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=$\int_{1}^{\frac{4}{3}}\dfrac{1}{x^{2}}\sqrt{1-\dfrac{1}{x}}dx$, $u=1-\dfrac{1}{x}$ και ισούται $\int_{0}^{\frac{1}{4}}\sqrt{u}du=\dfrac{1}{12}$.
1 σχόλιο:
=$\int_{1}^{\frac{4}{3}}\dfrac{1}{x^{2}}\sqrt{1-\dfrac{1}{x}}dx$, $u=1-\dfrac{1}{x}$ και ισούται
ΑπάντησηΔιαγραφή$\int_{0}^{\frac{1}{4}}\sqrt{u}du=\dfrac{1}{12}$.