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人気のある微分積分 >

integral from 0 to 7 of (2xsqrt(x^2-1))

解

\int_{0}^{7} (2 x x^{2} - 1) d x

解

1283 + i \frac{2}{3}

解答ステップ

\int_{0}^{7} 2 x x^{2} - 1 d x

定数を除く:

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

= 2 \cdot \int_{0}^{7} x x^{2} - 1 d x

u置換積分法を適用する

= 2 \cdot \int_{i}^{43} u^{2} d u

べき乗則を適用する

= 2 [\frac{u ^{3}}{3}]_{i}^{43}

限度を計算する:

643 + \frac{i}{3}

= 2 (643 + \frac{i}{3})

簡素化

= 1283 + i \frac{2}{3}

人気の例

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

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

integral from-1 to 3 of e^x-3

\int_{- 1}^{3} e^{x} - 3 d x

integral from 0 to 1 of (1-e^2)^{1/2}

\int_{0}^{1} (1 - e^{2})^{\frac{1}{2}} d x

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

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

integral from-pi to pi of (sin(x))^2

\int_{- π}^{π} (sin (x))^{2} d x