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人気のある微分積分 >

integral from 0 to 1 of tsqrt(1+36t^2)

解

\int_{0}^{1} t 1 + 36 t^{2} d t

解

\frac{37 37 - 1}{108}

+1

十進法表記

2.07465 \dots

解答ステップ

\int_{0}^{1} t 1 + 36 t^{2} d t

u置換積分法を適用する

= \int_{1}^{37} \frac{u}{72} d u

定数を除く:

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

= \frac{1}{72} \cdot \int_{1}^{37} u d u

べき乗則を適用する

= \frac{1}{72} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{37}

限度を計算する:

\frac{74 37}{3} - \frac{2}{3}

= \frac{1}{72} (\frac{74 37}{3} - \frac{2}{3})

簡素化

= \frac{37 37 - 1}{108}

グラフ

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