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人気のある微分積分 >

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

解

\int_{0}^{2} \frac{x}{4 4 - x ^{2}} d x

解

\frac{1}{2}

+1

十進法表記

0.5

解答ステップ

\int_{0}^{2} \frac{x}{4 4 - x ^{2}} d x

定数を除く:

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

= \frac{1}{4} \cdot \int_{0}^{2} \frac{x}{4 - x ^{2}} d x

u置換積分法を適用する

= \frac{1}{4} \cdot \int_{4}^{0} - \frac{1}{2 u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= \frac{1}{4} (- \int_{0}^{4} - \frac{1}{2 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} (- (- \frac{1}{2} \cdot \int_{0}^{4} \frac{1}{u} d u))

べき乗則を適用する

= \frac{1}{4} (- (- \frac{1}{2} [2 u^{\frac{1}{2}}]_{0}^{4}))

簡素化

\frac{1}{4} (- (- \frac{1}{2} [2 u^{\frac{1}{2}}]_{0}^{4})) : \frac{1}{8} [2 u]_{0}^{4}

= \frac{1}{8} [2 u]_{0}^{4}

限度を計算する:

4

= \frac{1}{8} \cdot 4

簡素化

= \frac{1}{2}

グラフ

人気の例

integral from-1 to 2 of (3x^2-2)

\int_{- 1}^{2} (3 x^{2} - 2) d x

integral from 0 to 3 of ye^{-xy}

\int_{0}^{3} y e^{- x y} d y

integral from 0 to 5 of (2x^2)/(25)

\int_{0}^{5} \frac{2 x ^{2}}{25} d x

integral from 0 to 1 of 1/((11x+1)^3)

\int_{0}^{1} \frac{1}{( 11 x + 1 ) ^{3}} d x

integral from-2 to 1 of-x^2+4x+5

\int_{- 2}^{1} - x^{2} + 4 x + 5 d x