integral from 0 to 2 of sqrt(8x^2+8)

Lösung

\int_{0}^{2} 8 x^{2} + 8 d x

210 + 2 ln (2 + 5)

Dezimale

8.36616 \dots

Schritte zur Lösung

\int_{0}^{2} 8 x^{2} + 8 d x

Trigonometrische Substitution anwenden

= \int_{0}^{a r c t a n (2)} 22 sec^{3} (u) d u

Entferne die Konstante:

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

= 22 \cdot \int_{0}^{a r c t a n (2)} sec^{3} (u) d u

Wende integrale Reduktion an

= 22 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (2)} sec (u) d u)

\int_{0}^{a r c t a n (2)} sec (u) d u = ln (2 + 5)

= 22 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5))

Vereinfache

22 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5)) : 22 [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (2)} + 2 ln (2 + 5)

= 22 [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (2)} + 2 ln (2 + 5)

Berechne die Grenzen:

5

= 225 + 2 ln (2 + 5)

Vereinfache

= 210 + 2 ln (2 + 5)

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