integral from 1 to 2 of ((x^2+1))/(x^2)

解

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

\frac{3}{2}

十進法表記

1.5

解答ステップ

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

\frac{x ^{2} + 1}{x ^{2}} = \frac{1}{x ^{2}} + 1

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

指数の規則を適用する:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{x ^{2}} = x^{- 2}

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

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

\int_{1}^{2} x^{- 2} d x = \frac{1}{2}

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

= \frac{1}{2} + 1

簡素化

\frac{1}{2} + 1 : \frac{3}{2}

= \frac{3}{2}