Question

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

Answer 1

5.50

Explanation:

Answer 2

The integral is approximated by the sum,


`\displaystyle\int_0^2f(x)\,\mathrm dx\approx\sum_(n=0)^4\frac12*\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=\frac12n`, 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(36)/(16)+(42)/(32)+(18)/(32)+\frac4{32}`

`\implies\displaystyle\int_0^2x^3\,\mathrm dx\approx\frac{17}4=4.25`