integral from 0 to 1 of 1/((1+2y)^3)

Lösung

\int_{0}^{1} \frac{1}{( 1 + 2 y ) ^{3}} d y

\frac{2}{9}

Dezimale

0.22222 \dots

Schritte zur Lösung

\int_{0}^{1} \frac{1}{( 1 + 2 y ) ^{3}} d y

Wende U-Substitution an

= \int_{1}^{3} \frac{1}{2 u ^{3}} d u

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int_{1}^{3} \frac{1}{u ^{3}} d u

Wende die Potenzregel an

= \frac{1}{2} [- \frac{1}{2 u ^{2}}]_{1}^{3}

Berechne die Grenzen:

\frac{4}{9}

= \frac{1}{2} \cdot \frac{4}{9}

Vereinfache

= \frac{2}{9}

\int_{0}^{0.8} \frac{1}{1 - x} d x

\int_{0}^{1} 1 \cdot x d x

\int_{- 1}^{1} \frac{6 x}{( x + 6 ) ^{2}} d x

\int_{3}^{9} 1 + x^{3} d x

\int_{- 2}^{+ 2} \frac{1}{x ^{3}} d x