Całki oznaczone cd...

[Robson1416]

Mam jeszcze 3 całki oznaczone otóż:


\(\int_{\frac{1}{4}}^{1} \frac{dx}{ \sqrt{x}(4-x) }\)


\(\int_{0}^{3} \sqrt{9-x^{2}}dx\)

Teraz całki oznaczone przez części


\(\int_{0}^{1} x^2e^2x\)


\(\int_{ \sqrt{e} }^{e} \frac{ln x}{x^2}\)

[slawekstudia6]

\(\int \frac{dx}{ \sqrt{x}(4-x)}= \begin{vmatrix} x=k^2\\dx=2kdx\end{vmatrix}=\int \frac{2kdk}{ k(4-k^2)}= 2\int \frac{dk}{4-x^2} =\\=2\int \left( \frac{1}{4(k+2)}-\frac{1}{4(k-2)} \right) dk= \frac{1}{2} \left(ln|k+2|-ln|k-2| \right)= \frac{1}{2} ln \left| \frac{k+2}{k-2} \right|=\\=\frac{1}{2} ln \left| \frac{ \sqrt{x} +2}{\sqrt{x} -2} \right|\)

\(\int_{\frac{1}{4}}^{1} \frac{dx}{ \sqrt{x}(4-x) }=\frac{1}{2} \left[ ln \left| \frac{ \sqrt{x} +2}{\sqrt{x} -2} \right|\right]^1_{ \frac{1}{4} }=\frac{1}{2} \left( ln \left| \frac{ 1 +2}{1 -2} \right|-ln \left| \frac{ \frac{1}{2} +2}{\frac{1}{2}-2} \right|\right)=\frac{1}{2} \left( ln3-ln \frac{5}{3} \right)=\frac{1}{2}ln \frac{9}{5}\)

[slawekstudia6]

\(\int\sqrt{9-x^{2}}dx= \begin{vmatrix} x=3sink\\dx=3coskdk\\k=arcsin \frac{x}{3} \\sink= \frac{x}{3}\\cosk= \frac{\sqrt{9-x^{2}}}{3} \end{vmatrix} =\int\sqrt{9-3sin^2k} \cdot 3coskdk=9\int cos^2kdk=\\= \frac{9}{2} \int(1+cos2k)dk= \frac{9}{2}k+ \frac{9}{2} \cdot \frac{1}{2} sin2k= \frac{9}{2} k+\frac{9}{2}sinkcosk=\\= \frac{9}{2} arcsin \frac{x}{3}+\frac{9}{2} \cdot \frac{x}{3} \cdot \frac{\sqrt{9-x^{2}}}{3}=\frac{9}{2}arcsin \frac{x}{3}+ \frac{1}{2} x\sqrt{9-x^{2}}\)

\(2cos^2k-1=cos2k\\cos^2k= \frac{1}{2} +\frac{1}{2}cos2k\)

\(\int_{0}^{3} \sqrt{9-x^{2}}dx= \frac{1}{2} \left[9 arcsin \frac{x}{3}+ x\sqrt{9-x^{2}} \right]^3_0=\\=\frac{1}{2} \left( 9 arcsin \frac{3}{3}+ 3\sqrt{9-3^{2}}-9 arcsin \frac{0}{3}- 0\sqrt{9-0^{2}}\right)= \frac{1}{2} \cdot 9 \cdot \frac{\pi}{2} = \frac{9\pi}{4}\)

[slawekstudia6]

\(\int x^2e^{2x}= \frac{1}{2} e^{2x} \cdot x^2-\int\frac{1}{2} e^{2x} \cdot 2xdx=\frac{1}{2} e^{2x} \cdot x^2-\int e^{2x} xdx=\\=\frac{1}{2} e^{2x} \cdot x^2-\frac{1}{2} e^{2x} \cdot x+\frac{1}{2} \int e^{2x} dx=\frac{1}{2} e^{2x} \cdot x^2-\frac{1}{2} e^{2x} \cdot x+\frac{1}{4} e^{2x}\)

\(\int_{0}^{1}x^{2}e^{2x}=\frac{1}{4}e^{2}-\frac{1}{4}\)

[slawekstudia6]

\(\int \frac{ln x}{x^2}=- \frac{1}{x} lnx+\int \frac{1}{x} \cdot \frac{1}{x} dx=- \frac{1}{x} lnx+\int \frac{1}{x^2}dx=- \frac{1}{x}lnx- \frac{1}{x}\)

\(\int_{\sqrt{e}}^{e}\frac{\ln x}{x^{2}}= - \frac{2}{e} +\frac{3}{2 \sqrt{e} }\)