Question

How do you evaluate the definite integral by the limit definition given $ \int {{x^3}dx} $ from $ [ - 1,1] $ ?

Answer 1

Hint:Here in this question we apply the integration to the function by the limit definition. The integral is a definite integral. In the definite integral the value of x ranges will be given. The formula for integration by the limit definition is defined as $ \int\limits_a^b {f(x)dx = \mathop {\lim }\limits_{n \to
\infty } } \sum\limits_{i = 1}^n {f({x_i})\Delta x} $

Complete step by step explanation:
Now consider the given integral $ \int\limits_{ - 1}^1 {{x^3}dx} $ , this is a definite integral.
The formula for integration by limit definition we have $ \int\limits_a^b {f(x)dx = \mathop {\lim
}\limits_{n \to \infty } } \sum\limits_{i = 1}^n {f({x_i})\Delta x} $ where $ \Delta x = \dfrac{{b - a}}{n} $
, a=-1 and b=1 and $ i = 1,2,3,...,n $ let $ {x_i} = a + i\Delta x $ , $ {x_i} $ are the endpoints of the
subintervals.
Now we will find the value of $ \Delta x $
 $
\Delta x = \dfrac{{b - a}}{n} = \dfrac{{1 - ( - 1)}}{n} \\
\Rightarrow \Delta x = \dfrac{{1 + 1}}{n} = \dfrac{2}{n} \\
 $
Now we will find the $ {x_i} $ and it is given by $ {x_i} = a + i\Delta x $
 $
{x_i} = - 1 + i\left( {\dfrac{2}{n}} \right) \\
\Rightarrow {x_i} = \dfrac{{2i}}{n} - 1 \\
 $
The value of $ f({x_i}) $ is determined by
 $ f({x_i}) = f\left( {\dfrac{{2i}}{n} - 1} \right) $ Here $ f(x) = {x^3} $ , so we have
 $ \Rightarrow f({x_i}) = {\left( {\dfrac{{2i}}{n} - 1} \right)^3} $
Use the formula $ {(a - b)^3} = {a^3} - {b^3} - 3ab(a - b) $
 $
\Rightarrow f({x_i}) = {\left( {\dfrac{{2i}}{n}} \right)^3} - {1^3} - 3\left( {\dfrac{{2i}}{n}} \right)\left( 1
\right)\left( {\dfrac{{2i}}{n} - 1} \right) \\

\Rightarrow f({x_i}) = \dfrac{{8{i^3}}}{{{n^3}}} - 1 - \dfrac{{6i}}{n}\left( {\dfrac{{2i}}{n} - 1} \right)
\\
\Rightarrow f({x_i}) = \dfrac{{8{i^3}}}{{{n^3}}} - 1 - \left( {\dfrac{{12{i^2}}}{{{n^2}}}} \right) +
\dfrac{{6i}}{n} \\
\Rightarrow f({x_i}) = \dfrac{{8{i^3}}}{{{n^3}}} - 1 - \dfrac{{12{i^2}}}{{{n^2}}} + \dfrac{{6i}}{n} \\
 $
Now we find the summation for $ f({x_i}) $ and multiply by $ \Delta x $ we have
 $ \sum\limits_{i = 1}^n {f({x_i})\Delta x} = \sum\limits_{i = 1}^n {\left( {\dfrac{{8{i^3}}}{{{n^3}}} - 1 -
\dfrac{{12{i^2}}}{{{n^2}}} + \dfrac{{6i}}{n}} \right)} \left( {\dfrac{2}{n}} \right) $
On multiplication,
 $ \Rightarrow \sum\limits_{i = 1}^n {f({x_i})\Delta x} = \sum\limits_{i = 1}^n {\left(
{16\dfrac{{{i^3}}}{{{n^4}}} - \dfrac{2}{n} - 24\dfrac{{{i^2}}}{{{n^3}}} + 12\dfrac{i}{{{n^2}}}} \right)} $
Apply the summation to each and every term we have
 $ \Rightarrow \sum\limits_{i = 1}^n {f({x_i})\Delta x} = - \dfrac{2}{n}\sum\limits_{i = 1}^n 1 +
\dfrac{{12}}{{{n^2}}}\sum\limits_{i = 1}^n i - \dfrac{{24}}{{{n^3}}}\sum\limits_{i = 1}^n {{i^2}} +
\dfrac{{16}}{{{n^4}}}\sum\limits_{i = 1}^n {{i^3}} $
Use the formula of summations
 $ \sum\limits_{i = 1}^n 1 = n $ , $ \sum\limits_{i = 1}^n i = \dfrac{{n(n + 1)}}{2} $ , $ \sum\limits_{i = 1}^n
{{i^2}} = \dfrac{{n(n + 1)(2n + 1)}}{6} $ and $ \sum\limits_{i = 1}^n {{i^4}} = \dfrac{{{n^2}{{(n +
1)}^2}}}{4} $
And substituting
 $ \Rightarrow \sum\limits_{i = 1}^n {f({x_i})\Delta x} = - \dfrac{2}{n}n + \dfrac{{12}}{{{n^2}}}\left(
{\dfrac{{n(n + 1)}}{2}} \right) - \dfrac{{24}}{{{n^3}}}\left( {\dfrac{{n(n + 1)(2n + 1)}}{6}} \right) +
\dfrac{{16}}{{{n^4}}}\left( {\dfrac{{{n^2}{{(n + 1)}^2}}}{4}} \right) $
Now take n as common and simplify
 $ \Rightarrow \sum\limits_{i = 1}^n {f({x_i})\Delta x} = - 2 + 6\left( {1 + \dfrac{1}{n}} \right) - 4\left(
{\left( {1 + \dfrac{1}{n}} \right)\left( {2 + \dfrac{1}{n}} \right)} \right) + 4{\left( {1 + \dfrac{1}{n}}
\right)^2} $
Now we apply limit to above equation we have
 $ \int\limits_a^b {f(x)dx = \mathop {\lim }\limits_{n \to \infty } } \sum\limits_{i = 1}^n {f({x_i})\Delta
x} $

 $ \Rightarrow \int\limits_{ - 1}^1 {{x^3}dx = \mathop {\lim }\limits_{n \to \infty } } \sum\limits_{i =
1}^n { - 2 + 6\left( {1 + \dfrac{1}{n}} \right) - 4\left( {\left( {1 + \dfrac{1}{n}} \right)\left( {2 +
\dfrac{1}{n}} \right)} \right) + 4{{\left( {1 + \dfrac{1}{n}} \right)}^2}} $
On applying limit we have
 $ \Rightarrow \int\limits_{ - 1}^1 {{x^3}dx} = - 2 + 6\left( {1 + 0} \right) - 4\left( {\left( {1 + 0}
\right)\left( {2 + 0} \right)} \right) + 4{\left( {1 + 0} \right)^2} $
As we know that any number divided by the infinity the answer will be zero.
On further simplification we have
 $
\Rightarrow \int\limits_{ - 1}^1 {{x^3}dx} = - 2 + 6 - 8 + 4 \\
\Rightarrow \int\limits_{ - 1}^1 {{x^3}dx} = 0 \\
 $

Hence we found the value of definite integral by the limit definition.

Therefore $ \int\limits_{ - 1}^1 {{x^3}dx} = 0 $

Note: We have two types of integrals one is definite integral and another is indefinite integral. The definite integral where the limit points of the integration are mentioned and whereas in the indefinite integral the limits points of integration is not mentioned. The formula for the integration by the limit definition is given by $ \int\limits_a^b {f(x)dx = \mathop {\lim }\limits_{n \to \infty } } \sum\limits_{i =
1}^n {f({x_i})\Delta x} $ . This is the lengthy procedure we can obtain the solution only in two to three steps.