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人気のある微分積分 >

integral from 0 to 2 of te^{2t}

解

\int_{0}^{2} t e^{2 t} d t

解

\frac{3 e ^{4} + 1}{4}

+1

十進法表記

41.19861 \dots

解答ステップ

\int_{0}^{2} t e^{2 t} d t

u置換積分法を適用する

= \int_{0}^{4} \frac{e ^{u} u}{4} 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_{0}^{4} e^{u} u d u

部分積分を適用する

= \frac{1}{4} [e^{u} u - \int e^{u} d u]_{0}^{4}

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

= \frac{1}{4} [e^{u} u - e^{u}]_{0}^{4}

限度を計算する:

3 e^{4} + 1

= \frac{1}{4} (3 e^{4} + 1)

簡素化

= \frac{3 e ^{4} + 1}{4}

グラフ

人気の例

(\partial)/(\partial x)(x^2e^{3xy})

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