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

Lösung

\int_{0}^{2} 1 + (2 y - 2)^{2} d y

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

Dezimale

2.95788 \dots

Schritte zur Lösung

\int_{0}^{2} 1 + (2 y - 2)^{2} d y

Wende U-Substitution an

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

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int_{- 2}^{2} 1 + u^{2} d u

Trigonometrische Substitution anwenden

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

Wende integrale Reduktion an

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

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

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

Vereinfache

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

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

Berechne die Grenzen:

25

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

Vereinfache

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

\int_{- l n (2)}^{l n (2)} x sinh (x) d x

\int_{- 2}^{0} 2 x + 4 d x

\int_{0}^{x} \frac{1}{( 1 - x ) ( 1 - x )} d x

\int_{0}^{1} x 4 - 5 x d x

\int_{0}^{2} \frac{1}{3 x + 1} d x