integral from 0 to 1/5 of 1/(1+25x^2)

Lösung

\int_{0}^{\frac{1}{5}} \frac{1}{1 + 25 x ^{2}} d x

\frac{π}{20}

Dezimale

0.15707 \dots

Schritte zur Lösung

\int_{0}^{\frac{1}{5}} \frac{1}{1 + 25 x ^{2}} d x

Wende integrale Substitution an

= \int_{0}^{1} \frac{1}{5 ( 1 + u ^{2} )} d u

Entferne die Konstante:

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

= \frac{1}{5} \cdot \int_{0}^{1} \frac{1}{1 + u ^{2}} d u

Nutze das gemeinsame Integral :

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

= \frac{1}{5} [arctan (u)]_{0}^{1}

Berechne die Grenzen:

\frac{π}{4}

= \frac{1}{5} \cdot \frac{π}{4}

Vereinfache

= \frac{π}{20}

\int_{2}^{4} x e^{x^{2}} d x

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

\int_{0}^{1} \frac{2 x}{( x ^{2} - 3 ) ^{3}} d x

\int_{0}^{3} 2 x (x - 4) d x

\int_{4}^{\frac{1}{x}} arctan (5 t) d t