Properties Of The Definite Integral

The definite integral inherits its algebra from the summation process. Sums add, constants factor out, and inequalities are preserved, so integrals do all three as well.

Property	Name	Condition
$\int_{a}^{b} [f + g] = \int_{a}^{b} f + \int_{a}^{b} g$	Sum Rule	Both integrals exist
$\int_{a}^{b} c f (x) d x = c \int_{a}^{b} f (x) d x$	Constant Multiple Rule	c constant, not a function of x
$\int_{a}^{b} c d x = c (b - a)$	Constant Function Rule	c constant
$\int_{a}^{c} f + \int_{c}^{b} f = \int_{a}^{b} f$	Additivity Rule	Holds whether c is inside or outside $[a, b]$
$f \geq 0 \Rightarrow \int_{a}^{b} f \geq 0$	Positivity Rule	$a < b$
$f \geq g \Rightarrow \int_{a}^{b} f \geq \int_{a}^{b} g$	Comparison Rule	$a < b$
$m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$	Bounding Rule	$m \leq f \leq M$ on $[a, b]$ , $a < b$

In this section, we prove some key properties of definite integrals that will help us evaluate them in an easy way. Most of these properties are analogous to the properties of the summation process.

The Four Algebraic Properties

  
      ∫
      
        a
      
        b
      
      [
      f
      (
      x
      )
      +
      g
      (
      x
      )
      ]
    
    d
    x
    =
    
      ∫
      
        a
      
        b
      
    f
    (
    x
    )
    
    d
    x
    +
    
      ∫
      
        a
      
        b
      
    g
    (
    x
    )
    
    d
    x
  
      ∫
      
        a
      
        b
      
    c
    f
    (
    x
    )
    
    d
    x
    =
    c
    
      ∫
      
        a
      
        b
      
    f
    (
    x
    )
    
    d
    x
  
, where c is a constant.

In particular, 
  
      ∫
      
        a
      
        b
      
    −
    f
    (
    x
    )
    
    d
    x
    =
    −
    
      ∫
      
        a
      
        b
      
    f
    (
    x
    )
    
    d
    x
  
.

      ∫
      
        a
      
        b
      
    c
    
    d
    x
    =
    c
    (
    b
    −
    a
    )
  
, where c is a constant.

      ∫
      
        a
      
        c
      
    f
    (
    x
    )
    
    d
    x
    +
    
      ∫
      
        c
      
        b
      
    f
    (
    x
    )
    
    d
    x
    =
    
      ∫
      
        a
      
        b
      
    f
    (
    x
    )
    
    d
    x
  
, whether c is between a and b or outside.

Property 1: The Sum Rule

Property (1) says that the integral of a sum is the sum of the separate integrals. The truth of this formula follows at once from the definition of the definite integral and the corresponding property of the summation process:

\sum_{i = 1}^{n} (a_{i} + b_{i}) = \sum_{i = 1}^{n} a_{i} + \sum_{i = 1}^{n} b_{i} .

Areas under f and under g stacking to give the area under f plus g — The figure shows $\int_{a}^{b} (f + g) = \int_{a}^{b} f + \int_{a}^{b} g$ .

Property 2: The Constant Multiple Rule

Property (2) says that we can move a constant factor outside the integral sign. Note that c must be a constant, not a function of x. To prove it, we use the definition of the definite integral and the following property of sums:

\sum_{i = 1}^{n} c a_{i} = c \sum_{i = 1}^{n} a_{i} .

Together, Properties 1 and 2 say that the definite integral is a linear operation. In practice, this is what lets us break a complicated integrand into manageable pieces.

Property 3: The Constant Function Rule

Property (3) says that the integral of a constant function $f (x) = c$ from a to b is $c (b - a)$ . This is expected, because if $a < b$ , the signed area of the shaded rectangle in the figure is (height) $\times$ (width) $= c (b - a)$ . If $c$ is negative, the rectangle lies below the axis and the signed area is negative, exactly as the formula reports.

Rectangle of height c between x equals a and x equals b representing the integral of a constant — $\int_{a}^{b} c d x$ is the signed area of the region between $y = c$ and the $x$ -axis from $x = a$ to $x = b$ , namely $c (b - a)$ .

Property 4: The Additivity Rule

Property (4) says that to integrate a function $f$ from a to b, we can integrate it once from a to c, once from c to b, and then add the results together.

If $a < c < b$ , the truth of Property (4) can be shown graphically. It is also easy to see from the definition of the definite integral that in this case

\int_{a}^{b} f (x) d x = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x .

Region under a curve split at an interior point c into two adjacent regions — Splitting the region at $x = c$ splits the integral.

What is less obvious is that the rule still holds when c lies outside $[a, b]$ . Suppose $a < b < c$ . From the case just proved,

\int_{a}^{c}f(x)\,dx=\int_{a}^{b}f(x)\,dx+\int_{b}^{c}f(x)\,dx. \tag{i}

Because we defined

\int_{b}^{c} f (x) d x = - \int_{c}^{b} f (x) d x,

we can rewrite (i) as

\int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x - \int_{c}^{b} f (x) d x .

Taking the last integral to the left, we get

\int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x = \int_{a}^{b} f (x) d x .

The truth of the other cases, for example if $c < a < b$ , can be shown similarly. This is the payoff of the reversal rule from the previous section: once signed intervals are allowed, additivity holds with no restriction on the position of c.

In an example in the previous section, we found $\int_{1}^{3} x^{2} d x = \frac{26}{3}$ . Use the properties of the definite integral to evaluate

\int_{1}^{3} (2 - 5 x^{2}) d x .

Solution

\begin{aligned} \int_{1}^{3}(2-5x^{2})\,dx &=\int_{1}^{3}2\,dx+\int_{1}^{3}-5x^{2}\,dx &&{\small(\text{Property 1})}\\ &=\int_{1}^{3}2\,dx-5\int_{1}^{3}x^{2}\,dx &&{\small(\text{Property 2})}\\ &=2(3-1)-5\times\frac{26}{3} &&{\small(\text{Property 3 and the given integral})}\\ &=4-\frac{130}{3}\\ &=-\frac{118}{3}. \end{aligned}

Find $\int_{8}^{9} f (x) d x$ , given that $\int_{- 1}^{9} f (x) d x = - 2$ and $\int_{- 1}^{8} f (x) d x = 4$ .

Solution

Using Property 4, we have

\underset{- 2}{\underset{⏟}{\int_{- 1}^{9} f (x) d x}} = \underset{4}{\underset{⏟}{\int_{- 1}^{8} f (x) d x}} + \int_{8}^{9} f (x) d x .

Therefore,

\int_{8}^{9} f (x) d x = - 2 - 4 = - 6.

The Inequality Properties

Property (5), the Positivity Rule, says that if the graph of $f$ does not lie below the $x$ -axis, the signed area of the region lying between the curve $y = f (x)$ and the $x$ -axis is nonnegative.

Rigorously: if $f (x) \geq 0$ , then each term in $\sum_{k = 1}^{n} f (x_{k}^{*}) (x_{k} - x_{k - 1})$ is nonnegative (because $x_{k} > x_{k - 1}$ ), and so is its limit as $n \to \infty$ .

A nonnegative function whose region under the curve lies entirely above the x-axis — If the graph of $f$ does not lie below the $x$ -axis on $[a, b]$ , the net signed area of the region under its curve is nonnegative.

Property (6), the Comparison Rule, follows immediately: because $f (x) - g (x) \geq 0$ , Property 5 gives $\int_{a}^{b} [f (x) - g (x)] d x \geq 0$ , and Properties 1 and 2 let us split that into $\int_{a}^{b} f - \int_{a}^{b} g \geq 0$ .

If 
  m
  ≤
  f
  (
  x
  )
  ≤
  M
 for 
  a
  <
  x
  <
  b
, then

  m
  (
  b
  −
  a
  )
  ≤
  
    ∫
    
      a
    
    
      b
    
  
  f
  (
  x
  )
  
  d
  x
  ≤
  M
  (
  b
  −
  a
  )
  .

Property (7), the Bounding Rule, is the Comparison Rule applied twice, once against the constant function $m$ and once against the constant function $M$ . Geometrically it says that the region under the curve contains the rectangle of height $m$ and is contained in the rectangle of height $M$ , both on the same base of width $b - a$ .

A curve trapped between the horizontal lines y equals m and y equals M with the corresponding rectangles — The region under $y = f (x)$ is trapped between two rectangles of heights $m$ and $M$ .

This property is genuinely useful: it produces numerical bounds for integrals we cannot evaluate at all.

Show that

π \sqrt{2} \leq \int_{0}^{π} \sqrt{2 + \sin x} d x \leq π \sqrt{3} .

Solution

Because $0 \leq \sin x \leq 1$ on the interval $[0, π]$ , the maximum value of $\sqrt{2 + \sin x}$ on $[0, π]$ is $\sqrt{2 + 1} = \sqrt{3}$ and the minimum is $\sqrt{2 + 0} = \sqrt{2}$ . That is,

\sqrt{2} \leq \sqrt{2 + \sin x} \leq \sqrt{3} .

It follows from Property (7), with $m = \sqrt{2}$ , $M = \sqrt{3}$ , $a = 0$ and $b = π$ , that

\underset{\approx 4.44288}{\underset{⏟}{\sqrt{2} (π - 0)}} \leq \int_{0}^{π} \sqrt{2 + \sin x} d x \leq \underset{\approx 5.4414}{\underset{⏟}{\sqrt{3} (π - 0)}} .

The actual value of the integral is

\int_{0}^{π} \sqrt{2 + \sin x} d x \approx 5.0922,

which indeed lies between the two bounds.

Numerical evaluation of the integral of the square root of 2 plus sine x from 0 to pi — A numerical evaluation confirming the estimate.

Use Property (7) to estimate the integral

\int_{0.5}^{2} 3^{1 / x} d x .

Solution

On $[0.5, 2]$ the exponent $1 / x$ runs from $1 / 2$ (at $x = 2$ ) up to $2$ (at $x = 0.5$ ), and $3^{t}$ is increasing in t. Hence

3^{1 / 2} \leq 3^{1 / x} \leq 3^{1 / 0.5} = 3^{2} (0.5 \leq x \leq 2),

so that

\sqrt{3} (2 - 0.5) \leq \int_{0.5}^{2} 3^{1 / x} d x \leq 9 (2 - 0.5),

2.59808 \leq \int_{0.5}^{2} 3^{1 / x} d x \leq 13.5 .

The actual value of this integral is approximately $4.57596$ . Notice how wide the bracket is: the Bounding Rule is cheap but crude, because it replaces the function by its extreme values across the whole interval.

Numerical evaluation of the integral of 3 to the power 1 over x from 0.5 to 2 — A numerical evaluation confirming the estimate.

Exercises

Given $\int_{0}^{2} f (x) d x = 5$ and $\int_{0}^{2} g (x) d x = - 3$ , evaluate

(a) $\int_{0}^{2} [2 f (x) - g (x)] d x$

(b) $\int_{0}^{2} [f (x) + 4] d x$

(c) $\int_{2}^{0} 3 g (x) d x$

Answer

(a) $13$ (b) $13$ (c) $9$

Solution

(a) By Properties 1 and 2, $2 (5) - (- 3) = 13$ .

(b) By Properties 1 and 3, $5 + 4 (2 - 0) = 5 + 8 = 13$ .

(c) First reverse the limits, then factor the constant: $\int_{2}^{0} 3 g = - 3 \int_{0}^{2} g = - 3 (- 3) = 9$ .

Given $\int_{1}^{5} f (x) d x = 10$ and $\int_{1}^{3} f (x) d x = 4$ , find $\int_{3}^{5} f (x) d x$ and $\int_{5}^{3} f (x) d x$ .

Answer

$6$ and $- 6$ respectively.

Solution

By the Additivity Rule, $\int_{1}^{5} f = \int_{1}^{3} f + \int_{3}^{5} f$ , so $10 = 4 + \int_{3}^{5} f$ and $\int_{3}^{5} f = 6$ . Reversing the limits changes the sign, giving $\int_{5}^{3} f = - 6$ .

Suppose $\int_{0}^{4} h (x) d x = 7$ and $\int_{0}^{6} h (x) d x = 2$ . Find $\int_{6}^{4} h (x) d x$ .

Answer

$5$

Solution

Additivity gives $\int_{0}^{6} h = \int_{0}^{4} h + \int_{4}^{6} h$ , so $2 = 7 + \int_{4}^{6} h$ and $\int_{4}^{6} h = - 5$ . Reversing the limits,

\int_{6}^{4} h (x) d x = - \int_{4}^{6} h (x) d x = 5.

Note that $\int_{4}^{6} h$ came out negative even though nothing was said about the sign of h; the properties do all the work without any picture.

Use the Bounding Rule to show that

\frac{1}{2} \leq \int_{0}^{1} \frac{d x}{1 + x^{2}} \leq 1,

and compare with the exact value.

Answer

The exact value is $\frac{π}{4} \approx 0.785$ , comfortably inside $[\frac{1}{2}, 1]$ .

Solution

On $[0, 1]$ the denominator satisfies $1 \leq 1 + x^{2} \leq 2$ , so the integrand satisfies

\frac{1}{2} \leq \frac{1}{1 + x^{2}} \leq 1.

With $m = \frac{1}{2}$ , $M = 1$ , and $b - a = 1$ , Property 7 gives $\frac{1}{2} \leq \int_{0}^{1} \frac{d x}{1 + x^{2}} \leq 1$ .

The exact value is $\arctan 1 - \arctan 0 = \frac{π}{4} \approx 0.7854$ .

Without evaluating either integral, decide which is larger:

\int_{0}^{1} x^{2} d x or \int_{0}^{1} x^{3} d x .

Then decide the same question on the interval $[1, 2]$ .

Answer

On $[0, 1]$ the integral of $x^{2}$ is larger; on $[1, 2]$ the integral of $x^{3}$ is larger.

Solution

For $0 \leq x \leq 1$ we have $x^{3} \leq x^{2}$ , since multiplying by a number less than 1 shrinks it. The Comparison Rule then gives $\int_{0}^{1} x^{3} d x \leq \int_{0}^{1} x^{2} d x$ .

For $x \geq 1$ the inequality reverses: $x^{3} \geq x^{2}$ , so $\int_{1}^{2} x^{3} d x \geq \int_{1}^{2} x^{2} d x$ .

(For confirmation, the four values are $\frac{1}{3}$ , $\frac{1}{4}$ , $\frac{15}{4}$ , and $\frac{7}{3}$ .)

Show that $\int_{0}^{1} \sqrt{1 + x^{4}} d x \geq 1$ , and find an upper bound for the same integral.

Answer

$1 \leq \int_{0}^{1} \sqrt{1 + x^{4}} d x \leq \sqrt{2} \approx 1.414$ .

Solution

On $[0, 1]$ we have $0 \leq x^{4} \leq 1$ , hence $1 \leq 1 + x^{4} \leq 2$ and, taking square roots (which preserves the inequality for nonnegative numbers),

1 \leq \sqrt{1 + x^{4}} \leq \sqrt{2} .

By Property 7 with $b - a = 1$ , the integral lies between $1$ and $\sqrt{2}$ .

The lower bound can also be obtained from the Comparison Rule directly: $\sqrt{1 + x^{4}} \geq \sqrt{1} = 1$ and $\int_{0}^{1} 1 d x = 1$ .

Explain why the following calculation is wrong, and give the correct value:

\int_{0}^{2} x \cdot x d x = x \int_{0}^{2} x d x = x \cdot 2 = 2 x .

Answer

The Constant Multiple Rule applies only to constants, not to x. The correct value is $\frac{8}{3}$ .

Solution

Property 2 permits pulling out a factor only if that factor does not depend on the variable of integration. Here the factor pulled out is $x$ itself, which varies over the interval, so the step is invalid. A second warning sign is that the "answer" $2 x$ still contains $x$ , whereas a definite integral is always a number.

Correctly, $x \cdot x = x^{2}$ and

\int_{0}^{2} x^{2} d x = {[\frac{x^{3}}{3}]}_{0}^{2} = \frac{8}{3} .

Suppose $f$ is continuous on $[a, b]$ , that $f (x) \geq 0$ there, and that $\int_{a}^{b} f (x) d x = 0$ with $a < b$ . What can you conclude about $f$ ?

Answer

$f (x) = 0$ for every $x$ in $[a, b]$ .

Solution

Suppose, to the contrary, that $f (x_{0}) > 0$ for some $x_{0}$ in $[a, b]$ . Because $f$ is continuous, there is a small interval $[c, d]$ around $x_{0}$ inside $[a, b]$ on which $f (x) \geq \frac{f (x_{0})}{2} > 0$ .

Splitting the integral with the Additivity Rule and using positivity on the outer pieces,

\int_{a}^{b} f = \underset{\geq 0}{\underset{⏟}{\int_{a}^{c} f}} + \int_{c}^{d} f + \underset{\geq 0}{\underset{⏟}{\int_{d}^{b} f}} \geq \int_{c}^{d} f \geq \frac{f (x_{0})}{2} (d - c) > 0,

where the last inequality is the Bounding Rule. This contradicts the assumption that the integral is zero. Hence no such $x_{0}$ exists and $f$ is identically zero.

Continuity is essential: a function that is zero everywhere except at a single point still has integral zero without being identically zero.

Prove the inequality

| \int_{a}^{b} f (x) d x | \leq \int_{a}^{b} | f (x) | d x (a < b) .

Answer

Apply the Comparison Rule to the double inequality $- | f | \leq f \leq | f |$ .

Solution

For every $x$ , the number $f (x)$ satisfies

- | f (x) | \leq f (x) \leq | f (x) | .

Integrating each part from a to b and using the Comparison Rule twice,

- \int_{a}^{b} | f (x) | d x \leq \int_{a}^{b} f (x) d x \leq \int_{a}^{b} | f (x) | d x,

where the left-hand integral was rewritten with Property 2. A quantity trapped between $- K$ and $K$ has absolute value at most $K$ , so

| \int_{a}^{b} f (x) d x | \leq \int_{a}^{b} | f (x) | d x .

Interpreted geometrically, this says the net area can never exceed the total area, which is the integral version of the triangle inequality.

A continuous function $f$ satisfies $2 \leq f (x) \leq 5$ on $[1, 7]$ . What are the smallest and largest possible values of $\int_{1}^{7} f (x) d x$ , and what does each extreme case require of $f$ ?

Answer

Between $12$ and $30$ ; the extremes occur only for the constant functions $f \equiv 2$ and $f \equiv 5$ .

Solution

Here $b - a = 7 - 1 = 6$ , so Property 7 gives

2 (6) \leq \int_{1}^{7} f (x) d x \leq 5 (6), that is 12 \leq \int_{1}^{7} f (x) d x \leq 30.

The upper bound is attained only if $f$ equals 5 throughout. Indeed, if $\int_{1}^{7} f = 30$ , then $\int_{1}^{7} [5 - f (x)] d x = 0$ with a continuous, nonnegative integrand, so by the previous exercise $5 - f (x) \equiv 0$ . The same argument with $f (x) - 2$ handles the lower bound.

The Four Algebraic Properties

Property 1: The Sum Rule

Property 2: The Constant Multiple Rule

Property 3: The Constant Function Rule

Property 4: The Additivity Rule

The Inequality Properties

Exercises

Quick Links

Social Media