ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 2- Basic Properties

Monday 4 August 2014

CHAPTER 2- Basic Properties

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:

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

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).

ONE MORE STEP TOWARD BETTER TOMORROW

Monday 4 August 2014

CHAPTER 2- Basic Properties

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