Question

Вычислить определенные интегралы:

Answer 1

$\int\limits^{ 1} _ {0} {x}^{2} {e}^{3x} dx \\ \\ \text{} \\ u = {x}^{2} \: \: \: \: \: \: du = 2xdx \\ dv = {e}^{3x} dx \: \: \: \: \: \: v = \frac{1}{3} \int\limits {e}^{3x} d(3x) = \frac{1}{3} e {}^{3x} \\ \\ \frac{ {x}^{2} }{3} {e}^{3x} | ^{ 1 } _ {0} - \int\limits^{ 1 } _ {0} \frac{2x}{3} {e}^{3x} dx = \\ = \frac{ {x}^{2} }{3} {e}^{3x}| ^{ 1} _ {0} - \frac{2}{3} \int\limits^{ 1 } _ {0} x{e}^{3x} dx \\ \\ u = x \: \: \: \: du = dx \\ dv = {e}^{3x} dx \: \: \: v = \frac{1}{3} {e}^{3x} \\ \\ \frac{ {x}^{2} }{3} {e}^{3x} | ^{ 1 } _ {0} - \frac{2}{3} ( \frac{x}{3} e {}^{3x} | ^{ 1 } _ {0} - \int\limits^{ 1 } _ {0} \frac{1}{3} {e}^{3x} dx) = \\ = \frac{ {x}^{2} }{3} {e}^{3x} - \frac{2x}{9} {e}^{3x} + \frac{2}{9} \times \frac{1}{3} {e}^{3x} )| ^{ 1 } _ {0} = \\ = {e}^{3x}( \frac{ {x}^{2} }{3} - \frac{2x}{9} + \frac{2}{27} )| ^{ 1 } _ {0} = \\ = {e}^{3} ( \frac{1}{3} - \frac{2}{9} + \frac{2}{27} ) - (0 + \frac{2}{27} ) = \\ = {e}^{3} \times \frac{9 - 6 + 2}{27} - \frac{2}{27} = \frac{5 {e}^{3} - 2}{27}$