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

Lösung

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

\frac{1}{8} ln (3)

Dezimale

0.13732 \dots

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{1}^{3} \frac{( u - 1 ) ^{2}}{8 u} d u

Entferne die Konstante:

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

= \frac{1}{8} \cdot \int_{1}^{3} \frac{( u - 1 ) ^{2}}{u} d u

Multipliziere aus

\frac{( u - 1 ) ^{2}}{u} : u - 2 + \frac{1}{u}

= \frac{1}{8} \cdot \int_{1}^{3} u - 2 + \frac{1}{u} d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{8} (\int_{1}^{3} u d u - \int_{1}^{3} 2 d u + \int_{1}^{3} \frac{1}{u} d u)

\int_{1}^{3} u d u = 4

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

\int_{1}^{3} \frac{1}{u} d u = ln (3)

= \frac{1}{8} (4 - 4 + ln (3))

Vereinfache

\frac{1}{8} (4 - 4 + ln (3)) : \frac{1}{8} ln (3)

= \frac{1}{8} ln (3)

\int_{3}^{10} \frac{1}{x} d x

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

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

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

\int_{0}^{1} 144 t^{2} + 224 t + 104 d t