с интегралами по выш мату , (Нужно решить первый второй и третий )

Ответы

gre4g3g4 10.03.2021 14:11

$\int\limits^{4} _ {0} \sqrt{9 + 4x}dx = \frac{1}{4} \int\limits^{4} _ {0} {(9 + 4x)}^{ \frac{1}{2} }d(4x) = \\ = \frac{1}{4} \int\limits^{4} _ {0} {(9 + 4x)}^{ \frac{1}{2} } d(9 + 4x) = \\ = \frac{1}{4} \times \frac{ {(9 + 4x)}^{ \frac{3}{2} } }{ \frac{3}{2} } | ^{4} _ {0} = \frac{1}{6} \sqrt{ {(9 + 4x)}^{3} } | ^{4} _ {0} = \\ = \frac{1}{6} ( \sqrt{{(9 + 16)}^{3}} - \sqrt{9^{3}} ) = \frac{1}{6} (125 - 27) = \frac{98}{6} = \frac{49}{3}$

$\int\limits^{ \frac{8}{3} } _ {0} \frac{xdx}{ \sqrt{3x + 1} } \\$

Замена:

$3x + 1 = t \\ 3dx = dt \\ dx = \frac{dt}{3} \\ x = \frac{t - 1}{3} \\ \\ t1 = 3 \times \frac{8}{3} + 1 = 9 \\ t2 = 3 \times 0 + 1 = 1$

$\int\limits^{9} _ {1} \frac{(t - 1)dt}{9 \sqrt{t} } = \frac{1}{9} \int\limits^{9} _ {1}( \frac{t}{ \sqrt{t} } - \frac{1}{ \sqrt{t} })dt = \\ = \frac{1}{9} \int\limits^{9} _ {1}( {t}^{ \frac{1}{2} } - {t}^{ - \frac{1}{2} })dt = \frac{1}{9} ( \frac{ {t}^{ \frac{3}{2} } }{ \frac{3}{2} } - \frac{ {t}^{ \frac{1}{2} } }{ \frac{1}{2} }) | ^{9} _ {1} = \\ = \frac{1}{9}( \frac{2}{3} t \sqrt{t} - 2 \sqrt{t}) | ^{9} _ {1} = \\ = \frac{1}{9} ( \frac{2}{3} \times 9 \times 3 - 2 \sqrt{9} - ( \frac{2}{3} -2 )) = \\ = \frac{1}{9} (18 - 6 + \frac{2}{3} + 2) = \\ = \frac{1}{9} (14 + \frac{2}{3} ) = \frac{1}{9} \times \frac{44}{3} = \frac{44}{27}$

$\int\limits^{ \frac{14}{3} } _ { \frac{7}{3} } \frac{xdx}{ \sqrt{3x + 2} } \\$

замена:

$3x + 2 = t \\ x = \frac{t - 2}{3} \\ 3dx = dt \\ dx = \frac{1}{3} dt \\ \\ t1 = 3 \times \frac{14}{3} + 2 = 16\\ t2 = 3 \times \frac{7}{3} + 2 = 9$

$\int\limits^{16} _ {9} \frac{(t - 2)dt}{9 \sqrt{t} } = \frac{1}{9} \int\limits^{16} _ {9}( \frac{t}{ \sqrt{t} } - \frac{2}{ \sqrt{t} })dt = \\ = \frac{1}{9} \int\limits^{16} _ {9}( {t}^{ \frac{1} {2} } - 2 {t}^{ - \frac{1}{2} } )dt = \frac{1}{9} ( \frac{2}{3}t \sqrt{t} - 4 \sqrt{t} ) | ^{16} _ {9} = \\ = \frac{1}{9} ( \frac{2}{3} \times 16 \times 4 - 4 \times 4 - ( \frac{2}{3} \times 9 \times 3 - 4 \times 3)) = \\ = \frac{1}{9} ( \frac{128}{3} - 16 - \frac{54}{3} + 12) = \\ = \frac{1}{3} ( - 4 + \frac{74}{3} ) = \frac{1}{9} \times \frac{74 - 12}{3} = \frac{62}{27}$