integral from 0 to 1 of tsqrt(t^2+1)

Lösung

\int_{0}^{1} t t^{2} + 1 d t

\frac{2 2 - 1}{3}

Dezimale

0.60947 \dots

Schritte zur Lösung

\int_{0}^{1} t t^{2} + 1 d t

Wende U-Substitution an

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

Wende die Potenzregel an

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

Berechne die Grenzen:

\frac{2 2}{3} - \frac{1}{3}

= \frac{2 2}{3} - \frac{1}{3}

Vereinfache

= \frac{2 2 - 1}{3}

\int_{- \infty}^{1} e^{3 x} d x

\int_{- 1}^{1} m x^{2} d x

\int_{2 x^{2}}^{c o s (x)} ln (5 t - 2) d t

\int_{- 1}^{2} (- x^{2} + 4) - (- x + 2) d x

\int_{0.5}^{1} \frac{ln ( x )}{4 x} d x