integral from 1 to 8 of (0.2x^2+1)

Lösung

\int_{1}^{8} (0.2 x^{2} + 1) d x

41.06666 \dots

Schritte zur Lösung

\int_{1}^{8} 0.2 x^{2} + 1 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}^{8} 0.2 x^{2} d x + \int_{1}^{8} 1 d x

\int_{1}^{8} 0.2 x^{2} d x = 34.06666 \dots

\int_{1}^{8} 1 d x = 7

= 34.06666 \dots + 7

Vereinfache

= 41.06666 \dots

\int_{2 y}^{3 - y} x + y d x

\int_{- 1}^{0} (x^{3} - 2 x + 3) d x

\int_{0}^{7} 3 x \cdot 3 x + 1 d x

\int_{1}^{e} (\frac{2 t ^{2} + 1}{t}) d t

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