Answer:
a. Trapezoidal Rule- 9.413607
b. Midpoint Rule-9.393861
c. Simpson's Rule -9.400407
Step-by-step explanation:
a. We use n=8 for this integral,
.
Therefore the integral range is ![[0,0.5,1.0,1.5...3.0,3.5,4.0]](https://tex.z-dn.net/?f=%5B0%2C0.5%2C1.0%2C1.5...3.0%2C3.5%2C4.0%5D)
#For the Trapezoidal Rule:
![\int\limits^a_b {\int(x)} \, dx \approx T_n=\frac{\bigtriangleup x}{2}[f(x_o)+2\int(x_1)+...+\int(x_n)]\\\\=\int\limits^4_0 {In(2+e^x)} \, dx \approx T_8\\\\=\frac{0.5}{2}[In(2+e^{0.0})+2In(2+e^{0.5})+2In(2+e^{1.0})+...+2In(2+e^{3.5})+In(2+e^{4.0})]\\\\=0.25[1.098612+2.588754+3.102889+3.737962+4.479090+5.304017+6.189846+7.117283+4.035976]\\\\\\\approx9.413607](https://tex.z-dn.net/?f=%5Cint%5Climits%5Ea_b%20%7B%5Cint%28x%29%7D%20%5C%2C%20dx%20%5Capprox%20T_n%3D%5Cfrac%7B%5Cbigtriangleup%20x%7D%7B2%7D%5Bf%28x_o%29%2B2%5Cint%28x_1%29%2B...%2B%5Cint%28x_n%29%5D%5C%5C%5C%5C%3D%5Cint%5Climits%5E4_0%20%7BIn%282%2Be%5Ex%29%7D%20%5C%2C%20dx%20%5Capprox%20T_8%5C%5C%5C%5C%3D%5Cfrac%7B0.5%7D%7B2%7D%5BIn%282%2Be%5E%7B0.0%7D%29%2B2In%282%2Be%5E%7B0.5%7D%29%2B2In%282%2Be%5E%7B1.0%7D%29%2B...%2B2In%282%2Be%5E%7B3.5%7D%29%2BIn%282%2Be%5E%7B4.0%7D%29%5D%5C%5C%5C%5C%3D0.25%5B1.098612%2B2.588754%2B3.102889%2B3.737962%2B4.479090%2B5.304017%2B6.189846%2B7.117283%2B4.035976%5D%5C%5C%5C%5C%5C%5C%5Capprox9.413607)
Hence, the approximate integral value using Trapezoidal rule is 9.413607
b.For the Midpoint Rule, we calculate the integral as:
![\int\limits^a_b {\int(x)} \, dx \approx M_n=\bigtriangleup x[\int(\bar x_1)+\int(\bar x_2)...\int(\bar x_n)]](https://tex.z-dn.net/?f=%5Cint%5Climits%5Ea_b%20%7B%5Cint%28x%29%7D%20%5C%2C%20dx%20%5Capprox%20M_n%3D%5Cbigtriangleup%20x%5B%5Cint%28%5Cbar%20x_1%29%2B%5Cint%28%5Cbar%20x_2%29...%5Cint%28%5Cbar%20x_n%29%5D)
is the midpoint of the
interval.
From our interval
, the applicable intervals are ![[0.25,0.75,1.25,1.75,2.25,2.75,3.25,3.75]](https://tex.z-dn.net/?f=%5B0.25%2C0.75%2C1.25%2C1.75%2C2.25%2C2.75%2C3.25%2C3.75%5D)
We therefore have the integral value as:
![\int\limits^4_0 {In(2+e^x)} \, dx \approx M_8\\\\\\=\frac{4}{8}[In(2+e^{0.25})+In(2+e^{0.75})+In(2+e^{1.25})+In(2+e^{1.75})+In(2+e^{2.25})+In(2+e^{2.75})+In(2+e^{3.25})+In(2+e^{3.75})]\\\\\\\\=0.5[1.189070+1.415125+1.702991+2.048287+2.441280+2.870318+3.324688+3.795963]\\\\\\\approx 9.393861](https://tex.z-dn.net/?f=%5Cint%5Climits%5E4_0%20%7BIn%282%2Be%5Ex%29%7D%20%5C%2C%20dx%20%5Capprox%20M_8%5C%5C%5C%5C%5C%5C%3D%5Cfrac%7B4%7D%7B8%7D%5BIn%282%2Be%5E%7B0.25%7D%29%2BIn%282%2Be%5E%7B0.75%7D%29%2BIn%282%2Be%5E%7B1.25%7D%29%2BIn%282%2Be%5E%7B1.75%7D%29%2BIn%282%2Be%5E%7B2.25%7D%29%2BIn%282%2Be%5E%7B2.75%7D%29%2BIn%282%2Be%5E%7B3.25%7D%29%2BIn%282%2Be%5E%7B3.75%7D%29%5D%5C%5C%5C%5C%5C%5C%5C%5C%3D0.5%5B1.189070%2B1.415125%2B1.702991%2B2.048287%2B2.441280%2B2.870318%2B3.324688%2B3.795963%5D%5C%5C%5C%5C%5C%5C%5Capprox%209.393861)
Hence, the approximate integral value using the Midpoint Rule is 9.393861
c. For the Simpson's Rule, we calculate the integral rule as follows:
![\int\limits^a_b {\int(x)} \, dx \approx S_n=\frac{\bigtriangleup x}{3}[\int(x_o)+4\int(x_1)+2\int(x_2)+4\int(x_3)+...+2\int(x_{n-2})+4\int(x_{n-1})+\int(x_n)]](https://tex.z-dn.net/?f=%5Cint%5Climits%5Ea_b%20%7B%5Cint%28x%29%7D%20%5C%2C%20dx%20%5Capprox%20S_n%3D%5Cfrac%7B%5Cbigtriangleup%20x%7D%7B3%7D%5B%5Cint%28x_o%29%2B4%5Cint%28x_1%29%2B2%5Cint%28x_2%29%2B4%5Cint%28x_3%29%2B...%2B2%5Cint%28x_%7Bn-2%7D%29%2B4%5Cint%28x_%7Bn-1%7D%29%2B%5Cint%28x_n%29%5D)
We then have:
![\int\limits^4_0 In(2+e^x)\, dx \approx S_8\\\\=\frac{0.5}{3}[In(2+e^0)+4In(2+e^{0.5})+2In(2+e^{1.0})+4In(2+e^{1.5})+2In(2+e^{2.0})+4In(2+e^{2.5})+2In(2+e^{3.0})+4In(2+e^{3.5})+In(2+e^{4.0})]\\\\=\frac{0.5}{3}[1.098612+5.177507+3.102889+7.475925+4.479090+10.608034+6.189846+14.234565+4.035976]\\\\\\\approx9.400407](https://tex.z-dn.net/?f=%5Cint%5Climits%5E4_0%20In%282%2Be%5Ex%29%5C%2C%20dx%20%5Capprox%20S_8%5C%5C%5C%5C%3D%5Cfrac%7B0.5%7D%7B3%7D%5BIn%282%2Be%5E0%29%2B4In%282%2Be%5E%7B0.5%7D%29%2B2In%282%2Be%5E%7B1.0%7D%29%2B4In%282%2Be%5E%7B1.5%7D%29%2B2In%282%2Be%5E%7B2.0%7D%29%2B4In%282%2Be%5E%7B2.5%7D%29%2B2In%282%2Be%5E%7B3.0%7D%29%2B4In%282%2Be%5E%7B3.5%7D%29%2BIn%282%2Be%5E%7B4.0%7D%29%5D%5C%5C%5C%5C%3D%5Cfrac%7B0.5%7D%7B3%7D%5B1.098612%2B5.177507%2B3.102889%2B7.475925%2B4.479090%2B10.608034%2B6.189846%2B14.234565%2B4.035976%5D%5C%5C%5C%5C%5C%5C%5Capprox9.400407)
Hence, the approximate integral value using the Simpson's Rule is 9.400407