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人気のある微分積分 >

integral from 0 to 1 of sqrt(1+2t^2+t^4)

解

\int_{0}^{1} 1 + 2 t^{2} + t^{4} d t

解

\frac{4}{3}

+1

十進法表記

1.33333 \dots

解答ステップ

\int_{0}^{1} 1 + 2 t^{2} + t^{4} d t

因数

1 + 2 t^{2} + t^{4} : (t^{2} + 1)^{2}

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

指数の規則を適用する:

a^{2} = ∣ a ∣

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

絶対数を排除する

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

指数の規則を適用する:

a^{2} = ∣ a ∣

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

絶対数を排除する

= \int_{0}^{1} t^{2} + 1 d t

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} t^{2} d t + \int_{0}^{1} 1 d t

\int_{0}^{1} t^{2} d t = \frac{1}{3}

\int_{0}^{1} 1 d t = 1

= \frac{1}{3} + 1

簡素化

\frac{1}{3} + 1 : \frac{4}{3}

= \frac{4}{3}

グラフ

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