Question

¿En qué consiste el método de aproximación del área por método de trapecio?​

Answer 1

Respuesta:

Ejemplo Editar

{\displaystyle \int _{0}^{2}3x\,dx}{\displaystyle \int _{0}^{2}3x\,dx} para n=6

Primero se obtiene h, de los límites de la integral que representan a y b y para n=6 queda: {\displaystyle h={\frac {b-a}{n}}}{\displaystyle h={\frac {b-a}{n}}} {\displaystyle ={\frac {2-0}{6}}={\frac {1}{3}}}{\displaystyle ={\frac {2-0}{6}}={\frac {1}{3}}}.

Y ahora se sustituye en la fórmula

{\displaystyle \int _{a}^{b}f(x)\,dx}{\displaystyle \int _{a}^{b}f(x)\,dx} = {\displaystyle {\frac {h}{2}}[f(a)+2f(a+h)+2f(a+2h)+...+f(b)]}{\displaystyle {\frac {h}{2}}[f(a)+2f(a+h)+2f(a+2h)+...+f(b)]}

y queda:

{\displaystyle \int _{0}^{2}3x\,dx}{\displaystyle \int _{0}^{2}3x\,dx} = {\displaystyle {\frac {1}{2}}\cdot {\frac {1}{3}}[3(0)+2[3(0+1\cdot {\frac {1}{3}})]+2[3(0+2\cdot {\frac {1}{3}})]+2[3(0+3\cdot {\frac {1}{3}})]+2[3(0+4\cdot {\frac {1}{3}})]+2[3(0+5\cdot {\frac {1}{3}})]+3(2)]=6}{\displaystyle {\frac {1}{2}}\cdot {\frac {1}{3}}[3(0)+2[3(0+1\cdot {\frac {1}{3}})]+2[3(0+2\cdot {\frac {1}{3}})]+2[3(0+3\cdot {\frac {1}{3}})]+2[3(0+4\cdot {\frac {1}{3}})]+2[3(0+5\cdot {\frac {1}{3}})]+3(2)]=6}