integral from 1 to 2 of x^2+6/2 x

Lösung

\int_{1}^{2} x^{2} + \frac{6}{2} x d x

\frac{41}{6}

Dezimale

6.83333 \dots

Schritte zur Lösung

\int_{1}^{2} x^{2} + \frac{6}{2} x d x

\frac{6}{2} x = 3 x

= \int_{1}^{2} x^{2} + 3 x d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{2} x^{2} d x + \int_{1}^{2} 3 x d x

\int_{1}^{2} x^{2} d x = \frac{7}{3}

\int_{1}^{2} 3 x d x = \frac{9}{2}

= \frac{7}{3} + \frac{9}{2}

Vereinfache

\frac{7}{3} + \frac{9}{2} : \frac{41}{6}

= \frac{41}{6}

\int_{0}^{4} x^{2} - 5 d x

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

\int_{1}^{4} \frac{1 + t}{t} d t