ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある微分積分 >

integral of ysqrt(10+8y-16y^2)

解

\int y 10 + 8 y - 16 y^{2} d y

解

- \frac{1}{48} (- 16 y^{2} + 8 y + 10)^{\frac{3}{2}} + \frac{11}{32} (arcsin (\frac{1}{11} (4 y - 1)) + \frac{1}{2} sin (2 arcsin (\frac{4 y - 1}{11}))) + C

解答ステップ

\int y 10 + 8 y - 16 y^{2} d y

y 10 + 8 y - 16 y^{2} = 2 y 5 + 4 y - 8 y^{2}

= \int 2 y 5 + 4 y - 8 y^{2} d y

定数を除く:

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

= 2 \cdot \int y 5 + 4 y - 8 y^{2} d y

平方完成する

5 + 4 y - 8 y^{2} : - 8 (y - \frac{1}{4})^{2} + \frac{11}{2}

= 2 \cdot \int y - 8 (y - \frac{1}{4})^{2} + \frac{11}{2} d y

u置換積分法を適用する

= 2 \cdot \int \frac{- 16 u ^{2} + 11}{2} (u + \frac{1}{4}) d u

\frac{- 16 u ^{2} + 11}{2} (u + \frac{1}{4}) = \frac{1}{2} - 16 u^{2} + 11 (u + \frac{1}{4})

= 2 \cdot \int \frac{1}{2} - 16 u^{2} + 11 (u + \frac{1}{4}) d u

定数を除く:

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

= 2 \frac{1}{2} \cdot \int - 16 u^{2} + 11 (u + \frac{1}{4}) d u

拡張

- 16 u^{2} + 11 (u + \frac{1}{4}) : u - 16 u^{2} + 11 + \frac{- 16 u ^{2} + 11}{4}

= 2 \frac{1}{2} \cdot \int u - 16 u^{2} + 11 + \frac{- 16 u ^{2} + 11}{4} 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 \frac{1}{2} (\int u - 16 u^{2} + 11 d u + \int \frac{- 16 u ^{2} + 11}{4} d u)

\int u - 16 u^{2} + 11 d u = - \frac{1}{48} (- 16 u^{2} + 11)^{\frac{3}{2}}

\int \frac{- 16 u ^{2} + 11}{4} d u = \frac{11}{32} (arcsin (\frac{4}{11} u) + \frac{1}{2} sin (2 arcsin (\frac{4}{11} u)))

= 2 \frac{1}{2} (- \frac{1}{48} (- 16 u^{2} + 11)^{\frac{3}{2}} + \frac{11}{32} (arcsin (\frac{4}{11} u) + \frac{1}{2} sin (2 arcsin (\frac{4}{11} u))))

代用を戻す

u = y - \frac{1}{4}

= 2 \frac{1}{2} - \frac{1}{48} (- 16 (y - \frac{1}{4})^{2} + 11)^{\frac{3}{2}} + \frac{11}{32} (arcsin (\frac{4}{11} (y - \frac{1}{4})) + \frac{1}{2} sin (2 arcsin (\frac{4}{11} (y - \frac{1}{4}))))

簡素化

2 \frac{1}{2} - \frac{1}{48} (- 16 (y - \frac{1}{4})^{2} + 11)^{\frac{3}{2}} + \frac{11}{32} (arcsin (\frac{4}{11} (y - \frac{1}{4})) + \frac{1}{2} sin (2 arcsin (\frac{4}{11} (y - \frac{1}{4})))) : - \frac{1}{48} (- 16 y^{2} + 8 y + 10)^{\frac{3}{2}} + \frac{11}{32} (arcsin (\frac{1}{11} (4 y - 1)) + \frac{1}{2} sin (2 arcsin (\frac{4 y - 1}{11})))

= - \frac{1}{48} (- 16 y^{2} + 8 y + 10)^{\frac{3}{2}} + \frac{11}{32} (arcsin (\frac{1}{11} (4 y - 1)) + \frac{1}{2} sin (2 arcsin (\frac{4 y - 1}{11})))

定数を解答に追加する

= - \frac{1}{48} (- 16 y^{2} + 8 y + 10)^{\frac{3}{2}} + \frac{11}{32} (arcsin (\frac{1}{11} (4 y - 1)) + \frac{1}{2} sin (2 arcsin (\frac{4 y - 1}{11}))) + C

グラフ

人気の例

derivative of sin(x+5)

\frac{d}{d x} (sin (x) + 5)

勾配 (3,4),(6,8)

s l o p e (3, 4), (6, 8)

integral from 0 to 1 of (x^2-x^4)

\int_{0}^{1} (x^{2} - x^{4}) d x

derivative of (3x/(4x^2+9))

\frac{d}{d x} (\frac{3 x}{4 x ^{2} + 9})

面積 x^2-17,16x

a re a x^{2} - 17, 16 x