Question

How do you use the Trapezoidal Rule with step size $n = 4$ to estimate $\int {\left( {{x^3} + x} \right)} dx$ with $\left[ {0,2} \right]$?

Answer 1

Hint: In the above question we need to estimate the given function for the given range with size $n = 4$ using the trapezoidal rule which states that \[\int\limits_a^b {f\left( x \right)} dx = \dfrac{{\Delta x}}{2}\left[ {f\left( {{x_0}} \right) + 2f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right) + ....... + 2f\left( {{x_{n - 1}}} \right) + f\left( {{x_n}} \right)} \right]\]. We would initiate by finding \[\Delta x\], then moving on to find the sub-intervals of the function given. Substituting all the values and getting the results by solving.
Formula used:
1.\[\int\limits_a^b {f\left( x \right)} dx = \dfrac{{\Delta x}}{2}\left[ {f\left( {{x_0}} \right) + 2f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right) + ....... + 2f\left( {{x_{n - 1}}} \right) + f\left( {{x_n}} \right)} \right]\]
2.\[\Delta x = \dfrac{{b - a}}{n}\]

Complete answer: We are given with the function $\int {\left( {{x^3} + x} \right)} dx$, for which we need to estimate the values using the intervals and size given.
Writing the function as $\int {f\left( x \right)} dx = \int {\left( {{x^3} + x} \right)} dx$.
Considering the intervals $\left[ {0,2} \right]$ to be written as $\left[ {a,b} \right]$, that gives us $a = 0$ and $b = 2$.
Since, we know that \[\Delta x = \dfrac{{b - a}}{n}\], substituting the values of a, b and n, we get:
\[ \Rightarrow \Delta x = \dfrac{{2 - 0}}{4} = \dfrac{2}{4} = \dfrac{1}{2}\]
We need to find the endpoints of the subintervals and that can be found with $a = 0$ adding \[\Delta x\] each step until, we reach \[{x_n} = b = 2\].
Therefore, we have:
\[ \Rightarrow {x_0} = a = 0\]
For next endpoint adding \[\Delta x\] to \[{x_0}\]:
\[ \Rightarrow {x_1} = 0 + \dfrac{1}{2} = \dfrac{1}{2}\]
\[ \Rightarrow {x_2} = \dfrac{1}{2} + \dfrac{1}{2} = 1\]
\[ \Rightarrow {x_3} = 1 + \dfrac{1}{2} = \dfrac{3}{2}\]
\[ \Rightarrow {x_4} = \dfrac{3}{2} + \dfrac{1}{2} = \dfrac{4}{2} = 2 = b\]
Since, we obtained 2 which is the end interval, so we ended.

We know that trapezoidal rule states the formula as:
\[\int\limits_a^b {f\left( x \right)} dx = \dfrac{{\Delta x}}{2}\left[ {f\left( {{x_0}} \right) + 2f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right) + ....... + 2f\left( {{x_{n - 1}}} \right) + f\left( {{x_n}} \right)} \right]\]
Substituting the values in the above equation, we get:
\[\int\limits_0^2 {f\left( x \right)} dx = \dfrac{{\dfrac{1}{2}}}{2}\left[ {f\left( 0 \right) + 2f\left( {\dfrac{1}{2}} \right) + 2f\left( 1 \right) + 2f\left( {\dfrac{3}{2}} \right) + f\left( 2 \right)} \right]\]
After simplifying, we get:
\[\int\limits_0^2 {\left( {{x^3} + x} \right)} dx = \dfrac{1}{4}\left[ {f\left( 0 \right) + 2f\left( {\dfrac{1}{2}} \right) + 2f\left( 1 \right) + 2f\left( {\dfrac{3}{2}} \right) + f\left( 2 \right)} \right]\] ….(1)
We need to find the values of all the endpoints of the interval:
Substituting \[f\left( 0 \right)\] in place of \[f\left( x \right)\]:
For \[f\left( 0 \right)\]:
\[f\left( 0 \right) = {0^3} + 0 = 0\]
Similarly, for others too:
For \[f\left( {\dfrac{1}{2}} \right)\]:
\[f\left( {\dfrac{1}{2}} \right) = {\left( {\dfrac{1}{2}} \right)^3} + \dfrac{1}{2} = \dfrac{1}{8} + \dfrac{1}{2} = \dfrac{{1 + 4}}{8} = \dfrac{5}{8}\]
For \[f\left( 1 \right)\]:
\[f\left( 1 \right) = {\left( 1 \right)^3} + 1 = 1 + 1 = 2\]

For \[f\left( {\dfrac{3}{2}} \right)\]:
\[f\left( {\dfrac{3}{2}} \right) = {\left( {\dfrac{3}{2}} \right)^3} + \dfrac{3}{2} = \dfrac{{27}}{8} + \dfrac{3}{2} = \dfrac{{27 + 12}}{8} = \dfrac{{39}}{8}\]
For \[f\left( 2 \right)\]:
\[f\left( 2 \right) = {\left( 2 \right)^3} + 2 = 8 + 2 = 10\]

Substituting all these values in the equation 1, we get:
\[ \Rightarrow \int\limits_0^2 {\left( {{x^3} + x} \right)} dx = \dfrac{1}{4}\left[ {0 + 2 \times \dfrac{5}{8} + 2 \times 2 + 2 \times \dfrac{{39}}{8} + 10} \right]\]
Solving the brackets, we get:
\[ \Rightarrow \int\limits_0^2 {\left( {{x^3} + x} \right)} dx = \dfrac{1}{4}\left[ {\dfrac{5}{4} + 4 + \dfrac{{39}}{4} + 10} \right]\]
Taking 4 as a common denominator, we get:
\[ \Rightarrow \int\limits_0^2 {\left( {{x^3} + x} \right)} dx = \dfrac{1}{4}\left[ {\dfrac{{5 + 16 + 39 + 40}}{4}} \right]\]
\[ \Rightarrow \int\limits_0^2 {\left( {{x^3} + x} \right)} dx = \dfrac{1}{4}\left[ {\dfrac{{5 + 16 + 39 + 40}}{4}} \right]\]
\[ \Rightarrow \int\limits_0^2 {\left( {{x^3} + x} \right)} dx = \dfrac{1}{4}\left[ {\dfrac{{100}}{4}} \right]\]
\[ \Rightarrow \int\limits_0^2 {\left( {{x^3} + x} \right)} dx = \dfrac{1}{4}\left[ {25} \right]\]
\[ \Rightarrow \int\limits_0^2 {\left( {{x^3} + x} \right)} dx = \dfrac{{25}}{4}\], which is the required solution;

Therefore, the value of $\int {\left( {{x^3} + x} \right)} dx$ for the interval $\left[ {0,2} \right]$ is $\dfrac{{25}}{4}$.

Note:
Remember, since we found the estimated value, it can be somehow different from the actual value obtained by directly integrating the functions and putting off the limits. The difference between the estimated and the real value is known as error, which can be obtained by subtracting the estimated value from the real value. Therefore, the integral found using the Trapezoidal rule is an approximate value not the real value.