Given that $f$ is a continuous function and that $\int_0^{x^2(1+x)} f(t) dt = x$, we need to calculate $f(2)$.

1. Problem Description

Given that $f$ is a continuous function and that $\int_0^{x^2(1+x)} f(t) dt = x$, we need to calculate $f(2)$.

2. Solution Steps

Let $g(x) = x^2(1+x) = x^2 + x^3$. Then we are given that

$$\int_0^{g(x)} f(t) dt = x.$$

Differentiating both sides with respect to $x$, using the Fundamental Theorem of Calculus and the Chain Rule, we have

$$f(g(x)) \cdot g'(x) = 1.$$

Thus,

$$f(g(x)) = \frac{1}{g'(x)}.$$

We want to find $f(2)$. Let $g(x) = 2$. Then we want to find $x$ such that $x^2(1+x) = 2$, or $x^3 + x^2 - 2 = 0$.

We can see that $x=1$ is a root of this equation since $1^3 + 1^2 - 2 = 1 + 1 - 2 = 0$. Thus, we can divide the polynomial by $(x-1)$.

$$(x^3 + x^2 - 2) \div (x-1) = x^2 + 2x + 2.$$

The roots of $x^2 + 2x + 2 = 0$ are $x = \frac{-2 \pm \sqrt{4 - 8}}{2} = \frac{-2 \pm \sqrt{-4}}{2} = -1 \pm i$. These are complex roots.

Since we are looking for a real value, we take $x = 1$. Thus, $g(1) = 1^2(1+1) = 2$.

So $f(g(1)) = f(2) = \frac{1}{g'(1)}$.

Now, we need to find $g'(x)$.

$g(x) = x^3 + x^2$.

$g'(x) = 3x^2 + 2x$.

So $g'(1) = 3(1)^2 + 2(1) = 3 + 2 = 5$.

Thus, $f(2) = \frac{1}{g'(1)} = \frac{1}{5}$.

3. Final Answer

$f(2) = \frac{1}{5}$