Question

Решите определенный интеграл
Очень нужно

Answer 1

б

$\int\limits^{ {e}^{2} } _ {e} \frac{(1 + ln(x)) }{x} dx \\ \\ 1 + ln(x) = t \\ \frac{dx}{x} = dt \\ t1 = 1 + ln(e {}^{2} ) = 3 \\ t2 = 1 + ln(e) = 2\\ \\ \int\limits^{ 2 } _ {1}tdt = \frac{ {t}^{2} }{2} | ^{ 2 } _ {1} = 2 - 0.5 = 1.5$

в

$\int\limits^{ \pi } _ { \frac{\pi}{4} }x \cos(6x) dx \\$

По частям:

$u = x \: \: \: \: \: \: \: \: du = dx \\ dv = \cos(6x) dx \: \: \: v = \frac{1}{6} \sin(6x) \\ \\ uv - \int\limits \: vdu = \\ = \frac{x}{6} \sin(6x) | ^{ \pi } _ { \frac{\pi}{4} } - \frac{1}{6} \int\limits^{ \pi } _ { \frac{}{4} } \sin(6x) dx = \\ = \frac{x}{6} \sin(6x) | ^{ \pi } _ { \frac{\pi}{4} } - \frac{1}{6} \times \frac{1}{6} \int\limits^{ \pi } _ { \frac{\pi}{4} } \sin(6x)d(6x) = \\ = ( \frac{x}{6} \sin(6x) + \frac{1}{36} \cos(6x) )| ^{ \pi } _ { \frac{\pi}{4} } = \\ = \frac{\pi}{6} \times 0 + \frac{1}{36} \times 1 - ( \frac{\pi}{24} \sin( \frac{3\pi}{2} ) + \frac{1}{ 36} \cos( \frac{3\pi}{2} ) ) = \\ = \frac{1}{36} + \frac{\pi}{24}$

г

$\int\limits^{ 2 } _ {0} \frac{dx}{ \sqrt{x} + 3} \\ \\ \sqrt{x} = t \\ \frac{dx}{2 \sqrt{x} } = dt \\ dx = 2tdt \\ t1 = \sqrt{2} \\ t2 = 0 \\ \\ \int\limits^{ \sqrt{2} } _ {0} \frac{2tdt}{t + 3} = 2 \int\limits^{ \sqrt{2} } _ {0} \frac{t + 3 - 3}{t + 3} dt = \\ = 2(\int\limits^{ \sqrt{2} } _ {0} \: dt - 3\int\limits^{ \sqrt{2} } _ {0} \frac{dt}{t + 3}) = \\ = (2t - 6 ln( |t + 3| ) )| ^{ \sqrt{2} } _ {0} = \\ = 2 \sqrt{2} - 6 ln(3 + \sqrt{2} ) - 0 + 6 ln(3) = \\ = 2 \sqrt{2} + 6 ln( \frac{3}{3 + \sqrt{2} } )$