<s>[INST] Finding the value of the definite integral $\int_0^2{x\int_x^2{\frac{dy}{\sqrt{1+y^3}}}}dx$ If $$f(x) = \int_x^2{\frac{dy}{\sqrt{1+y^3}}}$$
then find the value of $$\int_0^2{xf(x)}dx$$
I have no idea how to solve this question. Please help.
 [/INST] Using integration by parts
$$\int_0^2xf(x)dx=\frac{1}{2}\int_{x=0}^{x=2}f(x)d(x^2)$$
$$=\frac{1}{2}x^2f(x)\bigg|_0^2-\frac{1}{2}\int_0^2x^2f'(x)dx$$
$$=2f(2)+\frac{1}{2}\int_0^2x^2\frac{1}{\sqrt{1+x^3}}dx$$
$$=0+\frac{1}{3}\sqrt{1+x^3}\bigg|_0^2$$
$$=\frac{1}{3}(3-1)$$
$$=\frac{2}{3}$$
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