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New Zealand
Level 8 - NCEA Level 3

Properties of Definite Integrals


The definite integral arose from problems that can be conceived as problems of calculating an area. For example, in kinematics we calculate a displacement as an average velocity multiplied by the time interval. We write $s=\overline{v}t$s=vt and this can be pictured as in the following graph.

If the velocity varied in such a way that the graph was a straight line, there would be no difficulty in working out the average velocity. This would be the constant acceleration situation where the average velocity on some time interval is just the arithmetic mean of the initial and final velocities.

In this case, the acceleration is not constant and the rule does not apply. However, the rule does give the correct result over a very small time interval during which the acceleration is almost constant.

The idea that follows from this is that the overall time interval $(a,b)$(a,b) can be divided into many small sub-intervals, and the displacement calculated for each of them. In the limit, the sum of the many vanishingly small displacements will then be the total displacement. In symbols, we may write

$s=\lim_{\delta t\rightarrow0}\sum_a^bv(t)\delta t$s=limδt0bav(t)δt

This is equivalent to dividing the area between the curve and the horizontal axis into many thin strips. Each strip has height $v(t)$v(t) for a particular $t$t and it has width $\delta t$δt so that the area of a strip$v(t)\delta t$v(t)δt  is the displacement over that small time interval.

The notation used for this infinite sum of vanishingly small areas is the integral sign.

$s=\int_a^bv(t)\ \mathrm{d}t$s=bav(t) dt


The important discovery of Newton and Leibniz was that the area problem is the inverse of the gradient problem. In the case described above, the area under the velocity curve is related to the gradient of the displacement function. The relationship is known as the fundamental theorem of integral calculus:

  1. If a function $F$F has derivative $f$f then $F$F is called an anti-derivative for $f$f and $F(x)=\int f(x)\ \mathrm{d}x+c$F(x)=f(x) dx+c.
  2. The area under the graph of a function $f$f between $a$a and $b$b is given by $\int_a^bf(t)\ \mathrm{d}t=F(b)-F(a)$baf(t) dt=F(b)F(a)

The numbers $a$a and $b$b are called the terminals of the definite integral.


some properties

1.   We see that if the upper terminal is variable, say $b=x$b=x, then we have a function $A(x)=\int_a^xf(t)\ \mathrm{d}t=F(x)-F(a)$A(x)=xaf(t) dt=F(x)F(a).

The derivative of this function $A'(x)$A(x) must be the derivative of $F(x)-F(a)$F(x)F(a) which is just $f(x)$f(x).


2.   Since $\int_a^bf(t)\ \mathrm{d}t=F(b)-F(a)$baf(t) dt=F(b)F(a), we have $\int_b^af(t)\ \mathrm{d}t=F(a)-F(b)$abf(t) dt=F(a)F(b) and it follows that 

$\int_b^af(t)\ \mathrm{d}t=-\int_a^bf(t)\ \mathrm{d}t$abf(t) dt=baf(t) dt


3.   The definite integral of a sum is a sum of definite integrals:  $\int_a^b\left(f(t)+g(t)\right)\ \mathrm{d}t=\int_a^bf(t)\ \mathrm{d}t+\int_a^bg(t)\ \mathrm{d}t$ba(f(t)+g(t)) dt=baf(t) dt+bag(t) dt.

4.   The definite integral of a function, could be split into two sections and then we can find the sum of the two parts. 

$\int_a^b(f(t)\ \mathrm{d}t=\int_a^cf(t)\ \mathrm{d}t+\int_c^bf(t)\ \mathrm{d}t$ba(f(t) dt=caf(t) dt+bcf(t) dt


5.   Depending on the nature of the particular problem, it may be necessary to integrate the absolute values $|f(x)|$|f(x)|. This could be to avoid the effect of the function going below the horizontal axis. For example, a negative area may not be desirable.

If $f(x)$f(x) is positive between $a$a and $b$b but negative between $b$b and $c$c, we may wish to calculate $\int_a^c|f(x)|\ \mathrm{d}x$ca|f(x)| dx.  To do this, we calculate $\int_a^bf(x)\ \mathrm{d}x+\int_c^bf(x)\ \mathrm{d}x$baf(x) dx+bcf(x) dx. (The terminals have been reversed on the second integral.)



The velocity-time graph at the beginning of this chapter represents the function $v(t)=-0.75t+\frac{\cos3t}{3}+4$v(t)=0.75t+cos3t3+4 between $t=0$t=0 and $t=5$t=5. What is the displacement at $t=5$t=5 and what was the average velocity over the interval $(0,5)$(0,5)?

We require $s=\int_0^5\ \left(-0.75t+\frac{\cos3t}{3}+4\right)\ \mathrm{d}t$s=50 (0.75t+cos3t3+4) dt. This is, $\left[\frac{-0.75}{2}t^2+\frac{\sin3t}{9}+4t\right]_0^5\approx10.697$[0.752t2+sin3t9+4t]5010.697.

Given $s=\overline{v}t$s=vt, we have $10.697=\overline{v}\times5$10.697=v×5. Therefore $\overline{v}\approx2.139$v2.139.


Worked Examples

Question 1

Consider the function $f\left(x\right)=6x$f(x)=6x.

  1. Find the definite integral, $\int_4^8f\left(x\right)dx$84f(x)dx.

  2. Find the definite integral, $\int_8^4f\left(x\right)dx$48f(x)dx.

  3. Which property of definite integrals do parts (a) and (b) demonstrate?


Question 2

Consider the function $f\left(x\right)$f(x) as shown. The numbers in the shaded regions indicate the area of the region.

  1. Determine $\int_{-3}^0f\left(x\right)dx$03f(x)dx.

  2. Determine the area enclosed by the curve and the $x$x-axis for $x<0$x<0.

  3. Determine $\int_{-3}^2f\left(x\right)dx$23f(x)dx.

  4. Determine the area enclosed by the curve and the $x$x-axis.



Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods


Apply integration methods in solving problems

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