Calcular Integral
11 fev 2014, 21:45
sendo f(x) ...
- Anexos
-
- Foto0449.jpg (101.12 KiB) Visualizado 1101 vezes
Re: Calcular Integral
11 fev 2014, 22:12
\(\int_{0}^{2} f(x)dx=\int_{0}^{1} x^2dx+\int_{1}^{2} \sqrt{x}dx=...\)
Re: Calcular Integral
11 fev 2014, 22:23
tem como voce continuar a desenvolver? obrigado
Re: Calcular Integral
11 fev 2014, 22:46
\(\int_{0}^{2} f(x)dx=\int_{0}^{1} x^2dx+\int_{1}^{2} \sqrt{x}dx= \frac{(x^3)_{0}^{1}}{3} + \frac{2(\sqrt{x^3})_{1}^{2}}{3}\)
\(\frac{(x^3)_{0}^{1}}{3} + \frac{2(\sqrt{x^3})_{1}^{2}}{3} = \frac{1}{3}(1^3-0^3+2(\sqrt{8}-\sqrt{1}))=\frac{1}{3}(1^3+4\sqrt{2}-1))=\frac{4\sqrt{2}}{3}\)
\(\frac{(x^3)_{0}^{1}}{3} + \frac{2(\sqrt{x^3})_{1}^{2}}{3} = \frac{1}{3}(1^3-0^3+2(\sqrt{8}-\sqrt{1}))=\frac{1}{3}(1^3+4\sqrt{2}-1))=\frac{4\sqrt{2}}{3}\)