Use (a) the midpoint rule and (b) simpson's rule to approximate the below integral. ∫ x^2sin(x) dx with n = 8.

Answer:

midpoint rule =  5.93295663

simpson's rule = 5.869246855

Step-by-step explanation:

a) midpoint rule

[tex]\int\limits^b_a {(x)} \, dx[/tex]≈ Δ x (f(x₀+x₁)/2 + f(x₁+x₂)/2 + f(x₂+x₃)/2 +...+ f(x[tex]_{n}[/tex]_₂+x[tex]_{n}[/tex]_₁)/2 +f(x[tex]_{n}[/tex]_₁+x[tex]_{n}[/tex])/2)

Δx = (b − a) / n

We have that a = 0, b = π, n = 8

Therefore

Δx = (π − 0) / 8 = π/8

Divide the interval [0,π] into n=8 sub-intervals of length Δx = π/8 with the following endpoints:

a=0, π/8, π/4, 3π/8, π/2, 5π/8, 3π/4, 7π/8, π = b

Now, we just evaluate the function at these endpoints:

[tex]f(\frac{x_{0}+x_{1} }{2} ) = f(\frac{0+\frac{\pi}{8} }{2} ) = f(\frac{\pi }{16})=\frac{\pi^{2}sin(\frac{\pi }{16}) }{256} =[/tex] 0.00752134

[tex]f(\frac{x_{1}+x_{2} }{2} ) = f(\frac{\frac{\pi }{8} +\frac{\pi}{4} }{2} ) = f(\frac{3\pi }{16})=\frac{9\pi ^{2} sin(\frac{3\pi }{16}) }{256}[/tex] = 0.19277080

[tex]f(\frac{x_{2}+x_{3} }{2} ) = f(\frac{\frac{\pi }{4} +\frac{3\pi}{8} }{2} ) = f(\frac{5\pi }{16})=\frac{25\pi ^{2} sin(\frac{5\pi }{16}) }{256}[/tex] = 0.80139415

[tex]f(\frac{x_{3}+x_{4} }{2} ) = f(\frac{\frac{3\pi }{8} +\frac{\pi}{2} }{2} ) = f(\frac{7\pi }{16})=\frac{49\pi ^{2} sin(\frac{7\pi }{16}) }{256}[/tex] = 1.85280536

[tex]f(\frac{x_{4}+x_{5} }{2} ) = f(\frac{\frac{\pi }{2} +\frac{5\pi}{8} }{2} ) = f(\frac{9\pi }{16})=\frac{81\pi ^{2} sin(\frac{7\pi }{16}) }{256}[/tex] = 3.062800704

[tex]f(\frac{x_{5}+x_{6} }{2} ) = f(\frac{\frac{5\pi }{8} +\frac{3\pi}{4} }{2} ) = f(\frac{11\pi }{16})=\frac{121\pi ^{2} sin(\frac{5\pi }{16}) }{256}[/tex] = 3.878747709

[tex]f(\frac{x_{6}+x_{7} }{2} ) = f(\frac{\frac{3\pi }{4} +\frac{7\pi}{8} }{2} ) = f(\frac{13\pi }{16})=\frac{169\pi ^{2} sin(\frac{3\pi }{16}) }{256}[/tex] = 3.61980731

[tex]f(\frac{x_{7}+x_{8} }{2} ) = f(\frac{\frac{7\pi }{8} +\pi }{2} ) = f(\frac{15\pi }{16})=\frac{225\pi ^{2} sin(\frac{\pi }{16}) }{256}[/tex] = 1.69230261

Finally, just sum up the above values and multiply by Δx = π/8:

π/8 (0.00752134 +0.19277080+ 0.80139415 + 1.85280536 + 3.062800704 + 3.878747709 + 3.61980731 + 1.69230261) = 5.93295663

b) simpson's rule

[tex]\int\limits^b_a {(x)} \, dx[/tex]  ≈ (Δx)/3 (f(x₀) + 4f(x₁) + 2f(x₂) + 4f(x₃) + 2f(x₄) + ... + 2f([tex]x_{n-2}[/tex]) + 4f([tex]x_{n-1}[/tex]) + f([tex]x_{n}[/tex]))

where Δx = (b−a) / n

We have that a = 0, b = π, n = 8

Therefore

Δx = (π−0) / 8 = π/8

Divide the interval [0,π] into n = 8 sub-intervals of length Δx = π/8, with the following endpoints:

a = 0, π/8, π/4, 3π/8, π/2, 5π/8, 3π/4, 7π/8 ,π = b

Now, we just evaluate the function at these endpoints:  

f(x₀) = f(a) = f(0) = 0 = 0

[tex]4f(x_{1} ) = 4f(\frac{\pi }{8} )=\frac{\pi^{2}\sqrt{\frac{1}{2}-\frac{\sqrt{2} }{4} } }{16}[/tex] = 0.23605838

[tex]2f(x_{2} ) = 2f(\frac{\pi }{4} )=\frac{\sqrt{2\pi^{2} } }{16}[/tex] = 0.87235802

[tex]4f(x_{3} ) = 4f(\frac{3\pi }{8} )=\frac{9\pi^{2}\sqrt{\frac{\sqrt{2} }{4}-\frac{{1} }{2} } }{16}[/tex] = 5.12905809

[tex]2f(x_{4} ) = 2f(\frac{\pi }{2} )=\frac{\pi ^{2} }{2}[/tex] = 4.93480220

[tex]4f(x_{5} ) = 4f(\frac{5\pi }{8} )=\frac{25\pi^{2}\sqrt{\frac{\sqrt{2} }{4}-\frac{{1} }{2} } }{16}[/tex] = 14.24738359

[tex]2f(x_{6} ) = 2f(\frac{3\pi }{4} )=\frac{9\sqrt{2\pi^{2} } }{16}[/tex] = 7.85122222

[tex]4f(x_{7} ) = 4f(\frac{7\pi }{8} )=\frac{49\pi^{2}\sqrt{\frac{1}{2}-\frac{\sqrt{2} }{4} } }{16}[/tex] = 11.56686065

f(x₈) = f(b) = f(π) = 0 = 0

Finally, just sum up the above values and multiply by Δx/3 = π/24:

π/24 (0 + 0.23605838 + 0.87235802 + 5.12905809 + 4.93480220 + 14.24738359 + 7.85122222 + 11.56686065 = 5.869246855

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