Question

the instructions for the given integral have two parts, one for the trapezoidal rule and one for simpsons rule. complete the following parts
int_{-1}^{1}left(2 x^{2}+1
ight) d x
(type an exact answer. type an integer or a simplified fraction.)
left|e_{t}
ight|=0.17
(round to two decimal places as needed.)
c. use the formula ( left(left|e_{t}
ight| /(\text { true value })
ight) \times 100 ) to express ( left|e_{t}
ight| ) as a percentage of the integrals true value.
( 5 % )
(round to the nearest integer as needed.)
ii. using simpsons rule
a. estimate the integral with ( n=4 ) steps and find an upper bound for ( left|e_{s}
ight| ).
( s=square )
(type an exact answer. type an integer or a simplified fraction.)

Explanation:

Step1: Find the true value of the integral

We know that \(\int_{-1}^{1}(2x^{2}+1)dx=\int_{-1}^{1}2x^{2}dx+\int_{-1}^{1}1dx\).
Using the power rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)\), \(\int_{-1}^{1}2x^{2}dx=2\times\frac{x^{3}}{3}\big|_{-1}^{1}=\frac{2}{3}(1^{3}-(-1)^{3})=\frac{4}{3}\), and \(\int_{-1}^{1}1dx=x\big|_{-1}^{1}=1-(-1) = 2\).
So the true value \(I=\frac{4}{3}+2=\frac{4 + 6}{3}=\frac{10}{3}\).

Step2: Apply Simpson's rule formula

For \(n = 4\), \(a=-1\), \(b = 1\), \(\Delta x=\frac{b - a}{n}=\frac{1-(-1)}{4}=\frac{1}{2}\).
The \(x\) - values are \(x_{0}=-1,x_{1}=-\frac{1}{2},x_{2}=0,x_{3}=\frac{1}{2},x_{4}=1\).
\(y=f(x)=2x^{2}+1\), so \(y_{0}=2(-1)^{2}+1=3\), \(y_{1}=2(-\frac{1}{2})^{2}+1=\frac{1}{2}+1=\frac{3}{2}\), \(y_{2}=2(0)^{2}+1 = 1\), \(y_{3}=2(\frac{1}{2})^{2}+1=\frac{1}{2}+1=\frac{3}{2}\), \(y_{4}=2(1)^{2}+1=3\).
By Simpson's rule \(S=\frac{\Delta x}{3}(y_{0}+4y_{1}+2y_{2}+4y_{3}+y_{4})\).
Substitute \(\Delta x=\frac{1}{2}\), \(y_{0}=3\), \(y_{1}=\frac{3}{2}\), \(y_{2}=1\), \(y_{3}=\frac{3}{2}\), \(y_{4}=3\) into the formula:
\(S=\frac{1/2}{3}(3 + 4\times\frac{3}{2}+2\times1+4\times\frac{3}{2}+3)=\frac{1}{6}(3 + 6+2 + 6+3)=\frac{20}{6}=\frac{10}{3}\).

Answer:

\(\frac{10}{3}\)