integral from 0 to 1 of sech^2(5x)

Lösung

\int_{0}^{1} sech^{2} (5 x) d x

\frac{e ^{10} - 1}{5 ( e ^{10} + 1 )}

Dezimale

0.19998 \dots

Schritte zur Lösung

\int_{0}^{1} sech^{2} (5 x) d x

Wende U-Substitution an

= \int_{0}^{5} sech^{2} (u) \frac{1}{5} d u

Entferne die Konstante:

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

= \frac{1}{5} \cdot \int_{0}^{5} sech^{2} (u) d u

Nutze das gemeinsame Integral :

\int sech^{2} (u) d u = tanh (u)

= \frac{1}{5} [tanh (u)]_{0}^{5}

Berechne die Grenzen:

\frac{e ^{10} - 1}{e ^{10} + 1}

= \frac{1}{5} \cdot \frac{e ^{10} - 1}{e ^{10} + 1}

Vereinfache

= \frac{e ^{10} - 1}{5 ( e ^{10} + 1 )}

\int_{0}^{\infty} 25 e^{- 5 y} d y

\int_{0}^{l n (18)} tanh (x) d x

\int_{0}^{5} 2 (11 + 0.5 t) d t

\int_{10}^{100} \frac{1}{x} d x

\int_{21}^{32} k (x^{2} + y^{2}) d y