Question

2x%7D%20%20%20%2B%20%20%5Csqrt%7B1%20%2B%20%20%7Bx%7D%5E3%20%7D%20%5Cright%20%20%29dx%20%5C%5C%20" id="TexFormula1" title=" \rm \int_{0}^2 \left( \sqrt[3]{ {x}^{2} + 2x} + \sqrt{1 + {x}^3 } \right )dx \\ " alt=" \rm \int_{0}^2 \left( \sqrt[3]{ {x}^{2} + 2x} + \sqrt{1 + {x}^3 } \right )dx \\ " align="absmiddle" class="latex-formula">​

Answer 1

If f(x) has an inverse on [a, b], then integrating by parts (take u = f(x) and dv = dx), we can show

$\displaystyle \int_a^b f(x) \, dx = b f(b) - a f(a) - \int_{f(a)}^{f(b)} f^{-1}(x) \, dx$

Let $f(x) = \sqrt{1 + x^3}$. Compute the inverse:

$f\left(f^{-1}(x)\right) = \sqrt{1 + f^{-1}(x)^3} = x \implies f^{-1}(x) = \sqrt[3]{x^2-1}$

and we immediately notice that $f^{-1}(x+1)=\sqrt[3]{x^2+2x}$.

So, we can write the given integral as

$\displaystyle \int_0^2 f^{-1}(x+1) + f(x) \, dx$

Splitting up terms and replacing $x \to x-1$ in the first integral, we get

$\displaystyle \int_1^3 f^{-1}(x) \, dx + \int_0^2 f(x) \, dx = 2 f(2) - 0 f(0) = 2\times3+0\times1=\boxed{6}$