Question

Добрый вечер, можете 6,7,8 решить. Буду крайне благодарен.

Answer 1

Ответ:

6

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

7

$\int\limits^{ 7 } _ {6} \frac{2dx}{ {(x - 5)}^{2} } = 2\int\limits^{ 7} _ {6} {(x - 5)}^{ - 2} d(x - 5) = \\ = \frac{ {2(x - 5)}^{ - 1} }{ - 1} |^{ 7 } _ {6} = - \frac{2}{x - 5} |^{ 7 } _ {6} = \\ = - \frac{2}{2} + \frac{2}{1} = - 1 + 2 = 1$

8

$\int\limits^{ 6 } _ {0} \sqrt{2x + 4} = \frac{1}{2} \int\limits^{ 6 } _ {0} {(2x + 4)}^{ \frac{1}{2} } d(2x + 4) = \\ = \frac{1}{2} \times \frac{ {(2x + 4)}^{ \frac{3}{2} } }{ \frac{3}{2} } |^{ 6 } _ {0} = \frac{1}{3} \sqrt{ {(2x + 4)}^{3} } |^{ 6 } _ {0} = \\ = \frac{1}{3} ( \sqrt{ {16}^{3} } - \sqrt{ {4}^{3} } ) = \frac{16 \times 4 - 4 \times 2}{3} = \\ = \frac{64 - 8}{3} = \frac{56}{3}$