integral from 0 to 7 of sqrt(49+x^2)

Lösung

\int_{0}^{7} 49 + x^{2} d x

\frac{49 ( 2 + ln ( 1 + 2 ) )}{2}

Dezimale

56.24188 \dots

Schritte zur Lösung

\int_{0}^{7} 49 + x^{2} d x

Trigonometrische Substitution anwenden

= \int_{0}^{\frac{π}{4}} 49 sec^{3} (u) d u

Entferne die Konstante:

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

= 49 \cdot \int_{0}^{\frac{π}{4}} sec^{3} (u) d u

Wende integrale Reduktion an

= 49 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} \cdot \int_{0}^{\frac{π}{4}} sec (u) d u)

\int_{0}^{\frac{π}{4}} sec (u) d u = ln (1 + 2)

= 49 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

Vereinfache

49 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)) : 49 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

= 49 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

Berechne die Grenzen:

\frac{1}{2}

= 49 (\frac{1}{2} + \frac{1}{2} ln (1 + 2))

Vereinfache

= \frac{49 ( 2 + ln ( 1 + 2 ) )}{2}

\frac{d}{d x} ((x - 2) (3 x + 8))

\int \frac{x + 2}{5 x + 6} d x

\frac{d}{d x} (e^{x} cos (x) - sin (x) e^{x})

y^{^{''}} + 36 y^{^{'}} = 0

x \to \frac{π}{2} + lim (sec (x))