# Basic Properties of Definite Integrals

As explained in the chapter titled “**Integration Basics**”, the fundamental theorem of calculus tells us that to evaluate the area under a curve \(y = f\left( x \right)\) from \(x = a\,\,{\rm{to}}\,\,x = b\) , we first evaluate the anti-derivative \(g\left( x \right)\,\,{\rm{of}}\,f\left( x \right)\)

\[g\left( x \right)= \int {f\left( x \right)dx} \]

and then evaluate \(g\left( b \right) - g\left( a \right).\) That is, area under the curve *f*(*x*) from *\(x = a\)* to *\(x = b\)* is

\[\int\limits_a^b {f\left( x \right)dx = g\left( b \right) - g\left( a \right)} \]

Readers who have even the slightest doubt regarding the discussion above are advised to refer to the chapter on “**Integration Basics**” before reading on.

Definite integration is not all about just evaluating the anti-derivative and substituting the upper and lower limits. Working through this chapter, you will realise that a lot of techniques exist which help us in evaluating the definite integral without resorting to the (many times tedious) process of first determining the anti-derivative. We will develop all these techniques one by one from scratch, starting with some extremely basic properties in Section – 1

# Section - 1 BASIC PROPERTIES

**(1) ** Suppose that *f*(*x*) < 0 on some interval [*a*, *b*]. Then, the area under the curve *y* = *f*(*x*) from *\(x = a\)** *to *\(x = b\)** *will be negative in sign, i.e

\[\int\limits_a^b {f\left( x \right)dx < 0} \]

This is obvious once you consider how the definite integral was arrived at in the first place; as a limit of the sum of the *n* rectangles \(\left( {n \to \infty } \right)\) . Thus, if *f*(*x*) < 0 in some interval then the area of the rectangles in that interval will also be negative.

This property means that for example, if \(f(x)\) has the following form

then \(\int\limits_a^b {f\left( x \right)dx\,\,{\rm{will}}\,\,{\rm{equal}}\,{A_1} - {A_2} + {A_3} - {A_4}\,\,{\rm{and}}\,\,{\rm{not}}\,} {A_1} + {A_2} + {A_3} + {A_4}.\)

If we need to evaluate \({A_1} + {A_2} + {A_3} + {A_4}\) (the magnitude of the bounded area), we will have to calculate

\[\int\limits_a^x {f\left( x \right)dx + \left| {\int\limits_x^y {f\left( x \right)dx} } \right|} + \int\limits_y^z {f\left( x \right)dx + \left| {\int\limits_z^b {f\left( x \right)dx} } \right|} \]

From this, it should also be obvious that \(\left| {\int\limits_a^b {f\left( x \right)dx} } \right| \le \,\,\int\limits_a^b {\left| {f\left( x \right)} \right|dx} \)

**(2) ** The area under the curve *y* =* f*(*x*) from *x* = *a* to *x* = *b* is equal in magnitude but opposite in sign to the area under the same curve from *x* = *b* to *x* = *a*, i.e

\[\int\limits_a^b {f\left( x \right)dx} = - \int\limits_b^a {f\left( x \right)dx} \]

This property is obvious if you consider the Newton-Leibnitz formula. If \(g\left( x \right)\) is the anti-derivative of *x*(*f*), then \(\int\limits_a^b {f\left( x \right)} \,dx\) is \(g\left( b \right) - g\left( a \right)\,\,{\rm{while}}\,\,\int\limits_b^a {f\left( x \right)dx} \,\,{\rm{is}}\,\,g\left( a \right) - g\left( b \right)\) .

**(3) ** The area under the curve *y* = *f*(*x*) from *x* = *a* to *x* = *b* can be written as the sum of the area under the curve from *x* = *a* to *x* = *c* and from *x* = *c* to *x* = *b*, that is

\[\int\limits_a^b {f\left( x \right)dx = \int\limits_a^c {f\left( x \right)dx + \,} \int\limits_c^b {f\left( x \right)dx} } \]

Let us consider an example of this. Let \(c \in (a,\,\,b)\)

It is clear that the area under the curve from *x* = *a* to *x* = *b*, *A* is \({{A}_{1}}+{{A}_{2}}\)

Note that *c* need not lie between *a* and *b* for this relation to hold true. Suppose that *c* > *b*.

Observe that \(A = \int\limits_a^b {f\left( x \right)dx = \left( {A + {A_1}} \right) - {A_1}} \)

\[\begin{array}{l}= \int\limits_a^c {f\left( x \right)dx - \int\limits_b^c {f\left( x \right)dx} } \\= \int\limits_a^c {f\left( x \right)dx + \int\limits_c^b {f\left( x \right)dx} }\end{array}\]

Analytically, this relation can be proved easily using the Newton Leibnitz’s formula.

**(4) ** Let \(f\left( x \right) > g\left( x \right)\) on the interval [*a*, *b*]. Then,

\[\int\limits_a^b {f\left( x \right) > \int\limits_a^b {g\left( x \right)} \,dx.} \]

This is because the curve of *f*(*x*) lies above the curve of *g*(*x*), or equivalently, the curve of \(f\left( x \right) - g\left( x \right)\) lies above the *x*-axis for [*a*, *b*]

Similarly, if \(f\left( x \right) < g\left( x \right)\,\) on the interval [*a*, *b*], then

\[\int\limits_a^b {f\left( x \right)dx < } \,\int\limits_a^b {g\left( x \right)dx} \]

**(5) ** For the interval [*a*, *b*], suppose *m* < *f*(*x*) < *M*. That is, *m* is a lower-bound for *f*(*x*) while *M* is an upper bound.

Then,

\[m\left( {b - a} \right)\,\, < \int\limits_a^b {f\left( x \right)dx\,\, < \,\,\,M\left( {b - a} \right)} \]

This is obvious once we consider the figure below:

Observe that \({\rm{area}}\left( {{\rm{rect}}\,AXYB} \right) < \int\limits_a^b {f\left( x \right)dx < {\rm{area}}\left( {{\rm{rect}}\,DXYC} \right)} \)

**(6) ** Let us consider the integral of \({f_1}\left( x \right) + {f_2}\left( x \right)\,\,{\rm{from}}\,\,x = a\,\,{\rm{to}}\,\,x = b\) . To evaluate the area under \({f_1}\left( x \right) + {f_2}\left( x \right),\) we can separately evaluate the area under \({{f}_{1}}(x)\) and the area under \({{f}_{2}}(x)\) and add the two areas (algebraically). Thus:

\[\int\limits_a^b {\left( {{f_1}\left( x \right) + {f_2}\left( x \right)} \right)\,dx\,\, = } \,\,\int\limits_a^b {{f_1}\left( x \right)\,dx\,\, + } \,\,\int\limits_a^b {{f_2}\left( x \right)\,dx} \]

Now consider the integral of *kf*(*x*) from *x* = *a* to *x* = *b*. To evaluate the area under *kf*(*x*), we can first evaluate the area under *f*(*x*) and then multiply it by *k*, that is:

\[\int\limits_a^b {kf\left( x \right)dx\,\, = \,\,} k\int\limits_a^b {f\left( x \right)dx} \]

**(7) **Consider an odd function *f*(*x*), \(i.e,f\left( x \right) = - f\left( { - x} \right).\) This means that the graph of *f*(*x*) is symmetric about the origin.

From the figure, it should be obvious that \(\int\limits_{ - a}^a {f\left( x \right)dx = 0,} \) because the area on the left side and that on the right algebraically add to 0.

Similarly, if *f*(*x*) was even, \(i.e,\,\,f\left( x \right) = f\left( { - x} \right)\)

\(\int\limits_{ - a}^a {f\left( x \right)dx = \,\,2} \int\limits_0^a {f\left( x \right)\,\,dx} \) because the graph is symmetrical about the *y*-axis.

If you recall the discussion in the unit on functions, a function can also be even or odd about any arbitrary point *x* = *a .* Let us suppose that

*f*(

*x*) is odd about

*x*=

*a*, i.e

\[f\left( x \right) = - f\left( {2a - x} \right)\]

Suppose for example, that we need to calculate \(\int\limits_0^{2a} {f\left( x \right)dx.} \) It is obvious that this will be 0, since we are considering equal variation on either side of *x* = *a*, i.e. the area from *x* = 0 to *x* = *a* and the area from *x *= *a* to *x *= 2*a* will add algebraically to 0.

Similarly, if *f*(*x*) is even about *x* = *a*, i.e.

\[f\left( x \right) = f\left( {2a - x} \right)\]

then we have, for example

\[\int\limits_0^{2a} {f\left( x \right)dx = 2\int\limits_0^a {f\left( x \right)dx} } \]

From this discussion, you will get a general idea as to how to approach such issues regarding even/odd functions.

**(8) ** Let us consider a function *f*(*x*) on [*a*,* b*]

We want to somehow define the “average” value that *f*(*x*) takes on the interval [*a*, *b*]. What would be an appropriate way to define such an average?

Let \({f_{av}}\) be the average value that we are seeking. Let it be such that it is obtained at some \(x = c\,\, \in [a,\,\,b]\)

We can measure \({f_{av}}\) by saying that the area under *f*(*x*) from *x* = *a* to *x* = *b* should equal the area under the average value from *x* = *a* to *x* = *b*. This seems to be the only logical way to define the average (and this is how it is actually defined!). Thus

\[\begin{array}{l} \,\,\,\,\,\,\,\,{f_{av}}\left( {b - a} \right) = \int\limits_a^b {f\left( x \right)dx} \\\Rightarrow {f_{av}} = \frac{1}{{b - a}}\int\limits_a^b {f\left( x \right)dx}\end{array}\]

This value is attained for at least one \(c \in \left( {a,\,\,b} \right)\) (under the constraint that *f* is continuous, of course).