Exercise¶
Given the integral
$$ I=\int_{0}^{1}e^{x}dx, $$- Approximate its value by the composite Trapezoidal rule with two subintervals.
- Find an error bound for the error.
- How many intervals may guarantee an error lower than $10^{-6}$ using the composite trapezoidal formula?
The exact integral is
$$ I = \int_{0}^{1}e^{x}\,dx=\left[ e^x\right]_0^1=e^1-e^0=2.7183-1 = 1.7183 $$Composite Trapezoidal rule¶
The Trapezoidal rule is
$$\int_{a}^{b}f(x)dx\approx \frac{b-a}{2}\left( f\left( a\right) +f\left( b\right) \right)$$For two subintervals
$$h=\frac{b-a}{n}=\frac{1-0}{2}=0.5$$Let us compute the nodes
$$\begin{array}{lcl} x_0 & = & a = 0\\ x_1 & = & x_0 + h = 0 + 0.5 = 0.5\\ x_2 & = & x_1 + h = 0.5 + 0.5 = 1 = b \end{array} $$. $$ \approx\frac{h}{2}\left(f(x_0)+f(x_1)\right)+\frac{h}{2}\left(f(x_1)+f(x_2)\right)=\frac{h}{2}\left(f(x_0)+2f(x_1)+f(x_2)\right)$$
If $f(x)=e^x$
$$I_{trap} = \frac{0.5}{2}\left(e^0+2e^{0.5}+e^1\right)=0.25\left(1+2(1.65)+2.72\right)=1.75$$And the error is
$$e_a = \left|I-I_{trap}\right|= 1.75-1.72=0.03$$Error bound¶
The error for the composite Trapezoidal rule is
If $f(x) = e^x$ and $[a,b]=[0,1]$ then
$f^{\prime}(x)= e^x \quad \quad \mathrm{y} \quad \quad f^{\prime\prime}(x)= e^x$
$$\left|\;f^{\prime\prime}(c)\right| = e^c \lt e^1 = e \quad c\in (0,1)$$Thus
$$ E_h^T = \left|\;f^{\prime\prime}(c)\right| (b-a)\frac{h^2}{12}\lt \frac{e^1}{12}(b-a)h^2 = \frac{e}{12} (1-0) 0.5^2=0.06$$And, as it was to be expected, the error bound, $0.06$, is greater than the error, $0.03$.
Number of intervals¶
To approximate $I=\int_{0}^{1}e^{x}dx$ with an error less than $10^{-6}$
If we take the error in absolute value
$$E_h^T =\frac{h^2}{12} (b-a)\left|f^{\prime\prime}(c)\right|$$We can write
$$\frac{h^2}{12} (b-a)\left|f^{\prime\prime}(c)\right| \lt 10^{-6}$$and then
$$E_h^T \lt 10^{-6}$$As
$f^{\prime}(x)= e^x$ y $f^{\prime\prime}(x)= e^x$
$$\left|\;f^{\prime\prime}(c)\right| = e^c \lt e^1 = e \quad c\in (0,1)$$Then
$$ E_h^T = \left|\;f^{\prime\prime}(c)\right| (b-a)\frac{h^2}{12}\lt \frac{e}{12}(b-a)h^2 = \frac{e}{12} (1-0) h^2 \lt 10^{-6}$$That is
$$ \frac{e}{12} h^2 \lt 10^{-6}$$As
$$ h =\frac{b-a}{n}=\frac{1}{n} $$then $$ \frac{e}{12}\frac{1}{n^2} \lt 10^{-6}$$
and 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
$$\frac{e}{12}10^6 \lt n^2 $$And taking into account that if we have an increasing $g$ function
$x_1 \lt x_2 \Longrightarrow g(x_1) \lt g(x_2)$ and as the function $g(x)= \sqrt{x}$ is increasing
we have
$$\sqrt{\frac{e}{12}10^6} \lt n $$or
$$475.94 \lt n$$And a sufficient condition for the error to be less than $10^{-6}$ is that $\fbox{$n=476$}$.
In fact, running a script that approximates this integral with the composite Trapezoidal rule and 476 subintervals the results have been:
Aproximate value 1.7182824604330484Exact value 1.7182818284590453- The error is $6 \times 10^{-7} = 0.6 \times 10^{-6} \lt 10^{-6} $