\n\\int_{0}^{1} x^{2} e^{x^{3}-1} d x\nu=x^{3}-1\n

Explanation:

Step1: Use substitution

Let (u = x^{3}-1), then (du=3x^{2}dx), and (x^{2}dx=\frac{1}{3}du). When (x = 0), (u=0^{3}-1=-1); when (x = 1), (u=1^{3}-1 = 0). The integral (\int_{0}^{1}x^{2}e^{x^{3}-1}dx) becomes (\frac{1}{3}\int_{-1}^{0}e^{u}du).

Step2: Integrate (e^{u})

The antiderivative of (e^{u}) is (e^{u}). So (\frac{1}{3}\int_{-1}^{0}e^{u}du=\frac{1}{3}[e^{u}]_{-1}^{0}).

Step3: Evaluate the definite integral

(\frac{1}{3}(e^{0}-e^{-1})=\frac{1}{3}(1 - \frac{1}{e})).

Answer:

(\frac{1}{3}(1-\frac{1}{e}))