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integrala din 1/x^3 = -1/2x^2 .. cine imi poate explica ?

va salut cu respect!


Răspuns :

Salut,

[tex]\int x^ndx=\dfrac{1}{n+1}\cdot x^{n+1}+C\;(unde\;C\;este\;o\;constant\breve{a}\;real\breve{a}),\;pentru\;c\breve{a}\\\\\left(\dfrac{1}{n+1}\cdot x^{n+1}+C\right)^{'}=\dfrac{1}{n+1}\cdot(x^{n+1})^{'}+C^{'}=\dfrac1{n+1}\cdot(n+1)\cdot x^{n+1-1}+0=x^n.\\\\Pentru\;exerci\c{t}iul\;t\breve{a}u\;avem\;c\breve{a}\;n=-3,\;deci:\\\\\int x^{-3}dx=\dfrac{1}{-3+1}\cdot x^{-3+1}+C=-\dfrac{1}{2}\cdot x^{-2}+C=-\dfrac{1}{2\cdot x^2}+C.[/tex]

Green eyes.