integral from 1 to 3 of sqrt(1+(2x)^2)

Lösung

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

\frac{3 37 - 5}{2} + \frac{1}{4} ln (6 + 37) - \frac{1}{4} ln (2 + 5)

Dezimale

8.26814 \dots

Schritte zur Lösung

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

Trigonometrische Substitution anwenden

= \int_{a r c t a n (2)}^{a r c t a n (6)} \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_{a r c t a n (2)}^{a r c t a n (6)} sec^{3} (u) d u

Wende integrale Reduktion an

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

\int_{a r c t a n (2)}^{a r c t a n (6)} sec (u) d u = ln (6 + 37) - ln (2 + 5)

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (2)}^{a r c t a n (6)} + \frac{1}{2} (ln (6 + 37) - ln (2 + 5)))

Vereinfache

\frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (2)}^{a r c t a n (6)} + \frac{1}{2} (ln (6 + 37) - ln (2 + 5))) : \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (2)}^{a r c t a n (6)} + \frac{1}{2} (ln (6 + 37) - ln (2 + 5)))

= \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (2)}^{a r c t a n (6)} + \frac{1}{2} (ln (6 + 37) - ln (2 + 5)))

Berechne die Grenzen:

337 - 5

= \frac{1}{2} (337 - 5 + \frac{1}{2} (ln (6 + 37) - ln (2 + 5)))

Vereinfache

= \frac{3 37 - 5}{2} + \frac{1}{4} ln (6 + 37) - \frac{1}{4} ln (2 + 5)

\int_{- 11}^{- 3} \frac{1}{x} d x

\int_{1}^{e} x^{2} ln^{2} (x) d x

\int_{- 1.25}^{0} (5 x^{3} - 3 x + 6) d x

\int_{0}^{1} x cos (\frac{nπ}{2} x) d x

\int_{2}^{x} - \frac{2}{t ^{3}} d t