Answer:
(a) Trapezoid Rule
[tex]\int\limits^0_1 {e^{x^{2} } } \, dx=1.457[/tex]
(b) Simpson's Rule
[tex]\int\limits^0_1 {e^{x^{2} } } \, dx=1.463[/tex]
Step-by-step explanation:
(a) Trapezoid Rule
Trapezoid Rule says:
[tex]\int\limits^a_b {f(x)} \, dx =(b-a)[\frac{f(x)_{0}+2\sum\limits^{n-1}_{i=1}f(x_{n})+f(x)_{n}}{2n}[/tex]
1. DIvide the range in n=6 intervals:
[tex]P= [0, 0.176, 0.333, 0.5, 0.667, 0.833, 1][/tex]
2. Evaluate [tex]f(x)=\epsilon^{x^{2}}[/tex] for each interval (See the attachment 1)
3. Replace the values in Trapezoid Rule equation:
[tex]\int\limits^0_1 {e^{x^{2} } } \, dx =(1-0[\frac{f(0)+2[f(0.167)+f(0.133)+f(0.5)+f(0.667)+f(0.833)]+f(1)}{2(6)}\\\int\limits^0_1 {e^{x^{2} } } \, dx=1.457[/tex]
(b) Simpson Rule
Simpson Rule says:
[tex]\int\limits^a_b {f(x)} \, dx =(b-a)[\frac{f(x)_{0}+4\sum\limits^{n}_{i=1}f(x_{mi} )+2\sum\limits^{n-1}_{i=1}f(x_{i})+f(x_{n}) }{2n}[/tex]
1. DIvide the range in n=6 intervals:
[tex]P= [0, 0.176, 0.333, 0.5, 0.667, 0.833, 1][/tex]
2. Calculate the half points of each interval:
[tex]P_{m}= [0.083, 0.25, 0.417, 0.583, 0.75, 0.917][/tex]
3. Evaluate [tex]f(x)=\epsilon^{x^{2}}[/tex] for each interval of P and [tex]P_{m}[/tex] (See the attachment 2)
4. Replace the values in Trapezoid Rule equation:
[tex]\int\limits^0_1 {e^{x^{2} } } \, dx =(1-0)[\frac{f(0)+4[f(0.083)+f(0.25)+...+f(0.917)]+2[f(0.167)+f(0.333)+...+f(0.833)]+f(1]]}{6(6)} \\\int\limits^0_1 {e^{x^{2} } } \, dx =1.463[/tex]
