integral from 1 to 2 of y^3+1/(4y^3)

解

\int_{1}^{2} y^{3} + \frac{1}{4 y ^{3}} d y

\frac{123}{32}

十進法表記

3.84375

解答ステップ

\int_{1}^{2} y^{3} + \frac{1}{4 y ^{3}} d y

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} y^{3} d y + \int_{1}^{2} \frac{1}{4 y ^{3}} d y

\int_{1}^{2} y^{3} d y = \frac{15}{4}

\int_{1}^{2} \frac{1}{4 y ^{3}} d y = \frac{3}{32}

= \frac{15}{4} + \frac{3}{32}

簡素化

\frac{15}{4} + \frac{3}{32} : \frac{123}{32}

= \frac{123}{32}