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kaheart [24]
3 years ago
15

The estimated value of the integral from 0 to 2 of x cubed dx , using the trapezoidal rule with 4 trapezoids is

Mathematics
1 answer:
bulgar [2K]3 years ago
4 0
The integral is approximated by the sum,

\displaystyle\int_0^2f(x)\,\mathrm dx\approx\sum_{n=0}^4\frac12\times\frac{f(x_n)+f(x_{n+1})}2=\frac14\sum_{n=0}^3(f(x_n)+f(x_{n+1}))

where f(x)=x^3 and x_n=\dfrac12n, giving you

\displaystyle\frac14\sum_{n=0}^3\bigg(\left(\frac n2\right)^3+\left(\frac{n+1}2\right)^3\bigg)
\displaystyle\frac1{32}\sum_{n=0}^3(n^3+(n+1)^3)
\displaystyle\frac1{32}\sum_{n=0}^3(2n^3+3n^2+3n+1)

Faulhaber's formulas make short work of computing the sum. You have

\displaystyle\sum_{n=0}^k1=k+1
\displaystyle\sum_{n=0}^kn=\frac{k(k+1)}2
\displaystyle\sum_{n=0}^kn^2=\frac{k(k+1)(2k+1)}6
\displaystyle\sum_{n=0}^kn^3=\frac{k^2(k+1)^2}4

which gives

\displaystyle\frac1{16}\sum_{n=0}^3n^3+\frac3{32}\sum_{n=0}^3n^2+\frac3{32}\sum_{n=0}^3n+\frac1{32}\sum_{n=0}^31
\displaystyle\frac{36}{16}+\frac{42}{32}+\frac{18}{32}+\frac4{32}
\implies\displaystyle\int_0^2x^3\,\mathrm dx\approx\frac{17}4=4.25
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