integral from 0 to 1 of (10)/(t^2+1)

Lösung

\int_{0}^{1} \frac{10}{t ^{2} + 1} d t

\frac{5 π}{2}

Dezimale

7.85398 \dots

Schritte zur Lösung

\int_{0}^{1} \frac{10}{t ^{2} + 1} d t

Entferne die Konstante:

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

= 10 \cdot \int_{0}^{1} \frac{1}{t ^{2} + 1} d t

Nutze das gemeinsame Integral :

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

= 10 [arctan (t)]_{0}^{1}

Berechne die Grenzen:

\frac{π}{4}

= 10 \cdot \frac{π}{4}

Vereinfache

= \frac{5 π}{2}

\int_{0}^{t} \frac{1}{t + 10} d t

\int_{0}^{1} 4 x^{3} (3 x^{4} - 3)^{5} d x

\int_{0}^{1} (x + 3)^{3} d x

\int_{- 1}^{3} \frac{1}{( y + 2 ) ^{3}} d y

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