integral from-1 to 3 of sqrt(1+(6y+2)^2)

Lösung

\int_{- 1}^{3} 1 + (6 y + 2)^{2} d y

\frac{5 401 + 17}{3} + \frac{1}{12} ln (20 + 401) - \frac{1}{12} ln (- 4 + 17)

Dezimale

35.23136 \dots

Schritte zur Lösung

\int_{- 1}^{3} 1 + (6 y + 2)^{2} d y

Wende U-Substitution an

= \int_{- 4}^{20} \frac{1 + u ^{2}}{6} d u

Entferne die Konstante:

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

= \frac{1}{6} \cdot \int_{- 4}^{20} 1 + u^{2} d u

Trigonometrische Substitution anwenden

= \frac{1}{6} \cdot \int_{- a r c t a n (4)}^{a r c t a n (20)} sec^{3} (v) d v

Wende integrale Reduktion an

= \frac{1}{6} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{- a r c t a n (4)}^{a r c t a n (20)} + \frac{1}{2} \cdot \int_{- a r c t a n (4)}^{a r c t a n (20)} sec (v) d v)

\int_{- a r c t a n (4)}^{a r c t a n (20)} sec (v) d v = ln (20 + 401) - ln (- 4 + 17)

= \frac{1}{6} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{- a r c t a n (4)}^{a r c t a n (20)} + \frac{1}{2} (ln (20 + 401) - ln (- 4 + 17)))

Vereinfache

\frac{1}{6} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{- a r c t a n (4)}^{a r c t a n (20)} + \frac{1}{2} (ln (20 + 401) - ln (- 4 + 17))) : \frac{1}{6} ([\frac{1}{2} sec (v) tan (v)]_{- a r c t a n (4)}^{a r c t a n (20)} + \frac{1}{2} (ln (20 + 401) - ln (- 4 + 17)))

= \frac{1}{6} ([\frac{1}{2} sec (v) tan (v)]_{- a r c t a n (4)}^{a r c t a n (20)} + \frac{1}{2} (ln (20 + 401) - ln (- 4 + 17)))

Berechne die Grenzen:

10401 + 217

= \frac{1}{6} (10401 + 217 + \frac{1}{2} (ln (20 + 401) - ln (- 4 + 17)))

Vereinfache

= \frac{5 401 + 17}{3} + \frac{1}{12} ln (20 + 401) - \frac{1}{12} ln (- 4 + 17)

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

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

x \to 1 - lim (1 - x)

x \to 4 lim (\frac{x - 4}{2 - 8 - x})

d er i v a t i v e f (x) = 5 csc (x)