integral from 0 to 1 of 1/((1-x)^2)

Lösung

\int_{0}^{1} \frac{1}{( 1 - x ) ^{2}} d x

divergiert

Schritte zur Lösung

\int_{0}^{1} \frac{1}{( 1 - x ) ^{2}} d x

Wende U-Substitution an

= \int_{1}^{0} - \frac{1}{u ^{2}} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{1} - \frac{1}{u ^{2}} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \int_{0}^{1} \frac{1}{u ^{2}} d u)

Wende die Potenzregel an

= - (- [- \frac{1}{u}]_{0}^{1})

Vereinfache

= [- \frac{1}{u}]_{0}^{1}

Berechne die Grenzen:

divergiert

= divergiert

\int_{0}^{1} (x - 1) ln (x) d x

\int_{0}^{\frac{3 π}{2}} x sin (x) d x

\int_{1}^{7} ∣ sin (x) ∣ d x

\int_{0}^{1} (1 - x) sin (nπ x) d x

\int_{\frac{π}{2}}^{\frac{3 π}{2}} x sin (x) d x