integral from-1 to 1 of 7/(1+49t^2)

Lösung

\int_{- 1}^{1} \frac{7}{1 + 49 t ^{2}} d t

2 arctan (7)

Dezimale

2.85779 \dots

Schritte zur Lösung

\int_{- 1}^{1} \frac{7}{1 + 49 t ^{2}} d t

Entferne die Konstante:

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

= 7 \cdot \int_{- 1}^{1} \frac{1}{1 + 49 t ^{2}} d t

Wende integrale Substitution an

= 7 \cdot \int_{- 7}^{7} \frac{1}{7 ( 1 + u ^{2} )} d u

Entferne die Konstante:

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

= 7 \cdot \frac{1}{7} \cdot \int_{- 7}^{7} \frac{1}{1 + u ^{2}} d u

Nutze das gemeinsame Integral :

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

= 7 \cdot \frac{1}{7} [arctan (u)]_{- 7}^{7}

Vereinfache

7 \cdot \frac{1}{7} [arctan (u)]_{- 7}^{7} : [arctan (u)]_{- 7}^{7}

= [arctan (u)]_{- 7}^{7}

Berechne die Grenzen:

2 arctan (7)

= 2 arctan (7)

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