax + b)^n dx is (ax + b)^(n+1)/(a(n + 1)) + c. The extra division by a is the step candidates omit, so check by differentiating your answer back.
  • For a general substitution, let u be the inner function, find du/dx, replace dx completely, and change the limits to u values if the integral is definite. Never mix u and x in the same integral.
  • The area between a curve and the x-axis from x = a to x = b is the definite integral of y dx between those limits. Sketch the curve first so that you know where it crosses the axis.
  • A region below the x-axis gives a negative integral. Area is never negative, so split the integral at each root and take the absolute value of each piece before adding, otherwise the parts cancel and you report an area that is too small.
  • The area between two curves is the integral of (upper curve minus lower curve) between their points of intersection. Find the intersections first by setting the two equations equal to each other and solving.
  • The volume generated when the region under a curve is rotated completely about the x-axis is pi times the integral of y^2 dx. Square y before integrating, not after, and remember that the pi stays outside throughout.
  • To integrate a product of trigonometric functions, first use an identity to turn it into a sum. For instance sin^2 x is rewritten as (1 - cos 2x)/2, which you can integrate term by term.
  • The trapezium rule gives an approximate value when exact integration is impossible: area is approximately (h/2)[first ordinate + last ordinate + 2(sum of the remaining ordinates)], where h is the common width.

Key terms

Integration
The reverse process of differentiation, used to recover a function from its derivative or to accumulate a quantity.
Indefinite integral
An integral with no limits, whose answer is a family of functions differing by the arbitrary constant c.
Definite integral
An integral evaluated between two limits, giving a single number rather than a function.
Constant of integration
The arbitrary constant c added to every indefinite integral, because differentiating any constant gives zero.
Antiderivative
A function whose derivative is the given function.
Solid of revolution
The solid formed when a plane region is rotated completely about a straight line, usually one of the axes.

Formulae

  • Integral of ax^n dx = a x^(n+1)/(n + 1) + c, provided n is not -1
  • Integral of (1/x) dx = ln x + c
  • Integral of (ax + b)^n dx = (ax + b)^(n+1)/(a(n + 1)) + c
  • Integral of sin x dx = -cos x + c
  • Integral of cos x dx = sin x + c
  • Integral of sec^2 x dx = tan x + c
  • Integral of e^x dx = e^x + c, and integral of e^(kx) dx = e^(kx)/k + c
  • Definite integral from a to b of f(x) dx = F(b) - F(a)
  • Area under a curve = definite integral from a to b of y dx
  • Area between curves = definite integral of (upper - lower) dx
  • Volume of revolution about the x-axis = pi times integral of y^2 dx
  • sin^2 x = (1 - cos 2x)/2 and cos^2 x = (1 + cos 2x)/2
  • Trapezium rule: area is approximately (h/2)[y_first + y_last + 2(sum of other ordinates)]

Worked examples

Evaluate the definite integral of (3x^2 - 4x + 1) dx from x = 1 to x = 3.

  1. Integrate term by term: the integral of 3x^2 is x^3, the integral of -4x is -2x^2, and the integral of 1 is x.
  2. So the antiderivative is F(x) = x^3 - 2x^2 + x, written in square brackets with limits 1 and 3.
  3. Substitute the upper limit: F(3) = 27 - 2(9) + 3 = 27 - 18 + 3 = 12.
  4. Substitute the lower limit: F(1) = 1 - 2(1) + 1 = 1 - 2 + 1 = 0.
  5. Subtract: F(3) - F(1) = 12 - 0 = 12.
  6. Check the antiderivative by differentiating: d/dx(x^3 - 2x^2 + x) = 3x^2 - 4x + 1, which is the original integrand.

Find the area of the region enclosed between the curve y = x^2 and the line y = x + 2.

  1. Find the points of intersection by equating: x^2 = x + 2, so x^2 - x - 2 = 0.
  2. Factorise: (x - 2)(x + 1) = 0, so the limits are x = -1 and x = 2.
  3. Between these limits the line lies above the parabola, so the integrand is (x + 2) - x^2.
  4. Integrate: the antiderivative of (x + 2 - x^2) is x^2/2 + 2x - x^3/3.
  5. At x = 2: 4/2 + 4 - 8/3 = 2 + 4 - 2.6667 = 3.3333. At x = -1: 1/2 - 2 + 1/3 = 0.5 - 2 + 0.3333 = -1.1667.
  6. Subtract: 3.3333 - (-1.1667) = 4.5, which is 9/2 exactly.

The mistake to avoid

Omitting the constant of integration in an indefinite integral, which is an automatic loss of one mark on every such question. In area work, the standing error is integrating straight across a root so that the part below the axis cancels the part above it; split the integral at every crossing point and add the absolute values.

In the exam

Write the antiderivative inside square brackets with the limits attached before you substitute anything, because that notation alone earns a method mark. Sketch the curve whenever the question mentions area, since the sketch tells you the limits, tells you which function is on top, and warns you if any part of the region dips below the axis.