integral from 0 to 2 of sqrt(4y^2+4y+2)

Lösung

\int_{0}^{2} 4 y^{2} + 4 y + 2 d y

\frac{2 ( - 1 + 5 13 ) - ln ( 1 + 2 ) + ln ( 5 + 26 )}{4}

Dezimale

6.37798 \dots

Schritte zur Lösung

\int_{0}^{2} 4 y^{2} + 4 y + 2 d y

Vervollständige das Quadrat

4 y^{2} + 4 y + 2 : 4 (y + \frac{1}{2})^{2} + 1

= \int_{0}^{2} 4 (y + \frac{1}{2})^{2} + 1 d y

Wende U-Substitution an

= \int_{\frac{1}{2}}^{\frac{5}{2}} 4 u^{2} + 1 d u

Trigonometrische Substitution anwenden

= \int_{\frac{π}{4}}^{a r c t a n (5)} \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_{\frac{π}{4}}^{a r c t a n (5)} sec^{3} (v) d v

Wende integrale Reduktion an

= \frac{1}{2} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{\frac{π}{4}}^{a r c t a n (5)} + \frac{1}{2} \cdot \int_{\frac{π}{4}}^{a r c t a n (5)} sec (v) d v)

\int_{\frac{π}{4}}^{a r c t a n (5)} sec (v) d v = ln (5 + 26) - ln (1 + 2)

= \frac{1}{2} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{\frac{π}{4}}^{a r c t a n (5)} + \frac{1}{2} (ln (5 + 26) - ln (1 + 2)))

Vereinfache

\frac{1}{2} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{\frac{π}{4}}^{a r c t a n (5)} + \frac{1}{2} (ln (5 + 26) - ln (1 + 2))) : \frac{1}{2} ([\frac{1}{2} sec (v) tan (v)]_{\frac{π}{4}}^{a r c t a n (5)} + \frac{1}{2} (ln (5 + 26) - ln (1 + 2)))

= \frac{1}{2} ([\frac{1}{2} sec (v) tan (v)]_{\frac{π}{4}}^{a r c t a n (5)} + \frac{1}{2} (ln (5 + 26) - ln (1 + 2)))

Berechne die Grenzen:

\frac{5 13 - 1}{2}

= \frac{1}{2} (\frac{5 13 - 1}{2} + \frac{1}{2} (ln (5 + 26) - ln (1 + 2)))

Vereinfache

= \frac{2 ( - 1 + 5 13 ) - ln ( 1 + 2 ) + ln ( 5 + 26 )}{4}

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