integral from 1 to 4 of (sqrt(x))^2

Lösung

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

\frac{15}{2}

Dezimale

7.5

Schritte zur Lösung

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

Vereinfache

(x)^{2} : x

= \int_{1}^{4} x d x

Wende die Potenzregel an

= [\frac{x ^{2}}{2}]_{1}^{4}

Berechne die Grenzen:

\frac{15}{2}

= \frac{15}{2}

\int_{0}^{2} (t + 2)^{2} - 3 d t

\int_{0}^{1} (\frac{y}{1 + x y}) d x

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\int_{0}^{e} \frac{x ^{2}}{x ^{3} + e} d x