Explicație pas cu pas:
0 < 1 < e
[tex]\int_{0}^{1}f(x)dx + \int_{1}^{e}f(x)dx = \\ = \int_{0}^{1}(x^{2} - 3x + 2)dx + \int_{1}^{e}ln(x)dx \\ = \frac{x(2x^{2} - 9x + 12)}{6} |_{0}^{1} + (xln(x) - x) |_{1}^{e} \\ = \frac{5}{6} + 1 = \frac{11}{6} [/tex]
unde:
[tex]\int ln(x)dx = \int ln(x) \cdot 1dx \\ = \int ln(x) \cdot (x)'dx = xln(x) - x + C[/tex]
