Definite Integrals: Part 1

Evaluating definite integrals using antiderivatives and the Newton-Leibniz formula.

Learning goal Evaluate definite integrals accurately by finding an antiderivative and substituting the upper and lower limits.
Reminder This page is about evaluation. Interpretation as area and signed area belongs to Part 2.
Syllabus connection. This page supports the calculus progression from antiderivatives to definite integrals. In the AA syllabus this connects naturally to SL 5.10 and SL 5.11: finding antiderivatives first, then evaluating definite integrals analytically using \(F(b)-F(a)\). Area interpretation is treated separately in Part 2.

1Definite integral notation

After learning about antiderivatives, we now introduce definite integrals. A definite integral has limits and gives a single value.

\[ \int_a^b f(x)\,dx \]

Read this as: integrate \(f(x)\) with respect to \(x\), from \(x=a\) to \(x=b\).

\(b\)
Upper limit

The value substituted first: \(F(b)\).

\(a\)
Lower limit

The value subtracted second: \(F(a)\).

\(f(x)\)
Integrand

The function being integrated.

\(dx\)
With respect to \(x\)

The variable of integration is \(x\).

The expression \(\int_a^b f(x)\,dx\) is read as:

“the definite integral, from \(a\) to \(b\), of \(f(x)\), with respect to \(x\).”

where:

  • \(a\) is the lower limit.
  • \(b\) is the upper limit.
  • \(f(x)\) is the integrand.
  • \(dx\) tells us that the variable of integration is \(x\).

2The Newton-Leibniz formula

If \(F\) is an antiderivative, or primitive, of \(f\), then \(F'(x)=f(x)\).

Newton-Leibniz formula
\[ \begin{aligned} \int_a^b f(x)\,dx &=\left[F(x)\right]_a^b\\ &=F(b)-F(a) \end{aligned} \]

This is also called the evaluation theorem. The order matters: upper limit first, then subtract lower limit.

3The two-step method

Step 1. Find an antiderivative \(F(x)\) of the integrand \(f(x)\).

For definite integrals, you do not need to include the constant of integration \(C\). It cancels when calculating \(F(b)-F(a)\), as shown in Worked example 1 below.

Step 2. Evaluate \(F(b)-F(a)\).

4Worked examples

Calculations are set out vertically so each line shows one main algebraic step.

Worked example 1

Evaluate: \(\displaystyle \int_1^3(2x+1)\,dx\)
  1. Find the antiderivative: \[ \begin{aligned} F(x) &=\int(2x+1)\,dx\\ &=\int 2x\,dx+\int 1\,dx\\ &=x^2+x+C \end{aligned} \]
  2. Use the Newton-Leibniz formula: \[ \begin{aligned} \int_1^3(2x+1)\,dx &=\left[F(x)\right]_1^3\\ &=F(3)-F(1) \end{aligned} \]
  3. Evaluate \(F(3)\): \[ \begin{aligned} F(3) &=3^2+3+C\\ &=9+3+C\\ &=12+C \end{aligned} \]
  4. Evaluate \(F(1)\): \[ \begin{aligned} F(1) &=1^2+1+C\\ &=1+1+C\\ &=2+C \end{aligned} \]
  5. Subtract: \[ \begin{aligned} \int_1^3(2x+1)\,dx &=(12+C)-(2+C)\\ &=12+C-2-C\\ &=10 \end{aligned} \]
Notice that the constant \(C\) cancels. This will always happen for definite integrals.

Worked example 2

Evaluate: \(\displaystyle \int_0^2(x^2-4x+5)\,dx\)
  1. Find the antiderivative: \[ \begin{aligned} F(x) &=\int(x^2-4x+5)\,dx\\ &=\int x^2\,dx-\int 4x\,dx+\int 5\,dx\\ &=\frac{x^3}{3}-2x^2+5x \end{aligned} \]
  2. Apply the Newton-Leibniz formula: \[ \begin{aligned} \int_0^2(x^2-4x+5)\,dx &=\left[F(x)\right]_0^2\\ &=F(2)-F(0) \end{aligned} \]
  3. Evaluate \(F(2)\): \[ \begin{aligned} F(2) &=\frac{2^3}{3}-2(2)^2+5(2)\\ &=\frac{8}{3}-2(4)+10\\ &=\frac{8}{3}-8+10\\ &=\frac{8}{3}+2\\ &=\frac{8}{3}+\frac{6}{3}\\ &=\frac{14}{3} \end{aligned} \]
  4. Evaluate \(F(0)\): \[ \begin{aligned} F(0) &=\frac{0^3}{3}-2(0)^2+5(0)\\ &=0-0+0\\ &=0 \end{aligned} \]
  5. Subtract: \[ \begin{aligned} \int_0^2(x^2-4x+5)\,dx &=\frac{14}{3}-0\\ &=\frac{14}{3} \end{aligned} \]

Worked example 3

Evaluate: \(\displaystyle \int_0^\pi\sin x\,dx\)
  1. Find the antiderivative: \[ \begin{aligned} F(x) &=\int\sin x\,dx\\ &=-\cos x \end{aligned} \]
  2. Apply the Newton-Leibniz formula: \[ \begin{aligned} \int_0^\pi\sin x\,dx &=\left[-\cos x\right]_0^\pi\\ &=F(\pi)-F(0) \end{aligned} \]
  3. Evaluate \(F(\pi)\): \[ \begin{aligned} F(\pi) &=-\cos(\pi)\\ &=-(-1)\\ &=1 \end{aligned} \]
  4. Evaluate \(F(0)\): \[ \begin{aligned} F(0) &=-\cos(0)\\ &=-1 \end{aligned} \]
  5. Subtract: \[ \begin{aligned} \int_0^\pi\sin x\,dx &=1-(-1)\\ &=1+1\\ &=2 \end{aligned} \]

Worked example 4

Evaluate: \(\displaystyle \int_1^2x(x+3)\,dx\)
Note. Unlike differentiation, there is no simple product rule for integration. Expand first.
  1. Expand the integrand: \[ \begin{aligned} x(x+3) &=x^2+3x \end{aligned} \] So \[ \begin{aligned} \int_1^2x(x+3)\,dx &=\int_1^2(x^2+3x)\,dx \end{aligned} \]
  2. Find the antiderivative: \[ \begin{aligned} F(x) &=\int(x^2+3x)\,dx\\ &=\int x^2\,dx+\int 3x\,dx\\ &=\frac{x^3}{3}+\frac{3x^2}{2} \end{aligned} \]
  3. Evaluate \(F(2)\): \[ \begin{aligned} F(2) &=\frac{2^3}{3}+\frac{3(2)^2}{2}\\ &=\frac{8}{3}+\frac{3(4)}{2}\\ &=\frac{8}{3}+\frac{12}{2}\\ &=\frac{8}{3}+6\\ &=\frac{8}{3}+\frac{18}{3}\\ &=\frac{26}{3} \end{aligned} \]
  4. Evaluate \(F(1)\): \[ \begin{aligned} F(1) &=\frac{1^3}{3}+\frac{3(1)^2}{2}\\ &=\frac{1}{3}+\frac{3}{2}\\ &=\frac{2}{6}+\frac{9}{6}\\ &=\frac{11}{6} \end{aligned} \]
  5. Subtract: \[ \begin{aligned} \int_1^2x(x+3)\,dx &=F(2)-F(1)\\ &=\frac{26}{3}-\frac{11}{6}\\ &=\frac{52}{6}-\frac{11}{6}\\ &=\frac{41}{6} \end{aligned} \]

Worked example 5

Evaluate: \(\displaystyle \int_0^1 e^x\,dx\)
  1. Find the antiderivative: \[ \begin{aligned} F(x) &=\int e^x\,dx\\ &=e^x \end{aligned} \]
  2. Apply the Newton-Leibniz formula: \[ \begin{aligned} \int_0^1e^x\,dx &=\left[e^x\right]_0^1\\ &=F(1)-F(0) \end{aligned} \]
  3. Evaluate: \[ \begin{aligned} F(1) &=e^1\\ &=e \end{aligned} \] and \[ \begin{aligned} F(0) &=e^0\\ &=1 \end{aligned} \]
  4. Subtract: \[ \begin{aligned} \int_0^1e^x\,dx &=e-1 \end{aligned} \]
Leave \(e-1\) exact unless the question asks for a decimal approximation.

5Common mistakes

Wrong order

Use \(F(b)-F(a)\), not \(F(a)-F(b)\).

Missing brackets

Subtract the entire lower-limit value.

Adding \(+C\) at the end

The final value of a definite integral is not a family of functions.

Product expressions

For expressions like \(x(x+3)\), expand first.

6Practice

  1. Evaluate \(\int_0^4(3x+2)\,dx\).
    Show solution
    \[ \begin{aligned} \int_0^4(3x+2)\,dx &=\left[\frac{3x^2}{2}+2x\right]_0^4\\ &=\left(\frac{3(4)^2}{2}+2(4)\right)-0\\ &=\left(\frac{48}{2}+8\right)-0\\ &=32 \end{aligned} \]
  2. Evaluate \(\int_1^2(x^2+2x)\,dx\).
    Show solution
    \[ \begin{aligned} \int_1^2(x^2+2x)\,dx &=\left[\frac{x^3}{3}+x^2\right]_1^2\\ &=\left(\frac{2^3}{3}+2^2\right)-\left(\frac{1^3}{3}+1^2\right)\\ &=\left(\frac{8}{3}+4\right)-\left(\frac{1}{3}+1\right)\\ &=\frac{20}{3}-\frac{4}{3}\\ &=\frac{16}{3} \end{aligned} \]
  3. Evaluate \(\int_0^\pi\cos x\,dx\).
    Show solution
    \[ \begin{aligned} \int_0^\pi\cos x\,dx &=\left[\sin x\right]_0^\pi\\ &=\sin(\pi)-\sin(0)\\ &=0-0\\ &=0 \end{aligned} \]
  4. Evaluate \(\int_1^3x(2x-1)\,dx\).
    Show solution
    \[ \begin{aligned} x(2x-1) &=2x^2-x \end{aligned} \] \[ \begin{aligned} \int_1^3x(2x-1)\,dx &=\int_1^3(2x^2-x)\,dx\\ &=\left[\frac{2x^3}{3}-\frac{x^2}{2}\right]_1^3\\ &=\left(\frac{2(3)^3}{3}-\frac{3^2}{2}\right)-\left(\frac{2(1)^3}{3}-\frac{1^2}{2}\right)\\ &=\left(18-\frac{9}{2}\right)-\left(\frac{2}{3}-\frac{1}{2}\right)\\ &=\frac{27}{2}-\frac{1}{6}\\ &=\frac{81}{6}-\frac{1}{6}\\ &=\frac{40}{3} \end{aligned} \]
  5. Evaluate \(\int_0^2e^x\,dx\).
    Show solution
    \[ \begin{aligned} \int_0^2e^x\,dx &=\left[e^x\right]_0^2\\ &=e^2-e^0\\ &=e^2-1 \end{aligned} \]
  6. Evaluate \(\int_{-1}^{1}(x^3+x)\,dx\).
    Show solution
    \[ \begin{aligned} \int_{-1}^{1}(x^3+x)\,dx &=\left[\frac{x^4}{4}+\frac{x^2}{2}\right]_{-1}^{1}\\ &=\left(\frac{1^4}{4}+\frac{1^2}{2}\right)-\left(\frac{(-1)^4}{4}+\frac{(-1)^2}{2}\right)\\ &=\left(\frac{1}{4}+\frac{1}{2}\right)-\left(\frac{1}{4}+\frac{1}{2}\right)\\ &=0 \end{aligned} \]
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