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

Lösung

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

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

Dezimale

1.14779 \dots

Schritte zur Lösung

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

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_{0}^{5} π (2 x)^{2} d x

\int \frac{( sec ^{2} ( x ) )}{tan ^{2} ( x )} d x

\frac{d}{d x} y = (\frac{1}{cos ^{2} ( x )})

x \to - 9 lim (\frac{8 x + 72}{∣ x + 9 ∣})

in v erse l a pl a ce \frac{( 2 \cdot s )}{s ^{2} + 6 s + 13}