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

Lösung

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

\frac{16 - 4 2}{3}

Dezimale

3.44771 \dots

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{2}^{4} u d u

Wende die Potenzregel an

= [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{4}

Berechne die Grenzen:

\frac{16}{3} - \frac{4 2}{3}

= \frac{16}{3} - \frac{4 2}{3}

Vereinfache

= \frac{16 - 4 2}{3}

\int_{- 1}^{4} (3 x - 4) (x + 3) d x

\int_{3}^{6} x y d x

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\int_{0}^{\frac{π}{2}} 2 cos (t) d t