integral from-1 to 0 of 1/(4x^2+8x+8)

Lösung

\int_{- 1}^{0} \frac{1}{4 x ^{2} + 8 x + 8} d x

\frac{π}{16}

Dezimale

0.19634 \dots

Schritte zur Lösung

\int_{- 1}^{0} \frac{1}{4 x ^{2} + 8 x + 8} d x

\frac{1}{4 x ^{2} + 8 x + 8} = \frac{1}{4} \frac{1}{x ^{2} + 2 x + 2}

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

Entferne die Konstante:

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

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

Vervollständige das Quadrat

x^{2} + 2 x + 2 : (x + 1)^{2} + 1

= \frac{1}{4} \cdot \int_{- 1}^{0} \frac{1}{( x + 1 ) ^{2} + 1} d x

Wende U-Substitution an

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

Nutze das gemeinsame Integral :

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

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

Berechne die Grenzen:

\frac{π}{4}

= \frac{1}{4} \cdot \frac{π}{4}

Vereinfache

= \frac{π}{16}

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

\int_{- π}^{0} - x^{2} - π x d x

\int_{0}^{t} \frac{9}{1 - x ^{2}} d x

\int_{9}^{36} (x - 2) d x

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