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

Lösung

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

- \frac{3 - 3}{2}

Dezimale

- 0.63397 \dots

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{9}^{3} \frac{1}{4 u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{3}^{9} \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_{3}^{9} \frac{1}{u} d u

Wende die Potenzregel an

= - \frac{1}{4} [2 u^{\frac{1}{2}}]_{3}^{9}

Wende Radikal Regel an:

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

= - \frac{1}{4} [2 u]_{3}^{9}

Berechne die Grenzen:

6 - 23

= - \frac{1}{4} (6 - 23)

Vereinfache

= - \frac{3 - 3}{2}

\int_{1}^{3} \frac{x + 3}{x ^{2} + 6 x} d x

\int_{4}^{\infty} \frac{6}{x ^{4}} d x

\int_{0}^{1} (\frac{y}{e ^{2 y}}) d y

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

\int_{1}^{4} 4 x x + 1 d x