„Integrálás - Alapvető integrálok” változatai közötti eltérés
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#: c)$$\int_{0}^{\pi}\sin^{2}x\,dx=\int_{0}^{\pi}\left(\frac{1}{2}-\frac{\cos(2x)}{2}\right)dx=\left[\frac{x}{2}\right]^{\pi}_{0}-\left[\frac{\sin(2x)}{4}\right]^{\pi}_{0}=\frac{\pi}{2}$$ | #: c)$$\int_{0}^{\pi}\sin^{2}x\,dx=\int_{0}^{\pi}\left(\frac{1}{2}-\frac{\cos(2x)}{2}\right)dx=\left[\frac{x}{2}\right]^{\pi}_{0}-\left[\frac{\sin(2x)}{4}\right]^{\pi}_{0}=\frac{\pi}{2}$$ | ||
#: d)$$\int_{0}^{\pi}\cos^{2}x\,dx=\int_{0}^{\pi}\left(\frac{1}{2}+\frac{\cos(2x)}{2}\right)dx=\frac{\pi}{2}$$ | #: d)$$\int_{0}^{\pi}\cos^{2}x\,dx=\int_{0}^{\pi}\left(\frac{1}{2}+\frac{\cos(2x)}{2}\right)dx=\frac{\pi}{2}$$ | ||
| − | #: e)$$\int\sqrt{3x+2}dx=\int \left(3x+2\right)^{\frac{1}{2}}dx= | + | #: e)$$\int\sqrt{3x+2}dx=\int \left(3x+2\right)^{\frac{1}{2}}dx=\frac 29(3x+2)^{\frac{3}{2}}+C$$ |
#: f)$$\int\frac{1}{2x}dx=\frac{\ln (2x)}{2}+C$$ vagy $$\int\frac{1}{2x}dx=\frac{1}{2}\int\frac{1}{x}dx=\frac{1}{2}\ln x+C'$$ $$C'=C+\frac{\ln 2}{2}$$</wlatex> | #: f)$$\int\frac{1}{2x}dx=\frac{\ln (2x)}{2}+C$$ vagy $$\int\frac{1}{2x}dx=\frac{1}{2}\int\frac{1}{x}dx=\frac{1}{2}\ln x+C'$$ $$C'=C+\frac{\ln 2}{2}$$</wlatex> | ||
</noinclude> | </noinclude> | ||
A lap jelenlegi, 2013. szeptember 20., 08:56-kori változata
| Navigáció Pt·1·2·3 |
|---|
| Kísérleti fizika gyakorlat 1. |
| Gyakorlatok listája: |
| Integrálás |
Feladatok listája:
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| © 2012-2013 BME-TTK, TÁMOP4.1.2.A/1-11/0064 |
Feladat
- Határozzuk meg az alábbi integrálokat!
- a)
![\[\int \left(3+4x+5x^{2}\right)dx\]](/images/math/4/b/2/4b2db38419b9f18cd58c88b080ab5dc1.png)
- b)
![\[\int_{-1}^{2}\left(x^{2}-\sin(5x)\right)dx\]](/images/math/2/6/0/2600c8c15e99e8308ee54fff0df29d0c.png)
- c)
![\[\int_{0}^{\pi}\sin^{2}x\,dx\]](/images/math/4/8/b/48b05376492ca302010168d09383a173.png)
- d)
![\[\int_{0}^{\pi}\cos^{2}x\,dx\]](/images/math/9/8/d/98dc32b7de1d0b2f040ea4af278e614d.png)
- e)
![\[\int\sqrt{3x+2}dx\]](/images/math/4/8/3/483106f173cac054e038addbc0b6af43.png)
- f)
![\[\int\frac{1}{2x}dx\]](/images/math/1/6/b/16b1a9170bf5c75ce98e459e167fcaaa.png)
- a)
Megoldás
- a)
![\[\int \left(3+4x+5x^{2}\right)dx=3x+2x^{2}+\frac{5}{3}x^{3}+C\]](/images/math/e/7/3/e73b9ba7443b843b52dfd95ba8daeb40.png)
- b)
![\[\int_{-1}^{2}\left(x^{2}-\sin(5x)\right)dx=\left[\frac{x^{3}}{3}+\frac{\cos(5x)}{5}\right]^{2}_{-1}=\frac{8}{3}+\frac{\cos 10}{5}-\left(-\frac{1}{3}+\frac{\cos 5}{5}\right)\approx 2.78\]](/images/math/5/2/b/52be26af3238b3bd237bbf2b91dd4a3a.png)
- c)
![\[\int_{0}^{\pi}\sin^{2}x\,dx=\int_{0}^{\pi}\left(\frac{1}{2}-\frac{\cos(2x)}{2}\right)dx=\left[\frac{x}{2}\right]^{\pi}_{0}-\left[\frac{\sin(2x)}{4}\right]^{\pi}_{0}=\frac{\pi}{2}\]](/images/math/c/7/c/c7c55ef2b4179db4374a504fc01a0e30.png)
- d)
![\[\int_{0}^{\pi}\cos^{2}x\,dx=\int_{0}^{\pi}\left(\frac{1}{2}+\frac{\cos(2x)}{2}\right)dx=\frac{\pi}{2}\]](/images/math/f/a/3/fa3e1f7bf92bcb951b490b127939080e.png)
- e)
![\[\int\sqrt{3x+2}dx=\int \left(3x+2\right)^{\frac{1}{2}}dx=\frac 29(3x+2)^{\frac{3}{2}}+C\]](/images/math/9/7/6/9767771c81abae49bae297c6581ed304.png)
- f)vagy
![\[\int\frac{1}{2x}dx=\frac{\ln (2x)}{2}+C\]](/images/math/7/e/4/7e4383b6766ddcc171d9dc88e3d97bea.png)
![\[\int\frac{1}{2x}dx=\frac{1}{2}\int\frac{1}{x}dx=\frac{1}{2}\ln x+C'\]](/images/math/4/d/1/4d1313aed81f8575a6fa6d1bc77bd9eb.png)
![\[C'=C+\frac{\ln 2}{2}\]](/images/math/f/c/9/fc9fe836c909149bbefeae7aaa50fb78.png)
- a)
