Olá !
Vou-lhe resolver a segunda, porque a primeira também não consegui resolver.
[tex3]\int\limits_{0}^{8}(\sqrt{2x} + \sqrt[3]{x}) dx[/tex3]
[tex3]\sqrt{2}\int\limits_{0}^{8}\sqrt{x}dx\,+\,\int\limits_{0}^{8}\sqrt[^3]{x}dx[/tex3]
[tex3]\sqrt{2}\int\limits_{0}^{8} x^{\frac{1}{2}}dx\,+\,\int\limits_{0}^{8} x^{\frac{1}{3}} dx[/tex3]
[tex3]\sqrt{2}\int\limits_{0}^{8} x^{\frac{1}{2}+1}dx\,+\,\int \limits_{0}^{8} x^{\frac{1}{3}+1} dx[/tex3]
[tex3]\sqrt{2}\int\limits_{0}^{8} \frac{x^{\frac{3}{2}}}{\frac{3}{2}}dx\,+\,\int\limits_{0}^{8} \frac{x^{\frac{4}{3}}}{\frac{4}{3}} dx[/tex3]
[tex3]\frac{2}{3}\sqrt{2}\int\limits_{0}^{8} x^{\frac{3}{2}}dx\,+\,\int\limits_{0}^{8}\frac{3}{4}x^{\frac{4}{3}}dx[/tex3]
[tex3]\frac{2}{3}\sqrt{2}\(8^{\frac{3}{2}}\,-\,0^{\frac{3}{2}}\)\,+\;\frac{3}{4}\(8^{\frac{4}{3}}\,-\,0^{\frac{4}{3}}\)[/tex3]
[tex3]\frac{2}{3}\sqrt{2}\(\sqrt{8^3}\)\,+\;\frac{3}{4}\(\sqrt[3]{8^4}\)[/tex3]
[tex3]\frac{2}{3}\sqrt{2}\(\sqrt{512}\)\,+\;\frac{3}{4}\(\sqrt[3]{4096}\)[/tex3]
[tex3]\frac{2}{3}\sqrt{2}\(16\cdot \sqrt{2}\)\,+\;\(\frac{3}{4}\cdot 16\)[/tex3]
[tex3]\frac{2}{3}\(\sqrt{2^2}\cdot 16\)+\(\frac{3}{4}\cdot {16}\)[/tex3]
[tex3]\frac{2}{3}\cdot 2\cdot 16\;+\;\frac{3}{4}\cdot 16[/tex3]
[tex3]\frac{64}{3}\,+\,\frac{48}{4}[/tex3]
[tex3]\frac{(64\times4)\,+\,(48\times3)}{3\times 4}[/tex3]
[tex3]\frac{256\,+\,144}{12}[/tex3]
[tex3]\frac{400}{12}[/tex3]
[tex3]\boxed{\frac{100}{3}}[/tex3]
Quanto à segunda:
Numa calculadora TI-Nspire CX CAS dá:
- integral dupla-01.jpg (14.8 KiB) Exibido 2411 vezes
Se digitar como se segue: [tex3](int(0,8)int(x,3x)(3*sqrt(16-x^2)dy)dx[/tex3]
, no WolframAlpha, dará o valor de:[tex3]128.000000000...(1+3i\sqrt{3})\approx 128.\,+\,665.108i[/tex3]
Espero que de certa forma tenha ajudado