integral from 0 to 1 of 1/(9+16x^2)

Lösung

\int_{0}^{1} \frac{1}{9 + 16 x ^{2}} d x

\frac{1}{12} arctan (\frac{4}{3})

Dezimale

0.07727 \dots

Schritte zur Lösung

\int_{0}^{1} \frac{1}{9 + 16 x ^{2}} d x

Wende integrale Substitution an

= \int_{0}^{\frac{4}{3}} \frac{1}{12 ( u ^{2} + 1 )} d u

Entferne die Konstante:

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

= \frac{1}{12} \cdot \int_{0}^{\frac{4}{3}} \frac{1}{u ^{2} + 1} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= \frac{1}{12} [arctan (u)]_{0}^{\frac{4}{3}}

Berechne die Grenzen:

arctan (\frac{4}{3})

= \frac{1}{12} arctan (\frac{4}{3})

ma c l a u r in \frac{x}{( 3 - x ) ^{3}}

\int_{0}^{7} x e^{- x} d x

a re a x^{2} - 1, - x + 2, 0, 1

x \to \infty lim (x ln (x))

\frac{d}{d x} (\frac{- cos ( x )}{4 x ^{2} + 2 x - 3})