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人気のある微分積分 >

integral from-1 to 1 of (2x+|x|+1)^2

解

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

解

\frac{22}{3}

+1

十進法表記

7.33333 \dots

解答ステップ

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

絶対数を排除する

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

\int_{- 1}^{0} (2 x - x + 1)^{2} d x = \frac{1}{3}

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

= \frac{1}{3} + 7

簡素化

\frac{1}{3} + 7 : \frac{22}{3}

= \frac{22}{3}

グラフ

人気の例

integral from 3 to 9 of 1/(x(ln(x))^6)

\int_{3}^{9} \frac{1}{x ( ln ( x ) ) ^{6}} d x

integral from 0 to 1 of 2y^2(1-y)^{x-1}

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

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

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

integral from 6 to 9 of (6-x)(x-9)

\int_{6}^{9} (6 - x) (x - 9) d x

integral from-2 to 6 of |x|

\int_{- 2}^{6} ∣ x ∣ d x