Integrale definito

Question

[attach]103262[/attach] Svolgere senza X SOSTITUZIONE.

cmc · Accepted Answer

int _{-3}^3 ( frac {1}{x^2+9} , dx =$ $=  frac {1}{3}  int _{-3}^3 ( frac {1}{1+ ( frac {x}{3})^2} , dx =$ $=  left .  frac {1}{3} arctan( frac {x}{3}  right |_{-3}^3 = $ =  frac {1}{3} [arctan(1) - arctan(-1)] = $ $ =  frac {1}{3} [ 2arctan(1)] = $  $ =  frac {2}{3}  frac { pi }{4} = $   $ =   frac { pi }{6} $

Integrale definito

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