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Beliebt Rechnen >

integral from 0 to 1 of sqrt(1+4y^2)

Lösung

\int_{0}^{1} 1 + 4 y^{2} d y

Lösung

\frac{1}{2} (5 + \frac{1}{2} ln (2 + 5))

+1

Dezimale

1.47894 \dots

Schritte zur Lösung

\int_{0}^{1} 1 + 4 y^{2} d y

Trigonometrische Substitution anwenden

= \int_{0}^{a r c t a n (2)} \frac{1}{2} sec^{3} (u) d u

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int_{0}^{a r c t a n (2)} sec^{3} (u) d u

Wende integrale Reduktion an

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

\int_{0}^{a r c t a n (2)} sec (u) d u = ln (2 + 5)

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5))

Vereinfache

\frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5)) : \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5))

= \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5))

Berechne die Grenzen:

5

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

Graph

Beliebte Beispiele

integral of 18xcos(1/2 x)

\int 18 x cos (\frac{1}{2} x) d x

limit as x approaches 3-of f(x)x

x \to 3 - lim (f (x) x)

(\partial)/(\partial x)(ln(x+ct))

\frac{\partial}{\partial x} (ln (x + c t))

vereinfachen (1+2/x)^x

s im pl i f y (1 + \frac{2}{x})^{x}

integral of 3/(sqrt(x^2+4))

\int \frac{3}{x ^{2} + 4} d x