Question

Find the definite integral of $\left( {\dfrac{1}{{{x^2}}}} \right)$from 0 to 1.

Answer 1

Hint:We know that definite integral means or represents the area under a curve which has a boundary. It differs from indefinite integral on the fact that, in indefinite integral there are no specified boundary conditions. So by using the above definition of definite integral we can find the value of $\left( {\dfrac{1}{{{x^2}}}}
\right)$from 0 to 1.

Complete step by step solution:
Given $\left( {\dfrac{1}{{{x^2}}}} \right).........................................\left( i \right)$

Now we need to integrate $\left( {\dfrac{1}{{{x^2}}}} \right)$ from 0 to 1.

Also by the basic definition of definite integral we can write that:
For an interval $\left[ {a,b} \right]$ Definite integral is given by: $\int\limits_a^b {f\left( x \right)dx}
....................\left( {ii} \right)$

Where \[f\left( x \right)dx\] represents the area under the curve and $\left[ {a,b} \right]$represents the boundary of the curve.

So according to our question $\left[ {a,b} \right]$would be $\left[ {0,1} \right]$and $f\left( x
\right)$would be$\left( {\dfrac{1}{{{x^2}}}} \right)$.

So in order to find the definite integral of our question we need to substitute the above values in (ii).
Such that we get:
$
\int\limits_a^b {f\left( x \right)dx} = \int\limits_0^1 {\left( {\dfrac{1}{{{x^2}}}} \right)}
dx.................\left( {iii} \right) \\
\\
$
So in order to solve (iii) we need to first integrate $\left( {\dfrac{1}{{{x^2}}}} \right)$and then apply the upper and lower limits.

Now we know the power rule of integration which is:$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $

Therefore using power rule of integration to integrate $\left( {\dfrac{1}{{{x^2}}}} \right)$which can also
be written as$\left( {{x^{ - 2}}} \right)$, and then giving the upper and lower limits.
\[
\int\limits_0^1 {\left( {\dfrac{1}{{{x^2}}}} \right)dx\; = } \int\limits_0^1 {\left( {{x^{ - 2}}} \right)dx}
\\
= \left[ {\dfrac{{{x^{ - 2 + 1}}}}{{ - 2 + 1}}} \right]_0^1 \\
= \left[ {\dfrac{{{x^{ - 1}}}}{{ - 1}}} \right]_0^1 \\
= \left[ { - {x^{ - 1}}} \right]_0^1 \\
= \left[ { - \left( {\dfrac{1}{x}} \right)} \right]_0^1 \\
\]
Now giving the upper and lower limits we get:
\[
\left[ { - \left( {\dfrac{1}{x}} \right)} \right]_0^1 = \left[ {\left( { - \dfrac{1}{1}} \right) - \left( { -
\dfrac{1}{0}} \right)} \right] \\
= \left[ { - 1 + \left( {\dfrac{1}{0}} \right)} \right] \\
\]
But $\dfrac{1}{0} = \infty $
$ \Rightarrow \int\limits_0^1 {\left( {\dfrac{1}{{{x^2}}}} \right)dx = } \left[ { - 1 + \infty } \right] =
\infty .....................\left( {iv} \right)$

(iv) implies that the integral is infinity or in short the integral diverges.
So since the integral diverges it can be concluded that the definite integral doesn’t exist since by its definition the definite integral is an area and should be finite in nature.

Therefore $\int\limits_0^1 {\left( {\dfrac{1}{{{x^2}}}} \right)dx} $doesn’t exist.

Note: Since the basic definition a definite or indefinite integral simply implies the area under a curve the value of an integral must be finite or else the integral doesn’t exist as in the above question. Also the above integral can be called an improper integral since the value of the integral is not defined on one of the boundary limits.