tgt

## Monday, 4 August 2014

### CHAPTER 2- Basic Properties

1
Let us consider the integral of ${f_1}\left( x \right) + {f_2}\left( x \right)$ 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:
 $\int\limits_a^b {kf\left( x \right)dx\,\, = \,\,} k\int\limits_a^b {f\left( x \right)dx}$
2
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 $x = 2a$ 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.
(3) 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
 ${f_{av}}\left( {b - a} \right) = \int\limits_a^b {f\left( x \right)dx}$ $\Rightarrow\,\,\,\, {f_{av}} = \dfrac{1}{{b - a}}\int\limits_a^b {f\left( x \right)dx}$
This value is attained for at least one $c \in \left( {a,\,\,b} \right)$ (under the constraint that $f$ is continuous, of course).