Question

Помогите вычислить интегралы по формуле Ньютона-Лейбница. Даю 100 баллов

Answer 1

Ответ:

1.

$\int\limits^{ 2 } _ {}2 {x}^{2}dx = \frac{2 {x}^{3} }{3} | ^{ 2 } _ {0} = \frac{2}{3} ( {2}^{3} - 0) = \frac{2 \times 8}{3} = \frac{16}{3}$

2.

$\int\limits^{ 3 } _ {2}( {x}^{2} + 4x) dx = ( \frac{ {x}^{3} }{3} + \frac{4 {x}^{2} }{2}) | ^{ 3 } _ {2} = \\ =( \frac{ {x}^{3} }{3} + 2 {x}^{2}) | ^{ 3 } _ {2} = \\ = 9 + 2 \times 9 - \frac{8}{3} - 8 = 27 - 8 - \frac{8}{3} = 19 - \frac{8}{3} = \\ = 19 - 2 \frac{2}{3} = 16 \frac{1}{3}$

3.

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

4.

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

5.

$\int\limits^{ 1 } _ { - 2}(2 {x}^{3} + 4x)dx = ( \frac{2 {x}^{4} }{4} + \frac{4 {x}^{2} }{2}) | ^{ 1 } _ { - 2} = \\ = ( \frac{ {x}^{4} }{2} + 2 {x}^{2} ) | ^{ 1 } _ { - 2} = \\ = \frac{1}{2} + 2 - 8 - 8 = 2.5 - 16 = - 13.5$

6.

$\int\limits^{ 2 } _ { - 1}( \frac{4 {x}^{3} }{3} - \frac{3}{4} x {}^{2} + 5)dx =( \frac{4}{3} \times \frac{ {x}^{4} }{4} - \frac{3}{4} \times \frac{ {x}^{3} }{3} + 5x)| ^{ 2 } _ { - 1} = \\ = ( \frac{ {x}^{4} }{3} - \frac{ {x}^{3} }{4} + 5x) | ^{ 2 } _ { - 1} = \\ = \frac{16}{3} - 2 + 10 - ( \frac{1}{3} + \frac{1}{4} - 5) = \\ = \frac{16}{3} + 8 - \frac{1}{3} + 4.75 = 3 + 8 + 4.75 = 15 . 75$