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人気のある微分積分 >

integral from 0 to 2 of 7xe^{4-x^2}

解

\int_{0}^{2} 7 x e^{4 - x^{2}} d x

解

\frac{7 ( e ^{4} - 1 )}{2}

+1

十進法表記

187.59352 \dots

解答ステップ

\int_{0}^{2} 7 x e^{4 - x^{2}} d x

定数を除く:

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

= 7 \cdot \int_{0}^{2} x e^{4 - x^{2}} d x

u置換積分法を適用する

= 7 \cdot \int_{4}^{0} - \frac{e ^{u}}{2} d u

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

= 7 (- \int_{0}^{4} - \frac{e ^{u}}{2} d u)

定数を除く:

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

= 7 (- (- \frac{1}{2} \cdot \int_{0}^{4} e^{u} d u))

共通積分を使用する:

\int e^{u} d u = e^{u}

= 7 (- (- \frac{1}{2} [e^{u}]_{0}^{4}))

簡素化

7 (- (- \frac{1}{2} [e^{u}]_{0}^{4})) : \frac{7}{2} [e^{u}]_{0}^{4}

= \frac{7}{2} [e^{u}]_{0}^{4}

限度を計算する:

e^{4} - 1

= \frac{7}{2} (e^{4} - 1)

簡素化

= \frac{7 ( e ^{4} - 1 )}{2}

グラフ

人気の例

integral from 2 to 5 of xe^x

\int_{2}^{5} x e^{x} d x

integral from 4 to 9 of (2x-5)/4

\int_{4}^{9} \frac{2 x - 5}{4} d x

integral from-1 to 0 of 9(e^{1/x})/(x^3)

\int_{- 1}^{0} 9 \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

integral from 0 to pi of e^{2x}cos(x)

\int_{0}^{π} e^{2 x} cos (x) d x

integral from 1 to sqrt(e of) 1/x

\int_{1}^{e} \frac{1}{x} d x