Question

Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n. (Round your answers to six decimal places.) 4 0 ln(2 + ex) dx, n = 8

Answer 1

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, $\bigtriangleup t=\frac{4-0}{8}=0.5$.

Therefore the integral range is $[0,0.5,1.0,1.5...3.0,3.5,4.0]$

#For the Trapezoidal Rule:

$\int\limits^a_b {\int(x)} \, dx \approx T_n=\frac{\bigtriangleup x}{2}[f(x_o)+2\int(x_1)+...+\int(x_n)]\\\\=\int\limits^4_0 {In(2+e^x)} \, dx \approx T_8\\\\=\frac{0.5}{2}[In(2+e^{0.0})+2In(2+e^{0.5})+2In(2+e^{1.0})+...+2In(2+e^{3.5})+In(2+e^{4.0})]\\\\=0.25[1.098612+2.588754+3.102889+3.737962+4.479090+5.304017+6.189846+7.117283+4.035976]\\\\\\\approx9.413607$

Hence, the approximate integral value using Trapezoidal rule is 9.413607

b.For the Midpoint Rule, we calculate the integral as:

$\int\limits^a_b {\int(x)} \, dx \approx M_n=\bigtriangleup x[\int(\bar x_1)+\int(\bar x_2)...\int(\bar x_n)]$

$\bar x_i$ is the midpoint of the $i^{th}$ interval.

From our interval $[0,0.5,1.0,1.5...3.0,3.5,4.0]$, the applicable intervals are $[0.25,0.75,1.25,1.75,2.25,2.75,3.25,3.75]$

We therefore have the integral value as:

$\int\limits^4_0 {In(2+e^x)} \, dx \approx M_8\\\\\\=\frac{4}{8}[In(2+e^{0.25})+In(2+e^{0.75})+In(2+e^{1.25})+In(2+e^{1.75})+In(2+e^{2.25})+In(2+e^{2.75})+In(2+e^{3.25})+In(2+e^{3.75})]\\\\\\\\=0.5[1.189070+1.415125+1.702991+2.048287+2.441280+2.870318+3.324688+3.795963]\\\\\\\approx 9.393861$

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:

$\int\limits^a_b {\int(x)} \, dx \approx S_n=\frac{\bigtriangleup x}{3}[\int(x_o)+4\int(x_1)+2\int(x_2)+4\int(x_3)+...+2\int(x_{n-2})+4\int(x_{n-1})+\int(x_n)]$

We then have:

$\int\limits^4_0 In(2+e^x)\, dx \approx S_8\\\\=\frac{0.5}{3}[In(2+e^0)+4In(2+e^{0.5})+2In(2+e^{1.0})+4In(2+e^{1.5})+2In(2+e^{2.0})+4In(2+e^{2.5})+2In(2+e^{3.0})+4In(2+e^{3.5})+In(2+e^{4.0})]\\\\=\frac{0.5}{3}[1.098612+5.177507+3.102889+7.475925+4.479090+10.608034+6.189846+14.234565+4.035976]\\\\\\\approx9.400407$

Hence, the approximate integral value using the Simpson's Rule is 9.400407