Question

$\int\limits^1_0 {(1-x) e^{x} } \, dx$

Answer 1

= (1-x)e^x (de la 0 la 1) + $\int\limits^1_0 {e^x} \, dx$ = 0 -1 + e^x(de la 0 la 1) = -1 +e-e^0 = e--2
Am folosit formula integrarii prin parti si apoi formula Newton-Leibniz.
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Succes in continuare!