integral from 0 to 2 of sqrt(1+5x^2)

Lösung

\int_{0}^{2} 1 + 5 x^{2} d x

\frac{1}{5} (105 + \frac{1}{2} ln (25 + 21))

Dezimale

5.07524 \dots

Schritte zur Lösung

\int_{0}^{2} 1 + 5 x^{2} d x

Trigonometrische Substitution anwenden

= \int_{0}^{a r c t a n (5 \cdot 2)} \frac{1}{5} sec^{3} (u) d u

Entferne die Konstante:

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

= \frac{1}{5} \cdot \int_{0}^{a r c t a n (5 \cdot 2)} sec^{3} (u) d u

Wende integrale Reduktion an

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

\int_{0}^{a r c t a n (5 \cdot 2)} sec (u) d u = ln (25 + 21)

= \frac{1}{5} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (5 \cdot 2)} + \frac{1}{2} ln (25 + 21)

Vereinfache

\frac{1}{5} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (5 \cdot 2)} + \frac{1}{2} ln (25 + 21) : \frac{1}{5} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (25)} + \frac{1}{2} ln (25 + 21))

= \frac{1}{5} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (25)} + \frac{1}{2} ln (25 + 21))

Berechne die Grenzen:

105

= \frac{1}{5} (105 + \frac{1}{2} ln (25 + 21))

\int_{7}^{9} (x - 7)^{6} d x

\int_{2}^{3} \frac{2 x}{( 1 + x ^{2} )} d x

\int_{1}^{2} sin (π x) d x

\int_{0}^{\frac{π}{2}} cos (x + 2 y) d y

\int_{0}^{2} (3 - x) (x - 4) d x