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

Lösung

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

\frac{3 - 1}{2}

Dezimale

0.36602 \dots

Schritte zur Lösung

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

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Wende die Potenzregel an

= \frac{1}{4} [2 u^{\frac{1}{2}}]_{1}^{3}

Wende Radikal Regel an:

a^{\frac{1}{n}} = n a

= \frac{1}{4} [2 u]_{1}^{3}

Berechne die Grenzen:

23 - 2

= \frac{1}{4} (23 - 2)

Vereinfache

= \frac{3 - 1}{2}

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

\int_{3}^{5} \frac{x}{( x ^{2} + 7 ) ^{3}} d x

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

\int_{0}^{2} sin (2 x) d x

\int_{- 1}^{1} e^{y} d y