Question

How to find the definite integral using the limit definition? $\int_1^4 {\left( {{x^3} - 4} \right)dx}$

Answer 1

If a function $f(x)$ is continuous on the interval $[a,\;b]$ then we will divide the interval into “n” subintervals of equal width, $\Delta x$ , and in each interval we will choose a point, ${x_i}$ .

Then the definite integral will be given as
$\int_a^b {f(x)dx = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f({x_i})\Delta x} }$
Where $\Delta x$ is given as $\dfrac{{b - a}}{n}$
Solve this as $\int_1^4 {\left( {{x^3} - 4} \right)dx} = \int_0^4 {\left( {{x^3} - 4} \right)dx} - \int_0^1 {\left( {{x^3} - 4} \right)dx}$

Complete step by step solution:
To use the limit definition for finding the definite integral $\int_1^4 {\left( {{x^3} - 4} \right)dx}$ , we have to split it in two definite integrals as
$\int_1^4 {\left( {{x^3} - 4} \right)dx} = \int_0^4 {\left( {{x^3} - 4} \right)dx} - \int_0^1 {\left( {{x^3} - 4} \right)dx}$

And now we have to find which points in each interval we will use for ${x_i}$ . To make the calculation easier we will take appropriate endpoints of each interval.

We know that for $n$ number of subintervals, width is given as
$\Delta x = \dfrac{{b - a}}{n}$

We have ${b_1} = 4,\;{a_1} = 0\;{\text{and}}\;{b_2} = 1,\;{a_2} = 0$ so putting these values in order to find the width of subintervals
$\Rightarrow \Delta {x_1} = \dfrac{{4 - 0}}{n}\;{\text{and}}\;\Delta {x_2} = \dfrac{{1 - 0}}{n} \\\ \Rightarrow \Delta {x_1} = \dfrac{4}{n}\;{\text{and}}\;\Delta {x_2} = \dfrac{1}{n} \\\$

We got the width of subintervals then the subintervals can be written as following

1)}}{n},\;\dfrac{{4i}}{n}} \right],....,\;\left[ {\dfrac{{4(n - 1)}}{n},\;4} \right]\;{\text{and}}\;\left[ {0,\;\dfrac{1}{n}} \right],\;\left[ {\dfrac{1}{n},\;\dfrac{2}{n}} \right],....,\;\left[ {\dfrac{{i - 1}}{n},\;\dfrac{i}{n}} \right],....,\;\left[ {\dfrac{{n - 1}}{n},\;1} \right]$$ Where the right end of the $ {i^{th}} $ subinterval is given as $ {x_{{1_i}}} = \dfrac{{4i}}{n}\;{\text{and}}\;{x_{{2_i}}} = \dfrac{i}{n} $ Now we got the parameters, then the summation of the definite integral will be written as

= \sum\limits_{i = 1}^n {f({x_{{1_i}}})\Delta {x_1}} ;{\text{and}};\sum\limits_{i = 1}^n
{f({x_{{2_i}}})\Delta {x_2}} \\
= \sum\limits_{i = 1}^n {f\left( {\dfrac{{4i}}{n};} \right)\left( {\dfrac{4}{n}} \right)}
;{\text{and}};\sum\limits_{i = 1}^n {f\left( {\dfrac{i}{n}} \right);\left( {\dfrac{1}{n}} \right)} \\
= \sum\limits_{i = 1}^n {\left( {{{\left( {\dfrac{{4i}}{n}} \right)}^3} - 4} \right);\left( {\dfrac{4}{n}}
\right)} ;{\text{and}};\sum\limits_{i = 1}^n {\left( {{{\left( {\dfrac{i}{n}} \right)}^3} - 4} \right);\left(
{\dfrac{1}{n}} \right)} \\
= \sum\limits_{i = 1}^n {\left( {{{\dfrac{{64i}}{{{n^3}}}}^3} - 4} \right);\left( {\dfrac{4}{n}} \right)}
;{\text{and}};\sum\limits_{i = 1}^n {\left( {{{\dfrac{i}{{{n^3}}}}^3} - 4} \right);\left( {\dfrac{1}{n}}
\right)} \\
= \sum\limits_{i = 1}^n {\left( {{{\dfrac{{256i}}{{{n^4}}}}^3} - ;\dfrac{{16}}{n}} \right)}
;{\text{and}};\sum\limits_{i = 1}^n {\left( {{{\dfrac{i}{{{n^4}}}}^3} - ;\dfrac{4}{n}} \right)} \\

Now we need to evaluate the summation sign that is we have to use the formulae of summation Here we Know that $ n $ is a constant under the summation notation and $ i $ is a variable, as we go through the integers from $ 1 $ to $ n $ , only $ i $ will change. So considering this point and evaluating the summation

= \sum\limits_{i = 1}^n {f({x_{{1_i}}})\Delta {x_1}} ;{\text{and}};\sum\limits_{i = 1}^n
{f({x_{{2_i}}})\Delta {x_2}} \\
= \sum\limits_{i = 1}^n {\left( {{{\dfrac{{256i}}{{{n^4}}}}^3} - ;\dfrac{{16}}{n}} \right)}
;{\text{and}};\sum\limits_{i = 1}^n {\left( {{{\dfrac{i}{{{n^4}}}}^3} - ;\dfrac{4}{n}} \right)} \\

$$Using distributive property of summation$$

= \sum\limits_{i = 1}^n {{{\dfrac{{256i}}{{{n^4}}}}^3} - \sum\limits_{i = 1}^n {\dfrac{{16}}{n}} }
;{\text{and}};\sum\limits_{i = 1}^n {{{\dfrac{i}{{{n^4}}}}^3} - \sum\limits_{i = 1}^n {\dfrac{4}{n}} }
\\

= \dfrac{{256}}{{{n^4}}}\sum\limits_{i = 1}^n {{i^3} - \dfrac{1}{n}\sum\limits_{i = 1}^n {16} }
;{\text{and}};\dfrac{1}{{{n^4}}}\sum\limits_{i = 1}^n {{i^3} - \dfrac{1}{n}\sum\limits_{i = 1}^n 4 }
\\

We know that $ \sum\limits_{i = 1}^n {x_i^3 = } \dfrac{{{n^2}{{(n + 1)}^2}}}{4}\;{\text{and}}\;\sum\limits_{i = 1}^n x = nx $ using this to evaluate the summation

= \dfrac{{256}}{{{n^4}}} \times \dfrac{{{n^2}{{(n + 1)}^2}}}{4} - \dfrac{1}{n} \times 16 \times
n;{\text{and}};\dfrac{1}{{{n^4}}} \times \dfrac{{{n^2}{{(n + 1)}^2}}}{4} - \dfrac{1}{n} \times 4 \times
n \\
= \dfrac{{256}}{{{n^2}}} \times \left( {\dfrac{{{n^2} + 2n + 1}}{4}} \right) -
16;{\text{and}};\dfrac{1}{{{n^2}}} \times \left( {\dfrac{{{n^2} + 2n + 1}}{4}} \right) - 4 \\
= \dfrac{{256}}{4} + \dfrac{{256}}{{2n}} + \dfrac{{256}}{{4{n^2}}} - 16;{\text{and}};\dfrac{1}{4} +
\dfrac{1}{{2n}} + \dfrac{1}{{4{n^2}}} - 4 \\

$$Now computing the definite integral as$$

\int_a^b {f(x)dx = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f({x_i})\Delta x} }
\\
= \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{256}}{4} + \dfrac{{256}}{{2n}} +
\dfrac{{256}}{{4{n^2}}} - 16} \right);{\text{and}};\mathop {\lim }\limits_{n \to \infty } \left(
{\dfrac{1}{4} + \dfrac{1}{{2n}} + \dfrac{1}{{4{n^2}}} - 4} \right) \\

$$Using distributive property of limit,$$

= \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{256}}{4} + \dfrac{{256}}{{2n}} +
\dfrac{{256}}{{4{n^2}}} - 16} \right);{\text{and}};\mathop {\lim }\limits_{n \to \infty } \left(
{\dfrac{1}{4} + \dfrac{1}{{2n}} + \dfrac{1}{{4{n^2}}} - 4} \right) \\
= \mathop {\lim }\limits_{n \to \infty } \dfrac{{256}}{4} + \mathop {\lim }\limits_{n \to \infty }
\dfrac{{256}}{{2n}} + \mathop {\lim }\limits_{n \to \infty } \dfrac{{256}}{{4{n^2}}} - \mathop {\lim
}\limits_{n \to \infty } 16;{\text{and}};\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{4} + \mathop

{\lim }\limits_{n \to \infty } \dfrac{1}{{2n}} + \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{4{n^2}}}
- \mathop {\lim }\limits_{n \to \infty } 4 \\
= \dfrac{{256}}{4} + 0 + 0 - 16;{\text{and}};\dfrac{1}{4} + 0 + 0 - 4 \\
= 64 - 16;{\text{and}};\dfrac{1}{4} - 4 \\
= 48;{\text{and}};\dfrac{{ - 15}}{4} \\

That is $ \int_0^4 {\left( {{x^3} - 4} \right)dx} = 48\;{\text{and}}\;\int_0^1 {\left( {{x^3} - 4} \right)dx} = \dfrac{{ - 15}}{4} $ We know that $ \int_1^4 {\left( {{x^3} - 4} \right)dx} = \int_0^4 {\left( {{x^3} - 4} \right)dx} - \int_0^1 {\left( {{x^3} - 4} \right)dx} $ So putting the values to get the required answer $ = \int_1^4 {\left( {{x^3} - 4} \right)dx} \\\ = \int_0^4 {\left( {{x^3} - 4} \right)dx} - \int_0^1 {\left( {{x^3} - 4} \right)dx} \\\ = 48 - \left( {\dfrac{{ - 15}}{4}} \right) \\\ = 48 + \dfrac{{15}}{4} \\\ = \dfrac{{48 \times 4 + 15}}{4} \\\ = \dfrac{{192 + 15}}{4} \\\ = \dfrac{{207}}{4} \\\ $ **So the required answer is $ \dfrac{{207}}{4} $** **Note:** When the lower limit of a definite integral $ \ne 0 $ then split the limit into two parts and then solve. Learn some general summation formulae, this will help you. There are some more ways to solve this integral, solve them and check the answer.