2 до 0 sin2xdx...

November 2021 1 23 Report

1-вычислите интегралы 2,25 до 0,25 (dx)/ корень из x
2-вычислите интегралы пи/2 до 0 sin2xdx

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DimaPuchkov $\int\limits^{2,25}_{0,25} {\frac{dx}{\sqrt{x}}} = \int\limits^{2,25}_{0,25} {x^{-\frac{1}{2}}} \, dx = (\frac{x^{-\frac{1}{2} +1}}{-\frac{1}{2} + 1}) |^{2,25}_{0,25} = (\frac{x^{\frac{1}{2}}}{\frac{1}{2}}) |^{2,25}_{0,25} = ( 2\sqrt{x} ) |^{2,25}_{0,25} = \\ \\ = 2 \cdot ( \sqrt{2,25}- \sqrt{0,25} )= 2 \cdot ( \sqrt{2\frac{1}{4}}- \sqrt{\frac{1}{4}} ) = 2 \cdot ( \sqrt{\frac{9}{4}}- \frac{1}{2} ) =\\ \\ = 2 \cdot ( \frac{3}{2}}- \frac{1}{2} ) = 2 \cdot 1=2$

$\int\limits^{\frac{\pi}{2}}_0 {\sin 2x} \, dx = \frac{1}{2} \int\limits^{\frac{\pi}{2}}_0 {\sin 2x} \, d(2x)=-\frac{1}{2} \cdot (\cos 2x) |^{\frac{\pi}{2}}_0=-\frac{1}{2} \cdot (\cos \pi - \cos 0) = \\ \\ =-\frac{1}{2} \cdot (-1 - 1)=-\frac{1}{2} \cdot (-2) = 1$

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