Question

Помогите пожалуйста с решением хотя бы 3 примеров

Answer 1

Ответ:

1.

$\int\limits^{ 2 } _ { - 1} {( {x}^{2} - 1) }^{3}xdx = \frac{1}{2} \int\limits^{ 2 } _ { - 1} {( {x}^{2} - 1) }^{3}2xdx = \\ = \frac{1}{2} \int\limits^{ 2 } _ { - 1} {( {x}^{2} - 1) }^{3} d( {x}^{2} - 1) = \frac{ {( {x}^{2} - 1)}^{4} }{8} |^{ 2 } _ { - 1} = \\ = \frac{1}{8} ( {3}^{4} - 0) = \frac{81}{8}$

2.

$\int\limits^{ 5 } _ { - 2} \sqrt[3]{5x + 2} dx = \frac{1}{5} \int\limits^{ 5} _ { - 2} {(5 x + 2)}^{ \frac{1}{3} } d(5x + 2) = \\ = \frac{1}{5} \times \frac{ {(5x + 2)}^{ \frac{4}{3} } }{ \frac{4}{3} } |^{ 5 } _ { - 2} = \frac{3}{20} \sqrt[3]{ {(5x + 2)}^{4} } |^{ 5} _ { - 2} = \\ = \frac{3}{20} ( \sqrt[3]{ {27}^{4} } - \sqrt[3]{( - 8) {}^{4} } ) = \frac{3}{20} (81 - 16) = \frac{3 \times 65}{20} = \\ = \frac{3 \times 13}{4} = \frac{39}{4}$

3.

$\int\limits^{ 2\pi } _ { \frac{3\pi}{2} } \sqrt{1 - \cos(x) } \sin(x) dx \\ \\ 1 - \cos(x) = t \\ \sin(x) dx = dt \\ t1 = 1 - \cos(2\pi) = 1 - 1 = 0 \\ t2 = 1 - \cos( \frac{3\pi}{2} ) = 1 \\ \\ \int\limits^{ 0 } _ {1} \sqrt{t}dt = \frac{2}{3}t \sqrt{t} |^{ 0 } _ {1} = \\ = \frac{2}{3} (0 - 1) = - \frac{2}{3}$

4.

$\int\limits^{ \frac{\pi}{3} } _ { \frac{\pi}{2} } \frac{ \sin(x) }{ 1 + \cos(x) } dx \\ \\ 1 + \cos(x) = t \\ - \sin(x) dx = dt \\ t1 = 1 + \cos( \frac{\pi}{3} ) = \frac{3}{2} \\ t2 = 1 + \cos( \frac{\pi}{2} ) = 1 \\ \\ - \int\limits^{ \frac{3}{2} } _ {1} \frac{dt}{t} = - ln(t) |^{ \frac{3}{2} } _ {1} = \\ = - ln(1.5 + ln(1) ) = - ln(1.5)$

5.

$\int\limits^{ 1 } _ {0} {e}^{ {x}^{2} } xdx = \frac{1}{2} \int\limits^{ 1 } _ {0} {e}^{ {x}^{2} }2 xdx = \\ = \frac{1}{2} \int\limits^{ 1 } _ {0} {e}^{ {x}^{2} }d( {x}^{2} ) = \frac{1}{2} e {}^{ {x}^{2} } |^{ 1 } _ {0} = \\ = \frac{e - 1}{2}$