integral from 1 to 2 of-1/(16z)+1/(4z)

解

\int_{1}^{2} - \frac{1}{16 z} + \frac{1}{4 z} d z

\frac{3}{16} ln (2)

十進法表記

0.12996 \dots

解答ステップ

\int_{1}^{2} - \frac{1}{16 z} + \frac{1}{4 z} d z

総和規則を適用する:

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

= - \int_{1}^{2} \frac{1}{16 z} d z + \int_{1}^{2} \frac{1}{4 z} d z

\int_{1}^{2} \frac{1}{16 z} d z = \frac{1}{16} ln (2)

\int_{1}^{2} \frac{1}{4 z} d z = \frac{1}{4} ln (2)

= - \frac{1}{16} ln (2) + \frac{1}{4} ln (2)

簡素化

- \frac{1}{16} ln (2) + \frac{1}{4} ln (2) : \frac{3}{16} ln (2)

= \frac{3}{16} ln (2)