integral from 0 to 1 of te^{2t}

Lösung

\int_{0}^{1} t e^{2 t} d t

\frac{e ^{2} + 1}{4}

Dezimale

2.09726 \dots

Schritte zur Lösung

\int_{0}^{1} t e^{2 t} d t

Wende U-Substitution an

= \int_{0}^{2} \frac{e ^{u} u}{4} d u

Entferne die Konstante:

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

= \frac{1}{4} \cdot \int_{0}^{2} e^{u} u d u

Wende die partielle Integration an

= \frac{1}{4} [e^{u} u - \int e^{u} d u]_{0}^{2}

\int e^{u} d u = e^{u}

= \frac{1}{4} [e^{u} u - e^{u}]_{0}^{2}

Berechne die Grenzen:

e^{2} + 1

= \frac{1}{4} (e^{2} + 1)

Vereinfache

= \frac{e ^{2} + 1}{4}

t an g e n t tan (x), at x = - \frac{π}{3}

\int_{0}^{14} 10 e^{x} d x

\frac{\partial}{\partial x} (x y ln (\frac{x}{y}))

x \to - \frac{3 π}{2} - lim (tan (x))

\int_{1}^{b} \frac{1}{( x + 7 ) ^{2}} d x