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

Lösung

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

\frac{2 + ln ( 1 + 2 )}{2}

Dezimale

1.14779 \dots

Schritte zur Lösung

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

Trigonometrische Substitution anwenden

= \int_{0}^{\frac{π}{4}} sec^{3} (u) d u

Wende integrale Reduktion an

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} \cdot \int_{0}^{\frac{π}{4}} sec (u) d u

\int_{0}^{\frac{π}{4}} sec (u) d u = ln (1 + 2)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)

Vereinfache

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2) : [\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)

= [\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)

Berechne die Grenzen:

\frac{1}{2}

= \frac{1}{2} + \frac{1}{2} ln (1 + 2)

Vereinfache

= \frac{2 + ln ( 1 + 2 )}{2}

\int_{1}^{e} \frac{1}{x \cdot ln ( x )} d x

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

\int_{1}^{6} 2 x^{- 4} d x

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

\int_{0}^{2} 4 - r^{2} d r