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Beliebt Rechnen >

integral from 0 to 10 of sqrt(1+16x^2)

Lösung

\int_{0}^{10} 1 + 16 x^{2} d x

Lösung

\frac{1}{4} (201601 + \frac{1}{2} ln (40 + 1601))

+1

Dezimale

200.61026 \dots

Schritte zur Lösung

\int_{0}^{10} 1 + 16 x^{2} d x

Trigonometrische Substitution anwenden

= \int_{0}^{a r c t a n (40)} \frac{1}{4} sec^{3} (u) d u

Entferne die Konstante:

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

= \frac{1}{4} \cdot \int_{0}^{a r c t a n (40)} sec^{3} (u) d u

Wende integrale Reduktion an

= \frac{1}{4} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (40)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (40)} sec (u) d u)

\int_{0}^{a r c t a n (40)} sec (u) d u = ln (40 + 1601)

= \frac{1}{4} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (40)} + \frac{1}{2} ln (40 + 1601))

Vereinfache

\frac{1}{4} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (40)} + \frac{1}{2} ln (40 + 1601)) : \frac{1}{4} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (40)} + \frac{1}{2} ln (40 + 1601))

= \frac{1}{4} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (40)} + \frac{1}{2} ln (40 + 1601))

Berechne die Grenzen:

201601

= \frac{1}{4} (201601 + \frac{1}{2} ln (40 + 1601))

Graph

Beliebte Beispiele

integral of (3+x^2)^{3/2}

\int (3 + x^{2})^{\frac{3}{2}} d x

limit as x approaches 5 of 2/(5-x)

x \to 5 lim (\frac{2}{5 - x})

derivative of (1+8/x ^x)

\frac{d}{d x} ((1 + \frac{8}{x})^{x})

derivative of x+sqrt(1-x)

\frac{d}{d x} (x + 1 - x)

(dy)/(dt)+e^ty=-4e^t

\frac{d y}{d t} + e^{t} y = - 4 e^{t}