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人気のある微分積分 >

integral of 2xsqrt(10x-x^2)

解

\int 2 x 10 x - x^{2} d x

解

2 (- \frac{1}{3} (- x^{2} + 10 x)^{\frac{3}{2}} + \frac{125}{4} (2 arcsin (\frac{1}{5} x - 1) + sin (2 arcsin (\frac{1}{5} x - 1)))) + C

解答ステップ

\int 2 x 10 x - x^{2} d x

定数を除く:

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

= 2 \cdot \int x 10 x - x^{2} d x

平方完成する

10 x - x^{2} : - (x - 5)^{2} + 25

= 2 \cdot \int x - (x - 5)^{2} + 25 d x

u置換積分法を適用する

= 2 \cdot \int (u + 5) - u^{2} + 25 d u

拡張

(u + 5) - u^{2} + 25 : u - u^{2} + 25 + 5 - u^{2} + 25

= 2 \cdot \int u - u^{2} + 25 + 5 - u^{2} + 25 d u

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

= 2 (\int u - u^{2} + 25 d u + \int 5 - u^{2} + 25 d u)

\int u - u^{2} + 25 d u = - \frac{1}{3} (- u^{2} + 25)^{\frac{3}{2}}

\int 5 - u^{2} + 25 d u = \frac{125}{2} (arcsin (\frac{1}{5} u) + \frac{1}{2} sin (2 arcsin (\frac{1}{5} u)))

= 2 (- \frac{1}{3} (- u^{2} + 25)^{\frac{3}{2}} + \frac{125}{2} (arcsin (\frac{1}{5} u) + \frac{1}{2} sin (2 arcsin (\frac{1}{5} u))))

代用を戻す

u = x - 5

= 2 (- \frac{1}{3} (- (x - 5)^{2} + 25)^{\frac{3}{2}} + \frac{125}{2} (arcsin (\frac{1}{5} (x - 5)) + \frac{1}{2} sin (2 arcsin (\frac{1}{5} (x - 5)))))

簡素化

2 (- \frac{1}{3} (- (x - 5)^{2} + 25)^{\frac{3}{2}} + \frac{125}{2} (arcsin (\frac{1}{5} (x - 5)) + \frac{1}{2} sin (2 arcsin (\frac{1}{5} (x - 5))))) : 2 (- \frac{1}{3} (- x^{2} + 10 x)^{\frac{3}{2}} + \frac{125}{4} (2 arcsin (\frac{1}{5} x - 1) + sin (2 arcsin (\frac{1}{5} x - 1))))

= 2 (- \frac{1}{3} (- x^{2} + 10 x)^{\frac{3}{2}} + \frac{125}{4} (2 arcsin (\frac{1}{5} x - 1) + sin (2 arcsin (\frac{1}{5} x - 1))))

定数を解答に追加する

= 2 (- \frac{1}{3} (- x^{2} + 10 x)^{\frac{3}{2}} + \frac{125}{4} (2 arcsin (\frac{1}{5} x - 1) + sin (2 arcsin (\frac{1}{5} x - 1)))) + C

グラフ

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