integral from 0 to 8 of sqrt(1+t)

Lösung

\int_{0}^{8} 1 + t d t

\frac{52}{3}

Dezimale

17.33333 \dots

Schritte zur Lösung

\int_{0}^{8} 1 + t d t

Wende U-Substitution an

= \int_{1}^{9} u d u

Wende die Potenzregel an

= [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{9}

Berechne die Grenzen:

\frac{52}{3}

= \frac{52}{3}

\int_{- 1}^{3} 25 + 16 x d x

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

\int_{0}^{\frac{1}{2}} - 7 arccos (x) d x

\int_{0}^{3} 2^{4 x} d x

\int_{0}^{\frac{π}{4}} t sin (t) d t