Integrálás - Alapvető integrálok

Feladat

1. Határozzuk meg az alábbi integrálokat!

a)

$\[\int \left(3+4x+5x^{2}\right)dx\]$

b)

$\[\int_{-1}^{2}\left(x^{2}-\sin(5x)\right)dx\]$

c)

$\[\int_{0}^{\pi}\sin^{2}x\,dx\]$

d)

$\[\int_{0}^{\pi}\cos^{2}x\,dx\]$

e)

$\[\int\sqrt{3x+2}dx\]$

f)

$\[\int\frac{1}{2x}dx\]$

Megoldás

a)

$\[\int \left(3+4x+5x^{2}\right)dx=3x+2x^{2}+\frac{5}{3}x^{3}+C\]$

b)

$\[\int_{-1}^{2}\left(x^{2}-\sin(5x)\right)dx=\left[\frac{x^{3}}{3}+\frac{\cos(5x)}{5}\right]^{2}_{-1}=\frac{8}{3}+\frac{\cos 10}{5}-\left(-\frac{1}{3}+\frac{\cos 5}{5}\right)\approx 2.78\]$

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)

$\[\int_{0}^{\pi}\cos^{2}x\,dx=\int_{0}^{\pi}\left(\frac{1}{2}+\frac{\cos(2x)}{2}\right)dx=\frac{\pi}{2}\]$

e)

$\[\int\sqrt{3x+2}dx=\int \left(3x+2\right)^{\frac{1}{2}}dx=2(3x+2)^{\frac{3}{2}}+C\]$

f)

$\[\int\frac{1}{2x}dx=\frac{\ln (2x)}{2}+C\]$

vagy

$\[\int\frac{1}{2x}dx=\frac{1}{2}\int\frac{1}{x}dx=\frac{1}{2}\ln x+C'\]$

$\[C'=C+\frac{\ln 2}{2}\]$