integral from-1 to 2 of sqrt(4t^2-8t+5)

Lösung

\int_{- 1}^{2} 4 t^{2} - 8 t + 5 d t

\frac{5}{2} + 17 + \frac{1}{4} ln (2 + 5) - \frac{1}{4} ln (- 4 + 17)

Dezimale

6.12572 \dots

Schritte zur Lösung

\int_{- 1}^{2} 4 t^{2} - 8 t + 5 d t

Vervollständige das Quadrat

4 t^{2} - 8 t + 5 : 4 (t - 1)^{2} + 1

= \int_{- 1}^{2} 4 (t - 1)^{2} + 1 d t

Wende U-Substitution an

= \int_{- 2}^{1} 4 u^{2} + 1 d u

Trigonometrische Substitution anwenden

= \int_{- a r c t a n (4)}^{a r c t a n (2)} \frac{1}{2} sec^{3} (v) d v

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int_{- a r c t a n (4)}^{a r c t a n (2)} sec^{3} (v) d v

Wende integrale Reduktion an

= \frac{1}{2} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{- a r c t a n (4)}^{a r c t a n (2)} + \frac{1}{2} \cdot \int_{- a r c t a n (4)}^{a r c t a n (2)} sec (v) d v)

\int_{- a r c t a n (4)}^{a r c t a n (2)} sec (v) d v = ln (2 + 5) - ln (- 4 + 17)

= \frac{1}{2} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{- a r c t a n (4)}^{a r c t a n (2)} + \frac{1}{2} (ln (2 + 5) - ln (- 4 + 17)))

Vereinfache

\frac{1}{2} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{- a r c t a n (4)}^{a r c t a n (2)} + \frac{1}{2} (ln (2 + 5) - ln (- 4 + 17))) : \frac{1}{2} ([\frac{1}{2} sec (v) tan (v)]_{- a r c t a n (4)}^{a r c t a n (2)} + \frac{1}{2} (ln (2 + 5) - ln (- 4 + 17)))

= \frac{1}{2} ([\frac{1}{2} sec (v) tan (v)]_{- a r c t a n (4)}^{a r c t a n (2)} + \frac{1}{2} (ln (2 + 5) - ln (- 4 + 17)))

Berechne die Grenzen:

5 + 217

= \frac{1}{2} (5 + 217 + \frac{1}{2} (ln (2 + 5) - ln (- 4 + 17)))

Vereinfache

= \frac{5}{2} + 17 + \frac{1}{4} ln (2 + 5) - \frac{1}{4} ln (- 4 + 17)

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