Mathematics, 19.03.2020 18:02, diamondscott9297
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
Answers: 2
.![[0,0.5,1.0,1.5...3.0,3.5,4.0]](/tpl/images/0554/2973/662b8.png)
![\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](/tpl/images/0554/2973/2c645.png)
![\int\limits^a_b {\int(x)} \, dx \approx M_n=\bigtriangleup x[\int(\bar x_1)+\int(\bar x_2)...\int(\bar x_n)]](/tpl/images/0554/2973/8ef95.png)
is the midpoint of the 
interval.![[0.25,0.75,1.25,1.75,2.25,2.75,3.25,3.75]](/tpl/images/0554/2973/cf4ef.png)
![\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](/tpl/images/0554/2973/e209c.png)
![\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)]](/tpl/images/0554/2973/a3140.png)
![\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](/tpl/images/0554/2973/1dc04.png)
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Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral wi...
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