integral from 0 to 3 of (x^2-5)/(x+2)

解

\int_{0}^{3} \frac{x ^{2} - 5}{x + 2} d x

ln (2) - ln (5) - \frac{3}{2}

十進法表記

- 2.41629 \dots

解答ステップ

\int_{0}^{3} \frac{x ^{2} - 5}{x + 2} d x

\frac{x ^{2} - 5}{x + 2} = \frac{( x ^{2} - 5 ) ( x - 2 )}{( x ) ^{2} - ( 2 ) ^{2}}

= \int_{0}^{3} \frac{( x ^{2} - 5 ) ( x - 2 )}{x ^{2} - 2 ^{2}} d x

未定義の (特異) 点を求める:

x = 2

存在する場合は

b, a < b < c, f (b) =

未定義

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{0}^{2} \frac{( x ^{2} - 5 ) ( x - 2 )}{x ^{2} - 2 ^{2}} d x + \int_{2}^{3} \frac{( x ^{2} - 5 ) ( x - 2 )}{x ^{2} - 2 ^{2}} d x

\int_{0}^{2} \frac{( x ^{2} - 5 ) ( x - 2 )}{x ^{2} - 2 ^{2}} d x = - ln (2) - 2

\int_{2}^{3} \frac{( x ^{2} - 5 ) ( x - 2 )}{x ^{2} - 2 ^{2}} d x = 2 ln (2) - ln (5) + \frac{1}{2}

= - ln (2) - 2 + 2 ln (2) - ln (5) + \frac{1}{2}

簡素化

- ln (2) - 2 + 2 ln (2) - ln (5) + \frac{1}{2} : ln (2) - ln (5) - \frac{3}{2}

= ln (2) - ln (5) - \frac{3}{2}