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人気のある微分積分 >

integral from 0 to 2 of tsqrt(4t^2+1)

解

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

解

\frac{17 17 - 1}{12}

+1

十進法表記

5.75773 \dots

解答ステップ

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

u置換積分法を適用する

= \int_{1}^{17} \frac{u}{8} d u

定数を除く:

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

= \frac{1}{8} \cdot \int_{1}^{17} u d u

べき乗則を適用する

= \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{17}

限度を計算する:

\frac{34 17}{3} - \frac{2}{3}

= \frac{1}{8} (\frac{34 17}{3} - \frac{2}{3})

簡素化

= \frac{17 17 - 1}{12}

グラフ

人気の例

integral from 0 to 2 of x^2*3/4 x(2-x)

\int_{0}^{2} x^{2} \cdot \frac{3}{4} x (2 - x) d x

integral from 0 to 2pi of r

\int_{0}^{2 π} r d r

integral from 4 to 1 of-sqrt(x)

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

integral from 0 to 2 of 12x^2

\int_{0}^{2} 12 x^{2} d x

integral from 0 to y^2 of e^{x/(y^2)}

\int_{0}^{y^{2}} e^{\frac{x}{y ^{2}}} d x