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Beliebt Rechnen >

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

Lösung

\int_{0}^{π} 1 + x^{2} d x

Lösung

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

+1

Dezimale

6.10991 \dots

Schritte zur Lösung

\int_{0}^{π} 1 + x^{2} d x

Trigonometrische Substitution anwenden

= \int_{0}^{a r c t a n (π)} sec^{3} (u) d u

Wende integrale Reduktion an

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (π)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (π)} sec (u) d u

\int_{0}^{a r c t a n (π)} sec (u) d u = ln (π + π^{2} + 1)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (π)} + \frac{1}{2} ln (π + π^{2} + 1)

Vereinfache

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (π)} + \frac{1}{2} ln (π + π^{2} + 1) : [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (π)} + \frac{1}{2} ln (π + π^{2} + 1)

= [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (π)} + \frac{1}{2} ln (π + π^{2} + 1)

Berechne die Grenzen:

\frac{π π ^{2} + 1}{2}

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

Vereinfache

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

Graph

Beliebte Beispiele

integral from 3 to 6 of-5x+6

\int_{3}^{6} - 5 x + 6 d x

integral from 1 to 7 of 1/(5x)

\int_{1}^{7} \frac{1}{5 x} d x

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\int_{\frac{π}{6}}^{\frac{π}{3}} sin (θ) d θ

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\int_{- 3}^{0} 2 x + 3 d x

integral from 0 to x of 1/2 x

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