integral from 1 to 2 of 4ln(x)-x^2ln(x)

解

\int_{1}^{2} 4 ln (x) - x^{2} ln (x) d x

- \frac{29}{9} + \frac{16}{3} ln (2)

十進法表記

0.47456 \dots

解答ステップ

\int_{1}^{2} 4 ln (x) - x^{2} ln (x) 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} 4 ln (x) d x - \int_{1}^{2} x^{2} ln (x) d x

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

\int_{1}^{2} x^{2} ln (x) d x = - \frac{7}{9} + \frac{8}{3} ln (2)

= 4 (2 ln (2) - 1) - (- \frac{7}{9} + \frac{8}{3} ln (2))

簡素化

4 (2 ln (2) - 1) - (- \frac{7}{9} + \frac{8}{3} ln (2)) : - \frac{29}{9} + \frac{16}{3} ln (2)

= - \frac{29}{9} + \frac{16}{3} ln (2)