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

解

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

\frac{13}{12}

十進法表記

1.08333 \dots

解答ステップ

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

簡素化

(x + \frac{1}{2})^{2} : x^{2} + x + \frac{1}{4}

= \int_{0}^{1} x^{2} + x + \frac{1}{4} d x

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} x^{2} d x + \int_{0}^{1} x d x + \int_{0}^{1} \frac{1}{4} d x

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

\int_{0}^{1} x d x = \frac{1}{2}

\int_{0}^{1} \frac{1}{4} d x = \frac{1}{4}

= \frac{1}{3} + \frac{1}{2} + \frac{1}{4}

簡素化

\frac{1}{3} + \frac{1}{2} + \frac{1}{4} : \frac{13}{12}

= \frac{13}{12}