integral from 0 to 1 of ((-x+1)^3)/6

Lösung

\int_{0}^{1} \frac{( - x + 1 ) ^{3}}{6} d x

\frac{1}{24}

Dezimale

0.04166 \dots

Schritte zur Lösung

\int_{0}^{1} \frac{( - x + 1 ) ^{3}}{6} d x

Entferne die Konstante:

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

= \frac{1}{6} \cdot \int_{0}^{1} (- x + 1)^{3} d x

Multipliziere aus

(- x + 1)^{3} : - x^{3} + 3 x^{2} - 3 x + 1

= \frac{1}{6} \cdot \int_{0}^{1} - x^{3} + 3 x^{2} - 3 x + 1 d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{6} (- \int_{0}^{1} x^{3} d x + \int_{0}^{1} 3 x^{2} d x - \int_{0}^{1} 3 x d x + \int_{0}^{1} 1 d x)

\int_{0}^{1} x^{3} d x = \frac{1}{4}

\int_{0}^{1} 3 x^{2} d x = 1

\int_{0}^{1} 3 x d x = \frac{3}{2}

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

= \frac{1}{6} (- \frac{1}{4} + 1 - \frac{3}{2} + 1)

Vereinfache

\frac{1}{6} (- \frac{1}{4} + 1 - \frac{3}{2} + 1) : \frac{1}{24}

= \frac{1}{24}

\int_{1}^{3} h (x) d x

\int_{1}^{3} x e^{2 x} d x

\int_{4}^{x} \frac{1}{1 + t + t ^{2}} d t

\int_{0}^{1} e^{6 y} y d y

\int_{- π}^{π} cos^{2} (\frac{x}{2}) d x