Answer
$$\int^{1}_0x(\sqrt[3]x+\sqrt[4]x)dx=\frac{55}{63}$$
Work Step by Step
$$A=\int^{1}_0x(\sqrt[3]x+\sqrt[4]x)dx$$ $$A=\int^{1}_0x(x^{1/3}+x^{1/4})dx$$ $$A=\int^{1}_0(x^{4/3}+x^{5/4})dx$$According to Table 1, we have $$\int x^n dx=\frac{x^{n+1}}{n+1}+C$$ Therefore, $$A=(\frac{x^{7/3}}{\frac{7}{3}}+\frac{x^{9/4}}{\frac{9}{4}})\Bigg]^1_0$$ $$A=(\frac{3x^{7/3}}{7}+\frac{4x^{9/4}}{9})\Bigg]^1_0$$ $$A=(\frac{3\times1^{7/3}}{7}+\frac{4\times1^{9/4}}{9})-(\frac{3\times0^{7/3}}{7}+\frac{4\times0^{9/4}}{9})$$ $$A=\frac{3}{7}+\frac{4}{9}$$ $$A=\frac{55}{63}$$
