0:00 / 0:00

Error of Integral Approximations

Approximations aren't perfect - however, we are able to bound the error on them. By increasing nn (the number of rectangles) we can reduce the error created by our area approximations

Error for the Midpoint Approximation

Suppose ∣f′′(x)∣<K2|f''\left(x\right)|<K_2 for a≤x≤ba\le x\le b. Then the error using a definite integral Midpoint approximation is

∣EN∣=∣MN−Actual∣|E_N|=|M_N-\text{Actual}|
and the error can be bounded by

∣EN∣<K2(b−a)324n2\boxed{|E_N|<\frac{K_2\left(b-a\right)^3}{24n^2}}


PAGE BREAK

Error for the Trapezoid Approximation

Suppose ∣f′′(x)∣<K2|f''\left(x\right)|<K_2for a≤x≤ba\le x\le b. Then the Error using a definite integral Trapezoid approximation is

∣EN∣=∣TN−Actual∣|E_N|=|T_N-\text{Actual}|
and the error can be bounded by

∣EN∣<K2(b−a)312n2\boxed{|E_N|<\frac{K_2\left(b-a\right)^3}{12n^2}}

Error for the Simpson's Approximation

Suppose ∣f(4)(x)∣<K4|f^{(4)}\left(x\right)|<K_4 for a≤x≤ba\le x\le b. Then the Error using a definite integral Simpson's approximation is
∣EN∣=∣SN−Actual∣|E_N|=|S_N-\text{Actual}|
and the error can be bounded by

∣EN∣<K4(b−a)5180n4\boxed{|E_N|<\frac{K_4\left(b-a\right)^5}{180n^4}}

How large does n have to be to guarantee that the integral approximation of ∫02x5dx\displaystyle \int_0^2x^5dx
using the Midpoint rule is accurate to within 0.0001?