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人気のある微分積分 >

integral from 0 to 4 of x/(1+sqrt(2x+1))

解

\int_{0}^{4} \frac{x}{1 + 2 x + 1} d x

解

\frac{7}{3}

+1

十進法表記

2.33333 \dots

解答ステップ

\int_{0}^{4} \frac{x}{1 + 2 x + 1} d x

u置換積分法を適用する

= \int_{1}^{9} \frac{u - 1}{4 ( 1 + u )} d u

定数を除く:

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

= \frac{1}{4} \cdot \int_{1}^{9} \frac{u - 1}{1 + u} d u

\frac{u - 1}{1 + u} = u - 1

= \frac{1}{4} \cdot \int_{1}^{9} u - 1 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

= \frac{1}{4} (\int_{1}^{9} u d u - \int_{1}^{9} 1 d u)

\int_{1}^{9} u d u = \frac{52}{3}

\int_{1}^{9} 1 d u = 8

= \frac{1}{4} (\frac{52}{3} - 8)

簡素化

\frac{1}{4} (\frac{52}{3} - 8) : \frac{7}{3}

= \frac{7}{3}

グラフ

人気の例

integral from 5 to 7 of x^2-2x-15

\int_{5}^{7} x^{2} - 2 x - 15 d x

integral from 2 to 6 of x/((x^2+2)^3)

\int_{2}^{6} \frac{x}{( x ^{2} + 2 ) ^{3}} d x

integral from 1 to 2 of (e^x-x)

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

integral from 0 to 1 of (e^{3ax})/3

\int_{0}^{1} \frac{e ^{3 a x}}{3} d x

integral from-1 to 1 of 2xe^{x^2}

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