integral from 0 to 1 of 4/(2x+3)

Lösung

\int_{0}^{1} \frac{4}{2 x + 3} d x

2 (ln (5) - ln (3))

Dezimale

1.02165 \dots

Schritte zur Lösung

\int_{0}^{1} \frac{4}{2 x + 3} d x

Entferne die Konstante:

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

= 4 \cdot \int_{0}^{1} \frac{1}{2 x + 3} d x

Wende U-Substitution an

= 4 \cdot \int_{3}^{5} \frac{1}{2 u} d u

Entferne die Konstante:

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

= 4 \cdot \frac{1}{2} \cdot \int_{3}^{5} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{2} [ln ∣ u ∣]_{3}^{5}

Vereinfache

4 \cdot \frac{1}{2} [ln ∣ u ∣]_{3}^{5} : 2 [ln ∣ u ∣]_{3}^{5}

= 2 [ln ∣ u ∣]_{3}^{5}

Berechne die Grenzen:

ln (5) - ln (3)

= 2 (ln (5) - ln (3))

\int_{0}^{a} y a^{2} - y^{2} d y

\int_{- 3}^{3} (3 x^{3} - 2 x^{2} + x + 1) d x

\int_{- \infty}^{4} 3 e^{- 3 x} d x

\int_{- 2}^{2} (2 - 4 - x^{2}) d x

\int_{- 1}^{0} (2 x + 1)^{3} d x