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

Lösung

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

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

Dezimale

2.29558 \dots

Schritte zur Lösung

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

Vervollständige das Quadrat

x^{2} - 2 x + 2 : (x - 1)^{2} + 1

= \int_{0}^{2} (x - 1)^{2} + 1 d x

Wende U-Substitution an

= \int_{- 1}^{1} u^{2} + 1 d u

Trigonometrische Substitution anwenden

= \int_{- \frac{π}{4}}^{\frac{π}{4}} sec^{3} (v) d v

Wende integrale Reduktion an

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

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

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

Vereinfache

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

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

Berechne die Grenzen:

2

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

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