Question

вычислите интеграл
3 задание

Answer 1

Ответ:

3.

a)

$\int\limits^{ 4 } _ {1}(4 \sqrt{x} - 3 {x}^{2})dx = (4 \times \frac{ {x}^{ \frac{3}{2} } }{ \frac{3}{2} } - \frac{3 {x}^{3} }{3} ) | ^{4} _ {1} = \\ = ( \frac{8}{3} x \sqrt{x} - {x}^{3} )| ^{4} _ {1} = \\ = \frac{8}{3} \times 4 \times 2 - 64 - \frac{8}{3} + 1 = \\ = \frac{64 - 8}{3} - 63 = \frac{56}{3} - 63 = - \frac{133}{3}$

б)

$\int\limits^{ \frac{\pi}{?} } _ { \frac{\pi}{2} } \sin((2x - \frac{\pi}{4} )dx = \frac{1}{2} \int\limits^{ \frac{\pi}{?} } _ { \frac{\pi}{4} } \sin(2x - \frac{\pi}{4} )d(2x - \frac{\pi}{4}) = \\ = ( - \cos(2x - \frac{\pi}{4} )) | ^{ \frac{\pi}{2} } _ { \frac{\pi}{4} } = \\ = - \cos(\pi - \frac{\pi}{4} ) + \cos( \frac{\pi}{2} - \frac{\pi}{4} ) = \\ = - \cos( \frac{3\pi}{4} ) + \cos( \frac{\pi}{4} ) = \frac{ \sqrt{2} }{2} + \frac{ \sqrt{2} }{2} = \sqrt{2}$

а)

$\int\limits^{ 4} _ {1}(4 {x}^{3} - 3 \sqrt{x} )dx = ( \frac{4 {x}^{4} }{4} - 3 \times \frac{ {x}^{ \frac{3}{2} } }{ \frac{3}{2} }) | ^{4} _ {1}= \\ = ( {x}^{4} - 2x \sqrt{x} ) | ^{4} _ {1} = \\ = 256 - 2 \times 4 \times 2 - 1 + 2 = \\ = 256 - 16 + 1 = 241$

б)

$\int\limits^{ \frac{\pi}{4} } _ {0} \frac{1}{2} \cos(x + \frac{\pi}{4} ) dx = \frac{1}{2} \int\limits^{ \frac{\pi}{4} } _ {0} \cos(x + \frac{\pi}{4} ) d(x + \frac{\pi}{4}) = \\ = \frac{1}{2} \sin(x + \frac{\pi}{4} ) | ^{ \frac{\pi}{4} } _ {0} = \\ = \frac{1}{2} ( \sin( \frac{\pi}{2} ) - \sin( \frac{\pi}{4} ) ) = \\ = \frac{1}{2} (1 - \frac{ \sqrt{2} }{2} ) = \frac{2 - \sqrt{2} }{4}$