integral from-1 to 0 of (2t+2)^2

Lösung

\int_{- 1}^{0} (2 t + 2)^{2} d t

\frac{4}{3}

Dezimale

1.33333 \dots

Schritte zur Lösung

\int_{- 1}^{0} (2 t + 2)^{2} d t

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Wende die Potenzregel an

= \frac{1}{2} [\frac{u ^{3}}{3}]_{0}^{2}

Berechne die Grenzen:

\frac{8}{3}

= \frac{1}{2} \cdot \frac{8}{3}

Vereinfache

= \frac{4}{3}

\int_{1}^{\infty} x e^{- a x} d x

\int_{0}^{\infty} x + 3 x^{2} d x

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

\int_{0}^{4} \frac{y}{25 - y ^{2}} d y

\int_{π}^{3} csc^{2} (x) d x