Задача 2 (Визначений інтеграл)

Обчислити визначені інтеграли:

а)  \int_{1}^{4}{\left(4x^{3}-3x\sqrt{x} \right)dx} ;

б)  \int_{0}^{\frac{\pi }{2}}{cosxsinxdx} ;

в)  \int_{1}^{2}{\frac{e^{x}-x^{3}}{x^{3}e^{x}}dx} ;

г)  \int_{1}^{4}{\left(2\sqrt{x}-x \right)^{2}dx} ;

д)  \int_{1}^{4}{\left(\frac{3}{x} +x \right)dx} .

♦ а)  \int_{1}^{4}{\left(4x^{3}-3x\sqrt{x} \right)dx}=\int_{1}^{4}{4x^{3}-3x^{\frac{3}{2}}dx}=

 = \left(4\cdot \frac{x^{4}}{4}-3\cdot \frac{^{\frac{5}{2}}}{\frac{5}{2}} \right)|_{1}^{4}=

 = \left(x^{4}-\frac{6}{5}\sqrt{x^{5}} \right)|_{1}^{4}=4^{4}-\frac{6}{5}\left(\sqrt{4} \right)^{2}-\left(1^{4}-\frac{6}{5}\sqrt{1^{5}} \right)=

 = 256-\frac{6}{5}\cdot 32-1+\frac{6}{5}=255-31\cdot \frac{6}{5}=

 = 255-37\frac{1}{5}=217\frac{4}{5}=217,8  ;

б)  \int_{0}^{\frac{\pi }{2}}{cosxsinxdx}=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}{2cosxsinxdx}=  

 = \frac{1}{2}\int_{0}^{\frac{\pi }{2}}{sin2xdx}=-\frac{1}{2}\cdot \frac{1}{2}cos2x|_{0}^{\frac{\pi }{2}}=

 = -\frac{1}{4}cos2x|_{0}^{\frac{\pi }{2}}=-\frac{1}{4}cos\pi -(-\frac{1}{4}cos0)= -\frac{1}{4}(-1)+\frac{1}{4}=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2} ;

в)  \int_{1}^{2}{\frac{e^{x}-x^{3}}{x^{3}e^{x}}dx}=\int_{1}^{2}{\left(\frac{e^{x}}{x^{3}e^{x}}-\frac{x^{3}}{x^{3}e^{x}} \right)dx}=

 = \int_{1}^{2}{(x^{-3}-e^{-x})dx}=\left(\frac{x^{-2}}{-2}+e^{-x} \right)|_{1}^{2}=\left(-\frac{1}{2x^{2}}+e^{-x} \right)|_{1}^{2}=

 = -\frac{1}{2\cdot 2^{2}}+e^{-2}-\left(-\frac{1}{2\cdot 1^{2}} +e^{-1} \right)=-\frac{1}{8}+e^{-2}+\frac{1}{2}-e^{-1}=

 = \frac{3}{8}+\frac{1}{e^{2}}-\frac{1}{e}=\frac{3e^{2}-8e+8}{8e^{2}}  ;

г)  \int_{1}^{4}{\left(2\sqrt{x}-x \right)^{2}dx}=\int_{1}^{4}{\left(4x-4x\sqrt{x}+x^{2} \right)dx}=

 = \int_{1}^{4}{\left(4x-4x^{\frac{3}{2}}+x^{2} \right)dx}=\left(4\cdot \frac{x^{2}}{2}-4\cdot \frac{x\frac{5}{2}}{\frac{5}{2}}+\frac{x^{3}}{3} \right)|_{1}^{4}=

 = \left(2x^{2}-\frac{8}{5}\left(\sqrt{x} \right)^{5}+\frac{1}{3}x^{3} \right)|_{1}^{4}=2\cdot 4^{2}-\frac{8}{5}\left(\sqrt{4} \right)^{5}+\frac{1}{3}\cdot 4^{3}-

 - \left(2\cdot 1^{2}-\frac{8}{5}\left(\sqrt{1} \right)^{5}+\frac{1}{3}\cdot 1^{3} \right)=32-\frac{8}{5}\cdot 32+\frac{64}{3}-2+\frac{8}{5}-\frac{1}{3}=

 = -\frac{3}{5}\cdot 32+21-\frac{2}{5}=-\frac{98}{5}+21=\frac{7}{5}=1,4 ;

д)  \int_{1}^{4}{\left(\frac{3}{x} +x \right)dx}=\left(3ln\left|x \right|+\frac{x^{2}}{2} \right)|_{1}^{4}=

 = 3ln4+\frac{4^{2}}{2}-\left(3ln1+\frac{1}{2} \right)=

 = 3ln4+8-3ln1=3ln2^{2}+8-0-\frac{1}{2}=6ln2+7,5  .♦

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