integral from 1 to 4 of 3x^2+sqrt(x)

Lösung

\int_{1}^{4} 3 x^{2} + x d x

\frac{203}{3}

Dezimale

67.66666 \dots

Schritte zur Lösung

\int_{1}^{4} 3 x^{2} + 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}^{4} 3 x^{2} d x + \int_{1}^{4} x d x

\int_{1}^{4} 3 x^{2} d x = 63

\int_{1}^{4} x d x = \frac{14}{3}

= 63 + \frac{14}{3}

Vereinfache

63 + \frac{14}{3} : \frac{203}{3}

= \frac{203}{3}

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

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

\int_{4}^{5} (2 + 2 x)^{2} d x

\int_{0}^{\infty} x^{n} d x

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