Exercise¶
Given the integral
$$ I=\int_{0}^{3}(x^{3}+1)\,dx $$- Approximate its value using the Trapezoidal rule.
- Approximate its value using the Simpson's rule.
- Compare the approximate values to the exact value. Could you have predicted any of the errors?
- How many intervals may guarantee an error lower than $10^{-6}$ using the composite trapezoidal formula?
The exact integral is
$$ I = \int_{0}^{3}\left(x^3+1\right)\,dx=\left[ \frac{x^4}{4}+x\right]_0^3=\frac{3^4}{4}+3=23.25 $$Trapezoidal rule¶
The Trapezoidal rule is
$$\int_{a}^{b}f(x)dx\approx \frac{b-a}{2}\left( f\left( a\right) +f\left( b\right) \right)$$Applying it
$$\int_{0}^{3}f(x)dx\approx \frac{3-0}{2}\left( f\left( 0\right) +f\left( 3\right) \right)=\frac{3}{2}\left( \left( 0^3+1\right) +\left( 3^3+1\right) \right)=\frac{3}{2}\times 29=43.5$$The absolute error is
$$e_a = \left|I-I_{trap}\right| = \left|23.25-43.5\right|= 20.25$$
Simpson's rule¶
The Simpson's rule is
$$\int_{a}^{b}f(x)dx\approx \frac{b-a}{6}\left( f\left( a\right) +4f\left( \frac{a+b}{2}\right) +f\left( b\right) \right)$$. $$\int_{0}^{3}f(x)dx\approx \frac{3-0}{6}\left( f\left(0\right) +4f\left( 1.5\right) +f\left( 3\right) \right)=$$ . $$ =\frac{3}{6}\left( \left( 0^3+1\right) +4\left( 1.5^3+1\right) +\left( 3^3+1\right) \right)=23.25$$
An the absolute error
$$e_a = \left|I-I_{Simp}\right| = \left|23.25-23.25\right|= 0$$
Error¶
It could have been expected that the error would be zero for Simpson's rule because Simpson's degree of precision is 3 and therefore exact for polynomials of degree 3.
Number of intervals¶
Error of the Trapezoidal rule¶
Simple¶
The error for the Trapezoidal rule is $$ E^T = -f^{\prime\prime}(c_o) \frac{(b-a)^3}{12} \quad c_o \in (a,b)$$
Composite¶
If we split $[a,b]$ into $n$ subintervals of equal length
$$h=\frac{b-a}{n} \quad (1)$$and we apply the Trapezoidal rule in each of them, the error will be the sum of the errors for each of these $n$ subintervals
$$E_h^T = -f^{\prime\prime}(c_1) \frac{h^3}{12}-f^{\prime\prime}(c_2) \frac{h^3}{12}-\dots -f^{\prime\prime}(c_n) \frac{h^3}{12} $$that is
$$E_h^T = -\frac{h^3}{12}\left(f^{\prime\prime}(c_1) +f^{\prime\prime}(c_2) +\dots +f^{\prime\prime}(c_n) \right) $$if we multiply and divide by $n$
$$E_h^T = -\frac{n\,h^3}{12}\frac{f^{\prime\prime}(c_1) +f^{\prime\prime}(c_2) +\dots +f^{\prime\prime}(c_n)}{n} $$From the intermediate value theorem it can be deduced that there is a value of $c\in(a,b)$ so that $f^{\prime\prime}(c)$ is equal to this average that appears in the previous formula (as the average is an intermediate value). So
$$E_h^T = -\frac{n\,h^3}{12}f^{\prime\prime}(c) $$an due to (1), $nh = b-a$ the error is
$$E_h^T = - \frac{h^2}{12} (b-a)f^{\prime\prime}(c)$$Approximate $$I=\int_{0}^{3}(x^{3}+1)dx$$ with the Trapezoidal rule with error less than $10^{-6}$
If we use the error in absolute value
$$E_h^T =\frac{h^2}{12} (b-a)\left|f^{\prime\prime}(c)\right|$$If we propose
$$\frac{h^2}{12} (b-a)\left|f^{\prime\prime}(c)\right| \lt 10^{-6}$$then it will be true that
$$E_h^T \lt 10^{-6}$$If $f(x) = x^3+ 1$ and $[a,b]=[0,3]$ then
$$f^{\prime}(x)= 3x^2 \quad \quad \mathrm{y} \quad \quad f^{\prime\prime}(x)= 6x$$and we have
$$\left|\;f^{\prime\prime}(c)\right| = 6 c \lt 6 (3) = 18 \quad c\in (0,3)$$Thus, the error is
$$ E_h^T = \left|\;f^{\prime\prime}(c)\right| (b-a)\frac{h^2}{12}\lt \frac{18}{12}(b-a)h^2 = \frac{3}{2} (3-0) h^2 \lt 10^{-6}$$That is
$$ \frac{9}{2} h^2 \lt 10^{-6}$$As
$$ h =\frac{b-a}{n}=\frac{3}{n} $$we have
$$ \frac{9}{2}\left(\frac{3}{n}\right)^2 \lt 10^{-6}$$or
$$ \frac{81}{2n^2} \lt 10^{-6}$$and then, as
$$a\lt b, 0 \lt c \Longrightarrow ac \lt bc$$multiplying both sides of the inequality by $n^2$ and $10^6$ we have
$$\frac{81}{2}10^6 \lt n^2 $$Taking into account that if $h$ is an increasing function
$x_1 \lt x_2 \Longrightarrow h(x_1) \lt h(x_2)$ and $h(x)= \sqrt{x}$ is an increasing function
we have
$$\sqrt{\frac{81}{2}10^6} \lt n $$or
$$ 6363.96\lt n $$And taking $\fbox{$n = 6364$}$ we can guarantee that the error is less than $10^{-6}.$
In fact, running a script that approximates this integral with the compound trapezoid rule and 6364 subintervals the results have been:
- The approximate value is $23.250000499993877$
- The exact value is $23.25$
- The error is $5 \times 10^{-7} = 0.5 \times 10^{-6} \lt 10^{-6} $