integral from 1 to x of sqrt(t^2-t)

解

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

\frac{- ln 2 x + 2 x ( x - 1 ) - 1 + 2 ( 2 x - 1 ) x ( x - 1 )}{8}

解答ステップ

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

平方完成する

t^{2} - t : (t - \frac{1}{2})^{2} - \frac{1}{4}

= \int_{1}^{x} (t - \frac{1}{2})^{2} - \frac{1}{4} d t

u置換積分法を適用する

= \int_{\frac{1}{2}}^{x - \frac{1}{2}} \frac{4 u ^{2} - 1}{2} d u

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{\frac{1}{2}}^{x - \frac{1}{2}} 4 u^{2} - 1 d u

三角関数による置換を適用する

= \frac{1}{2} \cdot \int_{0}^{arcsec (2 x - 1)} \frac{1}{2} tan^{2} (v) sec (v) d v

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{arcsec (2 x - 1)} tan^{2} (v) sec (v) d v

三角関数の公式を使用して書き換える

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{arcsec (2 x - 1)} (- 1 + sec^{2} (v)) sec (v) d v

拡張

(- 1 + sec^{2} (v)) sec (v) : - sec (v) + sec^{3} (v)

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{arcsec (2 x - 1)} - sec (v) + sec^{3} (v) d v

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

= \frac{1}{2} \cdot \frac{1}{2} (- \int_{0}^{arcsec (2 x - 1)} sec (v) d v + \int_{0}^{arcsec (2 x - 1)} sec^{3} (v) d v)

\int_{0}^{arcsec (2 x - 1)} sec (v) d v = ln 2 x + 2 x (x - 1) - 1

\int_{0}^{arcsec (2 x - 1)} sec^{3} (v) d v = (2 x - 1) x (x - 1) + \frac{1}{2} ln 2 x + 2 x (x - 1) - 1

= \frac{1}{2} \cdot \frac{1}{2} (- ln 2 x + 2 x (x - 1) - 1 + (2 x - 1) x (x - 1) + \frac{1}{2} ln 2 x + 2 x (x - 1) - 1)

簡素化

\frac{1}{2} \cdot \frac{1}{2} (- ln 2 x + 2 x (x - 1) - 1 + (2 x - 1) x (x - 1) + \frac{1}{2} ln 2 x + 2 x (x - 1) - 1) : \frac{- ln 2 x + 2 x ( x - 1 ) - 1 + 2 ( 2 x - 1 ) x ( x - 1 )}{8}

= \frac{- ln 2 x + 2 x ( x - 1 ) - 1 + 2 ( 2 x - 1 ) x ( x - 1 )}{8}