12. Evaluate the definite integral. \[ \int_{0}^{1} x^{8} e^{-x^{9}} d x \] \( \square \)

To evaluate the integral \( \int_{0}^{1} x^{8} e^{-x^{9}} d x \), we can use the substitution \( u = x^{9} \), which gives \( du = 9x^{8} dx \) or \( dx = \frac{du}{9x^{8}} \). Changing the limits accordingly: when \( x = 0 \), \( u = 0 \) and when \( x = 1 \), \( u = 1 \). The integral becomes: \[ \int_{0}^{1} x^{8} e^{-x^{9}} d x = \frac{1}{9} \int_{0}^{1} e^{-u} du \] The integral \( \int e^{-u} du \) evaluates to \(-e^{-u}\), so: \[ \frac{1}{9} \left[-e^{-u}\right]_{0}^{1} = \frac{1}{9} \left[-e^{-1} + 1\right] = \frac{1}{9} \left(1 - \frac{1}{e}\right) \] Thus, the final result is: \[ \int_{0}^{1} x^{8} e^{-x^{9}} d x = \frac{1}{9} \left(1 - \frac{1}{e}\right) \]