Question

Три задания по функциям. 50 баллов + 50 за лучший

Answer 1

1.

$S = \int\limits^{ 2 } _ {1}( {x}^{2} - 1)dx = ( \frac{ {x}^{3} }{3} - x) |^{ 2 } _ {1} = \\ = \frac{8}{3} - 2 - \frac{1}{3} + 1 = \frac{7}{3} - 1 = \frac{4}{3}$

2.

1)

$\int\limits^{ 2\pi } _ {0}(2 \sin( \frac{x}{6} ) + \cos(5x)) dx = \\ =2 \times 6 \int\limits^{ 2\pi } _ {0} \sin( \frac{x}{6} )d( \frac{x}{6}) + \frac{1}{5} \int\limits^{ 2\pi } _ {0} \cos(5x) d(5x) = \\ = ( - 12 \cos( \frac{x}{6} ) + \frac{1}{5} \sin(5x)) |^{ 2 \pi} _ {0} = \\ = - 12 \cos( \frac{\pi}{3} ) + \frac{1}{5} \sin(10\pi) + 12 \cos(0) - \frac{1}{5} \sin(0) = \\ = - 12 \times \frac{1}{2} + 0 + 12 - 0 = 12 - 6 = 6$

2)

$\int\limits^{ 1} _ { \frac{1}{5} }(5x - 3) {}^{5}dx = \frac{1}{5} \int\limits^{ 1 } _ { \frac{1}{5} }(5x - 3) {}^{5}d(5x - 3) = \\ = \frac{ {(5x - 3)}^{6} }{30} |^{ 1} _ { \frac{1}{5} } = \\ = \frac{(5 - 3) {}^{6} - (1 - 3) {}^{6} }{30} = \frac{ {2}^{6} - ( - 2) {}^{6} }{30} = 0$

3.

1)

$y = \frac{ {x}^{5} }{ ln(x) }$

$y '= \frac{( {x}^{5} )' \times ln(x) - (ln(x) )' \times {x}^{5} }{ {ln}^{2}(x) } = \\ = \frac{5 {x}^{4} ln(x) - \frac{1}{x} \times {x}^{5} }{ {ln}^{2}(x) } = \frac{5 {x}^{4} ln(x) - {x}^{4} }{ {ln}^{2}(x) } = \\ = \frac{ {x}^{4} (5 ln(x) - 1)}{ {ln}^{2} (x)}$

2)

$y = {x}^{6} {e}^{6}$

$y' = 6 {x}^{5} {e}^{6}$