Aloha :)
a) Exakte Berechnung:$$I=\int\limits_0^1\frac{2}{1+x^2}\,dx=\int\limits_0^1f(x)\,dx\quad;\quad f(x)=\frac{2}{1+x^2}$$Das knacken wir mit Substitution:$$x=\tan y\;\;;\;\;\frac{dx}{dy}=\left(\frac{\sin y}{\cos y}\right)'=\frac{\cos y\cos y+\sin y\sin y}{\cos^2y}=1+\tan^2y=1+x^2$$$$y=\arctan(x)\;\;;\;\;y(0)=\arctan(0)=0\;\;;\;\;y(1)=\arctan(1)=\frac{\pi}{4}$$$$I=\int\limits_0^{\pi/4}\frac{2}{1+x^2}\,(1+x^2)dy=\int\limits_0^{\pi/4}2\,dy=\left[2y\right]_0^{\pi/4}$$$$I=\frac{\pi}{2}\approx1,570796$$
b1) Linksseitige Rechteckregel:$$I_{b1}\approx f(0)\cdot(1-0)=2\cdot1=2$$b2) Mittelpunktsregel:$$I_{b2}\approx f\left(0,5\right)\cdot(1-0)=f(0,5)=1,6$$b3) Trapezregel:$$I_{b3}\approx\frac{1}{2}\left(f(0)+f(1)\right)=\frac{1}{2}\left(2+1\right)=1,5$$b4) Simpson-Regel:$$I_{b4}\approx\frac{1}{6}\left(f(0)+4\cdot f(0,5)+f(1)\right)=\frac{1}{6}\left(2+4\cdot1,6+1\right)=1,566667$$b5) Newton 3/8-Regel:$$I_{b5}\approx\frac{1}{8}\left(f(0)+3\cdot f\left(\frac{1}{3}\right)+3\cdot f\left(\frac{2}{3}\right)+f(1)\right)$$$$\phantom{I_{b5}}=\frac{1}{8}\left(2+3\cdot1,8+3\cdot1,384615+1\right)=1,569231$$
