Question

Определенные интегралы​

Answer 1

Ответ:

4.

рисунок1

Найдем пределы:

${x}^{2} + 2x + 1 = x + 3 \\ {x}^{2} + x - 2 = 0 \\ d = 1 + 8 = 9 \\ x_1 = \frac{ - 1 + 3}{2} = 1 \\ x_2 = - 2$

$S= S_1 - S_2 = \int\limits^{ 1} _ { - 2}(x + 3)dx - \int\limits^{ 1 } _ { - 2}( {x}^{2} + 2x + 1) = \\ = \int\limits^{ 1} _ { - 2}(x + 3 - {x}^{2} - 2x - 1)dx = \int\limits^{ 1 } _ { - 2}( - {x}^{2} - x + 2) dx = \\ = ( - \frac{ {x}^{3} }{3} - \frac{ {x}^{2} }{2} + 2x) |^{ 1 } _ { - 2} = \\ = - \frac{1}{3} - \frac{1}{2} + 2 - ( \frac{8}{3} - 2 - 4) = \\ = - \frac{5}{6} + 2 - \frac{8}{3} + 6 = 8 - \frac{21}{6} = \\ = \frac{48 - 21}{6} = \frac{27}{6} = \frac{9}{2} = 4.5$

5.

рисунок2

$S = \int\limits^ \frac{\pi}{4} _0 \cos(x) dx = \sin(x) |^{ \frac{\pi}{4} } _ {0} = \\ = \sin( \frac{\pi}{4} ) - \sin(0) = \frac{ \sqrt{2} }{2}$