Question

СРОЧНО СРОЧНО СРОЧНО СРОЧНО

Answer 1

Ответ:

1)

$\int_{0}^{2} {2}^{x} dx = ( \frac{ {2}^{x} }{ ln(2) } )|_{0}^{2} = \frac{4}{ ln(2) } - \frac{1}{ ln(2) } = \frac{3}{ ln(2) }$

2)

$\sqrt{x} = 1 \\ x = 1$

$\int_{0}^{1} 1 - \sqrt{x} dx = (x - \frac{2}{3} x \sqrt{x} )|_{0}^{1} = 1 - \frac{2}{3} \times 1 \times \sqrt{1} - (0 - 0) = \frac{1}{3}$

3)

$\cos(x) = \sin(x) \\ \cot(x) = 1 \\ x = \frac{\pi}{4} + k\pi, \: k \in Z$

$\int _{0}^{ \frac{\pi}{4} } \cos(x) - \sin(x) dx = ( \sin(x) + \cos(x))|_{0}^{ \frac{\pi}{4} } = \sqrt{2} - 1$

$\int_{ \frac{\pi}{4} }^{ \frac{5\pi}{4} } \sin(x) - \cos(x) dx = ( - \cos(x) - \sin(x) )| _{ \frac{\pi}{4} }^{ \frac{5\pi}{4} } = 2 \sqrt{2}$

$\int_{ \frac{5\pi}{4} }^{2\pi} \cos(x) - \sin(x) dx = ( \sin(x) - \cos(x) )| _{ \frac{5\pi}{4} }^{2\pi} = 1 + \sqrt{2}$

$\sqrt{2} - 1 + 2 \sqrt{2} + 1 + \sqrt{2} = 4 \sqrt{2}$