integral from 0 to \sqrt[3]{pi of}t^2cos^2(t^3)sin(t^3)

Lösung

\int_{0}^{3 π} t^{2} cos^{2} (t^{3}) sin (t^{3}) d t

\frac{2}{9}

Dezimale

0.22222 \dots

Schritte zur Lösung

\int_{0}^{3 π} t^{2} cos^{2} (t^{3}) sin (t^{3}) d t

Wende U-Substitution an

= \int_{0}^{π} \frac{cos ^{2} ( u ) sin ( u )}{3} d u

Entferne die Konstante:

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

= \frac{1}{3} \cdot \int_{0}^{π} cos^{2} (u) sin (u) d u

Wende U-Substitution an

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

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

= \frac{1}{3} (- \int_{- 1}^{1} - v^{2} d v)

Entferne die Konstante:

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

= \frac{1}{3} (- (- \int_{- 1}^{1} v^{2} d v))

Wende die Potenzregel an

= \frac{1}{3} (- (- [\frac{v ^{3}}{3}]_{- 1}^{1}))

Vereinfache

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

Berechne die Grenzen:

\frac{2}{3}

= \frac{1}{3} \cdot \frac{2}{3}

Vereinfache

= \frac{2}{9}

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