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Definite Integrals

Definite integrals and definite integration give a little more meaning to integration.
Just as the derivative of a "speed/time" function lets us determine the "acceleration/time" function, integrating the same "speed/time" function will allow us to determine the distance travelled between two instance in time.

Another example could be: if a function \(f(t)\), where \(t\) is time in hours, is equal to the "net revenue per hour" of a company at any time \(t\), then definite integration will allow us to calculate the company's total net revenue between two instants in time \(t_1\) and \(t_2\).

Definite Integrals

Given a function \(f(x)\), continuous over a closed interval \(\begin{bmatrix}a,b \end{bmatrix}\), we define the definite integral as: \[\int_a^bf(x)dx = F(b)-F(a)\] Where \(F(x) = \int f(x) dx\) is the antiderivative of the function \(f(x)\).

We'll often write: \[\int_a^bf(x)dx = \begin{bmatrix}F(x) \end{bmatrix}_a^b\] Where \(\begin{bmatrix}F(x) \end{bmatrix}_a^b = F(b) - F(a) \).

This is explained and illustrated in Tutorial 1, below.

Tutorial 1 - Evaluating Definite Integrals

In the following tutorial we review and explain the formula we've just read for definite integrals and work through some examples to see how they are evaluated.

Exercise 1

Evaluate each of the following definite integrals:

  1. \(\int_1^2 3x^2 dx\)

  2. \(\int_0^2 \begin{pmatrix}4x - 3 \end{pmatrix}dx\)

  3. \(\int_1^4 3. \sqrt{x}dx\)

  4. \(\int_{-1}^1 \begin{pmatrix}x^2 - 4x + 2 \end{pmatrix}dx\)

  5. \(\int_0^{2\pi}2.cos(x)dx\)

  6. \(\int_1^2 \begin{pmatrix} x^2-1 \end{pmatrix}dx\)

  7. \(\int_1^2 \begin{pmatrix} 2x^3 - 4x \end{pmatrix}dx\)

  8. \(\int_0^{\pi} 5.sin(x)dx\)

  9. \(\int_1^4 \frac{3}{2\sqrt{x}}dx\)

  10. \(\int_1^e\frac{1}{2x}dx\)

Answers Without Working

  1. \(\int_1^2 3x^2 dx = 7\)

  2. \(\int_0^2 \begin{pmatrix} 4x - 3 \end{pmatrix}dx = 2\)

  3. \(\int_1^4 3.\sqrt{x}dx = 14\)

  4. \(\int_{-1}^1 \begin{pmatrix}x^2 - 4x + 2 \end{pmatrix} dx = \frac{14}{3}\)

  5. \(\int_0^{2\pi}2.cos(x)dx = 0\)

  6. \(\int_1^2 \begin{pmatrix} x^2 - 1 \end{pmatrix}dx = \frac{4}{3}\)

  7. \(\int_1^2 \begin{pmatrix} 2x^3 - 4x \end{pmatrix} dx = \frac{3}{2}\)

  8. \(\int_0^{\pi}5.sin(x)dx = 10\)

  9. \(\int_1^4 \frac{3}{2\sqrt{x}}dx = 3\)

  10. \(\int_1^e\frac{1}{2x}dx = \frac{1}{2}\)

Solution With Working

  1. We evaluate \(\int_1^2 3x^2 dx\) as follows.
    We start by finding the antiderivative: \[\begin{aligned} F(x) & = \int 3x^2 dx \\ & = \frac{3}{2+1}x^{2+1}+c \\ & = \frac{3}{3}x^3 + c \\ F(x) & = x^3 + c \end{aligned} \] We can ignore the constant of integration, \(c\), when evaluating definite integrals, so: \[F(x) = x^3\] We now evaluate the definite integral: \[\begin{aligned} \int_1^2 3x^2 dx & = F(2)-F(1) \\ & = \begin{bmatrix} x^3 \end{bmatrix}_1^2 \\ & = 2^3 - 1^3 \\ & = 8 - 1 \\ \int_1^2 3x^2 dx & = 7 \end{aligned}\]

  2. We evaluate \(\int_0^2 \begin{pmatrix} 4x - 3 \end{pmatrix}dx\) as follows.
    We start by finding the antiderivative: \[\begin{aligned} F(x) & = \int \begin{pmatrix} 4x - 3 \end{pmatrix}dx \\ & = \int \begin{pmatrix} 4x^1 - 3x^0 \end{pmatrix}dx \\ & = \frac{4}{1+1}x^{1+1}-\frac{3}{0+1}x^{0+1}+c \\ & = \frac{4}{2}x^2 - \frac{3}{1}x^1 + c \\ F(x) & = 2x^2 - 3x + c \end{aligned}\] When evaluating definite integrals, we can ignore the constant of integration so: \[F(x) = 2x^2 - 3x\] We now evaluate the definite integral: \[\begin{aligned} \int_0^2 \begin{pmatrix} 4x - 3 \end{pmatrix}dx & = F(2)-F(0)\\ & = \begin{bmatrix} 2x^2-3x \end{bmatrix}_0^2 \\ & = 2\times 2^2 - 3 \times 2 - \begin{pmatrix} 2\times 0^2 - 3 \times 0 \end{pmatrix} \\ & = 2\times 4 - 6 - 0 \\ & = 8 - 6 \\ \int_0^2 \begin{pmatrix} 4x - 3 \end{pmatrix}dx & = 2 \end{aligned}\]

  3. We evaluate \(\int_1^4 3.\sqrt{x}dx\) as follows.
    We start by finding the antiderivative: \[\begin{aligned} F(x) & = \int 3. \sqrt{x}dx \\ & = \int 3.x^{\frac{1}{2}} dx \\ & = \frac{3}{\frac{1}{2}+1}.x^{\frac{1}{2}+1}+c \\ & = \frac{3}{\frac{3}{2}}.x^{\frac{3}{2}}+c \\ & = 3\times \frac{2}{3}.x^{\frac{3}{2}}+c \\ & = 2.x^{\frac{3}{2}}+c \\ F(x) & = 2.\sqrt{x^3}+c \end{aligned}\] When we evaluate definite integrals we ignore the constant of integration, so we write: \[F(x) = 2.\sqrt{x^3}\] We now evaluate the definite integral: \[\begin{aligned} \int_1^4 3.\sqrt{x}dx & = F(4)-F(1)\\ & = \begin{bmatrix} 2.\sqrt{x^3} \end{bmatrix}_1^4 \\ & = 2.\sqrt{4^3} - 2.\sqrt{1^3} \\ & = 2.\sqrt{64} - 2.\sqrt{1} \\ & = 2\times 8 - 2\times 1 \\ & = 16 - 2 \\ \int_1^4 3.\sqrt{x}dx & = 14 \end{aligned} \]

  4. We evaluate \(\int_{-1}^1 \begin{pmatrix}x^2 - 4x + 2 \end{pmatrix}dx\) as follows.
    We start by finding the antiderivative: \[\begin{aligned} F(x) & = \int \begin{pmatrix}x^2 - 4x + 2 \end{pmatrix}dx \\ & = \frac{1}{2+1}x^{2+1} - \frac{4}{1+1}x^{1+1} + 2x + c \\ & = \frac{1}{3}x^3 - \frac{4}{2}x^2 + 2x + c \\ F(x) & = \frac{x^3}{3} - 2x^2 + 2x + c \end{aligned}\] When evaluating definite integrals, we ignore the constant of integration: \[F(x) = \frac{x^3}{3} - 2x^2 + 2x\] We now evaluate the definite integral: \[\begin{aligned} \int_{-1}^1 \begin{pmatrix}x^2 - 4x + 2 \end{pmatrix} dx & = F(1) - F(-1) \\ & = \begin{bmatrix} \frac{x^3}{3} - 2x^2 + 2x \end{bmatrix}_{-1}^1 \\ & = \frac{1^3}{3} - 2\times 1^2 + 2 \times 1 - \begin{pmatrix} \frac{(-1)^3}{3} - 2\times (-1)^2 + 2\times (-1) \end{pmatrix} \\ & = \frac{1}{3} - 2\times 1 + 2 \times 1 - \begin{pmatrix} \frac{-1}{3} - 2\times 1 + 2\times (-1) \end{pmatrix} \\ & = \frac{1}{3} - 2 + 2 - \begin{pmatrix} \frac{-1}{3} - 2 - 2 \end{pmatrix} \\ & = \frac{1}{3} - \begin{pmatrix} \frac{-1}{3} - 4 \end{pmatrix} \\ & = \frac{1}{3} + \frac{1}{3} + 4 \\ & = \frac{2}{3} + 4 \\ & = \frac{2 + 12}{3} \\ \int_{-1}^1 \begin{pmatrix}x^2 - 4x + 2 \end{pmatrix}dx & = \frac{14}{3} \end{aligned}\]

  5. We evaluate \(\int_0^{2\pi}2.cos(x)dx\) as follows.
    We start by finding the antiderivative: \[\begin{aligned} F(x) & = \int 2.cos(x)dx \\ F(x) & = 2.sin(x)+c \end{aligned}\] Since we ignore the constant of integration when working with definite integrals we write: \[F(x) = 2.sin(x)\] We now evaluate the definite integral: \[\begin{aligned} \int_0^{2\pi}2.cos(x)dx & = F\begin{pmatrix} 2\pi \end{pmatrix} - F(0) \\ & = \begin{bmatrix}2.sin(x) \end{bmatrix}_0^{2\pi} \\ & = 2.sin\begin{pmatrix} 2\pi \end{pmatrix} - 2.sin\begin{pmatrix} 0 \end{pmatrix} \\ & = 2 \times 0 - 2 \times 0 \\ \int_0^{2\pi}2.cos(x)dx & = 0 \end{aligned}\]

  6. We evaluate \(\int_1^2 \begin{pmatrix}x^2 - 1 \end{pmatrix}dx\) as follows.
    We start by finding the antiderivative: \[\begin{aligned} F(x) & = \int \begin{pmatrix} x^2 - 1 \end{pmatrix} dx \\ F(x) & = \frac{x^3}{3} - x + c \end{aligned}\] Since we can ignore the constant of integration, when working with definite integrals, we write: \[F(x) = \frac{x^3}{3} - x\] We now evaluate the definite integral: \[\begin{aligned} \int_1^2 \begin{pmatrix}x^2 - 1 \end{pmatrix}dx & = F(2) - F(1) \\ & = \begin{bmatrix} \frac{x^3}{3} - x \end{bmatrix}_1^2 \\ & = \frac{2^3}{3} - 2 - \begin{pmatrix} \frac{1^3}{3} - 1 \end{pmatrix} \\ & = \frac{8}{3} - 2 - \begin{pmatrix} \frac{1^3}{3} - 1 \end{pmatrix} \\ & = \frac{8}{3} - 2 - \frac{1}{3} + 1 \\ & = \frac{7}{3} - 1 \\ & = \frac{7-3}{3} \\ \int_1^2 \begin{pmatrix}x^2 - 1 \end{pmatrix}dx & = \frac{4}{3} \end{aligned}\]

  7. We evaluate \(\int_1^2 \begin{pmatrix} 2x^3 - 4x \end{pmatrix}dx\) as follows.
    We start by finding the antiderivative: \[\begin{aligned} F(x) & = \int \begin{pmatrix} 2x^3 - 4x \end{pmatrix}dx \\ & = \frac{2}{3+1}x^{3+1} - \frac{4}{2}x^2 + c \\ &= \frac{2}{4}x^4 - 2x^2 + c \\ &= \frac{1}{2}x^4 - 2x^2 + c \\ F(x) &= \frac{x^4}{2} - 2x^2 + c \end{aligned} \] When evaluating definite integrals, we can ignore the constant of integration so we write: \[F(x) = \frac{x^4}{2} - 2x^2 \] We now evaluate the definite integral: \[\begin{aligned} \int_1^2 \begin{pmatrix} 2x^3 - 4x \end{pmatrix}dx & = F(2) - F(1) \\ & = \begin{bmatrix} \frac{x^4}{2} - 2x^2 \end{bmatrix}_1^2 \\ & = \frac{2^4}{2} - 2\times 2^2 - \begin{pmatrix} \frac{1^4}{2} - 2\times 1^2 \end{pmatrix} \\ & = \frac{16}{2} - 2\times 4 - \begin{pmatrix} \frac{1}{2} - 2\times 1 \end{pmatrix} \\ & = \frac{16}{2} - 2\times 4 - \begin{pmatrix} \frac{1}{2} - 2\times 1 \end{pmatrix} \\ & = 8 - 8 - \begin{pmatrix} \frac{1}{2} - 2 \end{pmatrix} \\ & = 0 - \begin{pmatrix} \frac{1-4}{2} \end{pmatrix} \\ & = - \begin{pmatrix} \frac{-3}{2} \end{pmatrix} \\ \int_1^2 \begin{pmatrix} 2x^3 - 4x \end{pmatrix}dx & = \frac{3}{2} \end{aligned}\]

  8. We evaluate \(\int_0^{\pi} 5.sin(x)dx\) as follows.
    We start by finding the antiderivative: \[\begin{aligned} F(x) & = \int 5.sin(x)dx \\ F(x) & = -5.cos(x) + c \end{aligned}\] Since we can ignore the constant of integration when we evaluate definite integrals, we write: \[F(x) = -5.cos(x)\] We now evaluate the definite integral: \[\begin{aligned} \int_0^{\pi} 5.sin(x)dx & = F\begin{pmatrix} \pi \end{pmatrix} - F(0) \\ & = \begin{bmatrix} -5.cos(x)\end{bmatrix}_0^{\pi} \\ & = -5.cos \begin{pmatrix} \pi \end{pmatrix} - \begin{pmatrix} - 5.cos(0) \end{pmatrix} \\ & = -5 \times (-1)- \begin{pmatrix} -5 \times 1 \end{pmatrix} \\ & = 5 - (-5) \\ & = 5 + 5 \\ \int_0^{\pi} 5.sin(x)dx & = 10 \end{aligned}\]

  9. We evaluate \(\int_1^4 \frac{3}{4\sqrt{x}}dx\) as follows.
    We start by finding the antiderivative: \[\begin{aligned} F(x) & = \int \frac{3}{2\sqrt{x}}dx \\ & = \int \frac{3}{2}.x^{-\frac{1}{2}}dx \\ & = \frac{3}{2}\int x^{- \frac{1}{2}} dx \\ & = \frac{3}{2} \times \frac{1}{-\frac{1}{2}+ 1}.x^{-\frac{1}{2}+1}+c \\ & = \frac{3}{2} \times \frac{1}{\frac{1}{2}}.x^{\frac{1}{2}}+c \\ & = \frac{3}{2} \times 2.x^{\frac{1}{2}}+c \\ & = 3.x^{\frac{1}{2}}+c \\ F(x) & = 3.\sqrt{x}+ c \end{aligned}\] Since we can ignore the constant of integration, when working with definite integrals, we write: \[F(x) = 3.\sqrt{x}\] We now evaluate the definite integral: \[\begin{aligned} \int_1^4 \frac{3}{4\sqrt{x}}dx & = F(4)-F(1) \\ & = \begin{bmatrix} 3.\sqrt{x} \end{bmatrix}_1^4 \\ & = 3.\sqrt{4} - 3.\sqrt{1} \\ & = 3 \times 2 - 3\times 1 \\ & = 6 - 3 \\ \int_1^4 \frac{3}{4\sqrt{x}}dx & = 3 \end{aligned}\]

  10. We evaluate \(\int_1^e \frac{1}{2x}dx\) as follows.
    We start by finding the antiderivative: \[\begin{aligned} F(x) & = \int \frac{1}{2x}dx \\ & = \frac{1}{2}\frac{1}{x}dx \\ F(x) & = \frac{1}{2}.ln \begin{vmatrix} x \end{vmatrix} + c \\ \end{aligned}\] Since we can ignore the constant of integration, when evaluating definite integrals, we write: \[F(x) = \frac{1}{2}.ln \begin{vmatrix}x \end{vmatrix}\] We now evaluate the definite integral: \[\begin{aligned} \int_1^e \frac{1}{2x}dx & = F(e) - F(1) \\ & = \begin{bmatrix} \frac{1}{2}.ln \begin{vmatrix} x \end{vmatrix} \end{bmatrix}_1^e \\ & = \begin{bmatrix} \frac{1}{2}.ln (x) \end{bmatrix}_1^e \\ & = \frac{1}{2}.ln (e) - \frac{1}{2}.ln (1) \\ & = \frac{1}{2}\times 1 - \frac{1}{2} \times 0 \\ & = \frac{1}{2} - 0 \\ \int_1^e \frac{1}{2x}dx & = \frac{1}{2} \end{aligned}\]

Properties of Definite Integrals

Consider a continuous function \(f(x)\) over the closed interval \([a,b]\) and \(\alpha \) a real number, \(a \in \mathbb{R}\).

  • Any scalar \(\alpha \) multiplying the integrand can be taken out of the integral: \[\int_a^b \alpha.f(x) dx = \alpha \int_a^b f(x) dx\]

  • If \(c\) is a real number such that \(c \in \begin{bmatrix}a,b \end{bmatrix}\), then: \[\int_a^c f(x)dx + \int_c^b f(x)dx = \int_a^b f(x)dx\]

  • Given two functions \(f(x)\) and \(g(x)\), both continuous over \(\begin{bmatrix}a,b \end{bmatrix}\): \[\int_a^b f(x) dx + \int_a^b g(x) dx = \int_a^b \begin{pmatrix} f(x) \pm g(x) \end{pmatrix}dx\]

Exercise 2

Using the properties, listed above, calculate each of the following definite integrals:

  1. \(\int_0^2 \frac{x^3}{7}dx\)

  2. \(\int_2^4 x dx - \int_2^4 3x^2 dx\)

  3. \(\int_1^4 \frac{3}{5 \sqrt{x}}dx\)

  4. \( \int_1^2 6x^2 dx + \int_2^3 6x^2 dx\)

  5. \(\int_1^e \frac{5}{3x}dx\)

Solution Without Working

Solution With Working

Interpretation of Definite Integrals

When we write: \[\int_a^b f(x) dx\] it should be read/understood as follows:

  • \(\int_a^b\) : means "the sum from \(a\) to \(b\), included" (in fact the symbol \(\int \) is just an 'S' that's been stretched in the vertical direction).

  • \(f(x)dx\) : refers to the area of the rectangle of height \(f(x)\) and base width \(dx\), where \(dx\) is infinitely small.
So, when we read \(\int_a^b f(x) dx\) we should read: "the sum of the areas of all the infinitely thin rectangles, of area \(f(x)dx\), from \(a\) to \(b\)".

In other words:

\(\int_a^b f(x) dx\) is equal to the area enclosed by the curve and the \(x\)-axis, between the two values \(x = a\) and \(x = b\).

Note: when \(f(x)\) is negative, \(f(x)<0\), so is the area \(f(x).dx\) This is further discussed and explained in Tutorial 2 below. Spend a few minutes to watch it now.

Tutorial 2 - Interpretation of Definite Integrals

In the following tutorial we review and explain the formula we've just read for definite integrals and work through some examples to see how they are evaluated.

Exercise 3

Try answering each of the following questions without using a calculator, other than for checking your final answer.

    1. Solve \(sin(x)=0\), for \(0 < x < 2\pi\).
    2. Evaluate \(\int_0^{2\pi} sin(x)dx\).

    3. Explain the result obtained in question 1.

  1. A compnay's revenue per hour is modelled by the function: \[r(t) = \frac{2}{5} t^2-4t+8\] where \(r(t)\) is "revenue" in thousands of dollars per hour, \(t\) is time in hours and \(0\leq t \leq 8\).
    1. Explain, in a few words, what any negative value of \(r(t)\) represents.

    2. Solve \(r(t)=0\).

    3. Evaluate \(\int_1^6 r(t)dt\).

    4. In a few words, state the meaning of the value obtained in question 3.

    5. Evalaute \(\int_1^4 r(t)dt\).

    6. Explain, in a few words, why \(\int_1^4 r(t)dt > \int_1^6 r(t)dt\).

  2. A particle, moving in a straight line, has velocity function:

Solution Without Working

Solution With Working