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人気のある微分積分 >

integral from 0 to 1 of 5(1+sqrt(x))^6

解

\int_{0}^{1} 5 (1 + x)^{6} d x

解

\frac{3845}{28}

+1

十進法表記

137.32142 \dots

解答ステップ

\int_{0}^{1} 5 (1 + x)^{6} d x

定数を除く:

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

= 5 \cdot \int_{0}^{1} (1 + x)^{6} d x

拡張

(1 + x)^{6} : 1 + 6 x + 15 x + 20 x^{\frac{3}{2}} + 15 x^{2} + 6 x^{\frac{5}{2}} + x^{3}

= 5 \cdot \int_{0}^{1} 1 + 6 x + 15 x + 20 x^{\frac{3}{2}} + 15 x^{2} + 6 x^{\frac{5}{2}} + x^{3} 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

= 5 (\int_{0}^{1} 1 d x + \int_{0}^{1} 6 x d x + \int_{0}^{1} 15 x d x + \int_{0}^{1} 20 x^{\frac{3}{2}} d x + \int_{0}^{1} 15 x^{2} d x + \int_{0}^{1} 6 x^{\frac{5}{2}} d x + \int_{0}^{1} x^{3} d x)

\int_{0}^{1} 1 d x = 1

\int_{0}^{1} 6 x d x = 4

\int_{0}^{1} 15 x d x = \frac{15}{2}

\int_{0}^{1} 20 x^{\frac{3}{2}} d x = 8

\int_{0}^{1} 15 x^{2} d x = 5

\int_{0}^{1} 6 x^{\frac{5}{2}} d x = \frac{12}{7}

\int_{0}^{1} x^{3} d x = \frac{1}{4}

= 5 (1 + 4 + \frac{15}{2} + 8 + 5 + \frac{12}{7} + \frac{1}{4})

簡素化

5 (1 + 4 + \frac{15}{2} + 8 + 5 + \frac{12}{7} + \frac{1}{4}) : \frac{3845}{28}

= \frac{3845}{28}

グラフ

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