Question

С подробным решением !

Answer 1

$\boxed {1.} \displaystyle \int^{27}_0 \frac{dx}{\sqrt[3]{x^2}}=\int^{27}_0 \frac{1}{x^{2/3}}dx}=\int^{27}_0 x^{-2/3}dx= \frac{x^{-2/3+1}}{-\frac{2}{3}+1}=$
$\displaystyle = \frac{x^{1/3}}{\frac{1}{3}}=3 \cdot x ^{1/3}= 3 \sqrt[3] x \bigg|^{27}_0= 3 \sqrt[3] {27}-0=3 \cdot 3 =9$

$\boxed{2.} \displaystyle \int^{3 \pi/2}_{\pi/2} sin \frac{x}{2}dx= \frac{1}{ \frac{1}{2}}\int^{3 \pi/2}_{\pi/2} sin \frac{x}{2}dx=2 \cdot (-cos \frac{x}{2})=-2cos\frac{x}{2} \bigg|^{3 \pi/2}_{\pi/2}=$
$\displaystyle = -2cos \frac{3 \pi}{2} \cdot \frac{1}{2}-(-2cos \frac{ \pi}{2} \cdot \frac{1}{2})=-2cos \frac{3 \pi}{4}-(-2cos \frac{\pi}{4})=$
$\displaystyle = -(-2 \cdot \frac{ \sqrt 2}{2})-(-2 \cdot \frac{ \sqrt 2}{2})=\sqrt2 + \sqrt 2 = 2\sqrt2$