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人気のある微分積分 >

integral from 0 to x of t^6sqrt(1+t^7)

解

\int_{0}^{x} t^{6} 1 + t^{7} d t

解

\frac{2 ( x ^{7} + 1 ) ^{\frac{3}{2}} - 2}{21}

解答ステップ

\int_{0}^{x} t^{6} 1 + t^{7} d t

u置換積分法を適用する

= \int_{0}^{x^{6}} \frac{u ^{\frac{1}{6}} u ^{\frac{7}{6}} + 1}{6} d u

定数を除く:

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

= \frac{1}{6} \cdot \int_{0}^{x^{6}} u^{\frac{1}{6}} u^{\frac{7}{6}} + 1 d u

u置換積分法を適用する

= \frac{1}{6} \cdot \int_{0}^{(x^{6})^{\frac{1}{6}}} 6 v^{6} v^{7} + 1 d v

定数を除く:

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

= \frac{1}{6} \cdot 6 \cdot \int_{0}^{(x^{6})^{\frac{1}{6}}} v^{6} v^{7} + 1 d v

u置換積分法を適用する

= \frac{1}{6} \cdot 6 \cdot \int_{1}^{(x^{6})^{\frac{7}{6}} + 1} \frac{w}{7} d w

定数を除く:

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

= \frac{1}{6} \cdot 6 \cdot \frac{1}{7} \cdot \int_{1}^{(x^{6})^{\frac{7}{6}} + 1} w d w

べき乗則を適用する

= \frac{1}{6} \cdot 6 \cdot \frac{1}{7} [\frac{2}{3} w^{\frac{3}{2}}]_{1}^{(x^{6})^{\frac{7}{6}} + 1}

簡素化

\frac{1}{6} \cdot 6 \cdot \frac{1}{7} [\frac{2}{3} w^{\frac{3}{2}}]_{1}^{(x^{6})^{\frac{7}{6}} + 1} : \frac{1}{7} [\frac{2}{3} w^{\frac{3}{2}}]_{1}^{x^{7} + 1}

= \frac{1}{7} [\frac{2}{3} w^{\frac{3}{2}}]_{1}^{x^{7} + 1}

限度を計算する:

\frac{2}{3} (x^{7} + 1)^{\frac{3}{2}} - \frac{2}{3}

= \frac{1}{7} (\frac{2}{3} (x^{7} + 1)^{\frac{3}{2}} - \frac{2}{3})

簡素化

= \frac{2 ( x ^{7} + 1 ) ^{\frac{3}{2}} - 2}{21}

人気の例

integral from 4 to 1 of 4x+3x^2-2x^3

\int_{4}^{1} 4 x + 3 x^{2} - 2 x^{3} d x

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\int_{- a}^{a} (a^{2} - t^{2}) d t

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\int_{0}^{4} (2 t + 1) d t

integral from 0 to 20 of x+sin(x)

\int_{0}^{20} x + sin (x) d x

integral from 1 to 4 of 4x-x^2

\int_{1}^{4} 4 x - x^{2} d x