integral from 0 to 4 of xsqrt(x^2+1)

Lösung

\int_{0}^{4} x x^{2} + 1 d x

\frac{17 17 - 1}{3}

Dezimale

23.03093 \dots

Schritte zur Lösung

\int_{0}^{4} x x^{2} + 1 d x

Wende U-Substitution an

= \int_{1}^{17} u^{2} d u

Wende die Potenzregel an

= [\frac{u ^{3}}{3}]_{1}^{17}

Berechne die Grenzen:

\frac{17 17}{3} - \frac{1}{3}

= \frac{17 17}{3} - \frac{1}{3}

Vereinfache

= \frac{17 17 - 1}{3}

\int_{2}^{3} \frac{2}{( 3 x - 6 ) ^{2}} d x

\int_{- 1}^{2} (2 - 5 x + \frac{x ^{2}}{2}) d x

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

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

\int_{0}^{1} \frac{3}{5} x d x