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Mathematics: Analysis and Approaches

SL 5.11 Definite integrals and areas

Definite integrals are evaluated as F(b) minus F(a), and areas need care when a curve dips below the axis or when the region lies between two curves. Study guide, 3 practice questions with markschemes and 2 flashcards. Free.

Evaluating definite integrals analytically · Area under a curve including regions below the axis · Area between two curves · Using technology for hard integrals · Properties of definite integrals

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Study guide

Evaluating definite integrals

The fundamental idea: if \(F\) is any antiderivative of \(f\), then

\[\int_a^b f(x)\,dx = \big[F(x)\big]_a^b = F(b) - F(a).\]

The constant \(C\) cancels, so it is left out.

Example. \(\displaystyle\int_1^3 (3x^2 - 2x)\,dx = \big[x^3 - x^2\big]_1^3 = (27 - 9) - (1 - 1) = 18.\)

Signed area

A definite integral counts area above the \(x\)-axis as positive and area below as negative. So the integral is a net (signed) area, not always the geometric area. To find the total area:

  1. Find where the curve crosses the axis.
  2. Integrate each piece separately.
  3. Add the absolute values.

Worked example

Find the area enclosed by \(y = x^2 - 4\), the \(x\)-axis and the lines \(x = 0\) and \(x = 3\).

The curve crosses the axis at \(x = 2\): below on \([0, 2]\), above on \([2, 3]\). Using \(F(x) = \frac{x^3}{3} - 4x\):

Total area \(= \frac{16}{3} + \frac{7}{3} = \frac{23}{3}\). Note that \(\int_0^3 (x^2 - 4)\,dx = -3\), which is the net value and not the area.

On a GDC you can get the area in one step by integrating the absolute value of the function.

Area between two curves

If \(f(x) \ge g(x)\) on \([a, b]\), the area between them is

\[A = \int_a^b \big(f(x) - g(x)\big)\,dx,\]

"top minus bottom". It does not matter whether the curves are above or below the axis. The limits are usually the \(x\)-coordinates of the intersection points.

Example. \(y = x\) and \(y = x^2\) meet at \(x = 0\) and \(x = 1\), and \(x \ge x^2\) in between. Area \(= \int_0^1 (x - x^2)\,dx = \frac12 - \frac13 = \frac16\). If the curves cross inside the interval, split at the crossing and swap top and bottom.

Properties worth knowing

These let you answer "given \(\int_1^5 f = 8\), find \(\int_1^5 (2f(x) + 3)\,dx\)": the answer is \(16 + 12 = 28\).

Common mistakes

Exam tips

Cheat sheet

ItemFormula
Evaluate\(\int_a^b f\,dx = F(b) - F(a)\)
Area, curve below axis\(A = \left\lvert \int_a^b f\,dx \right\rvert\)
Area, mixed signssplit at roots, add absolute values
Between curves\(A = \int_a^b (f - g)\,dx\), \(f \ge g\)
Reverse limits\(\int_b^a f = -\int_a^b f\)
Join intervals\(\int_a^b f + \int_b^c f = \int_a^c f\)
Zero width\(\int_a^a f = 0\)
Add a constant\(\int_a^b (f + k) = \int_a^b f + k(b-a)\)

Key definitions

Definite integral

The number \(\int_a^b f(x)\,dx = F(b) - F(a)\), the net signed area between the curve and the \(x\)-axis.

Limits of integration

The values \(a\) and \(b\) marking the ends of the interval.

Signed area

Area counted positive above the \(x\)-axis and negative below it.

Area between curves

\(\int_a^b (\text{top} - \text{bottom})\,dx\) over the interval where one curve stays above the other.

Fundamental theorem of calculus

The result linking integration and differentiation that allows \(\int_a^b f\,dx = F(b) - F(a)\).

Practice questions

SLPaper 1Challenging

Question 1

The line \(y = 2x + 3\) and the curve \(y = x^2\) enclose a region \(R\).

(a) [2]

Find the \(x\)-coordinates of the points of intersection.

Markscheme
  • 1 Sets \(x^2 = 2x + 3\) and factorises: \((x - 3)(x + 1) = 0\)
  • 1 \(x = -1\) and \(x = 3\)
(b) [4]

Find the area of \(R\).

Markscheme
  • 1 Correct integral: \(\int_{-1}^{3} (2x + 3 - x^2)\,dx\)
  • 1 Correct antiderivative: \(x^2 + 3x - \frac{x^3}{3}\)
  • 1 Substitutes limits: \(9 - \left(-\frac{5}{3}\right)\)
  • 1 \(\frac{32}{3}\)
SLPaper 1Challenging

Question 2

Let \(f(x) = x^3 - 6x^2 + 8x\).

(a) [2]

Show that \(f(x) = x(x - 2)(x - 4)\).

Markscheme
  • 1 Takes out factor \(x\): \(x(x^2 - 6x + 8)\)
  • 1 Factorises quadratic: \(x(x - 2)(x - 4)\)
(b) [2]

Find \(\int_0^4 f(x)\,dx\).

Markscheme
  • 1 Antiderivative \(\frac{x^4}{4} - 2x^3 + 4x^2\)
  • 1 \(0\)
(c) [4]

Explain why your answer to (b) is not the area enclosed between the curve and the \(x\)-axis, and find that area.

Markscheme
  • 1 Explains the curve is below the axis for \(2 < x < 4\), so that integral is negative and cancels the positive part
  • 1 \(\int_0^2 f(x)\,dx = 4\)
  • 1 \(\int_2^4 f(x)\,dx = -4\)
  • 1 Total area \(= 8\)
SLPaper 1Very hard

Question 3

The line \(y = kx\), where \(k > 0\), and the curve \(y = x^2\) enclose a region with area 36.

(a) [4]

Show that the area enclosed is \(\frac{k^3}{6}\).

Markscheme
  • 1 Intersections at \(x = 0\) and \(x = k\)
  • 1 Correct integral: \(\int_0^k (kx - x^2)\,dx\)
  • 1 Antiderivative: \(\frac{kx^2}{2} - \frac{x^3}{3}\)
  • 1 Substitutes limits to obtain \(\frac{k^3}{2} - \frac{k^3}{3} = \frac{k^3}{6}\)
(b) [2]

Hence find the value of \(k\).

Markscheme
  • 1 \(k^3 = 216\)
  • 1 \(k = 6\)

Key facts

Area between two curves from \(x = a\) to \(x = b\)

\(\int_a^b (\text{top} - \text{bottom})\,dx\), with \(a\) and \(b\) usually the intersection points.

Common mistake: why can \(\int_a^b f(x)\,dx\) be smaller than the area between the curve and the \(x\)-axis?

Parts below the axis give negative integrals and cancel positive parts. Split at the roots or use \(\int_a^b |f(x)|\,dx\).