integral from 2 to 5 of 2/(3x+1)

Lösung

\int_{2}^{5} \frac{2}{3 x + 1} d x

\frac{2}{3} (4 ln (2) - ln (7))

Dezimale

0.55111 \dots

Schritte zur Lösung

\int_{2}^{5} \frac{2}{3 x + 1} d x

Entferne die Konstante:

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

= 2 \cdot \int_{2}^{5} \frac{1}{3 x + 1} d x

Wende U-Substitution an

= 2 \cdot \int_{7}^{16} \frac{1}{3 u} d u

Entferne die Konstante:

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

= 2 \cdot \frac{1}{3} \cdot \int_{7}^{16} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{3} [ln ∣ u ∣]_{7}^{16}

Vereinfache

2 \cdot \frac{1}{3} [ln ∣ u ∣]_{7}^{16} : \frac{2}{3} [ln ∣ u ∣]_{7}^{16}

= \frac{2}{3} [ln ∣ u ∣]_{7}^{16}

Berechne die Grenzen:

4 ln (2) - ln (7)

= \frac{2}{3} (4 ln (2) - ln (7))

\int_{0}^{l n (2)} e^{x} - e^{- x} d x

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

\int_{- 2}^{0} 9 x^{2} (2 x^{3} + 5)^{2} d x

\int_{0}^{x} e^{- \frac{t ^{2}}{2}} d t

\int_{- 1}^{1} (x^{4}) d x