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人気のある微分積分 >

integral from 0 to 2 of sqrt(-x^2+4)

解

\int_{0}^{2} - x^{2} + 4 d x

解

π

+1

十進法表記

3.14159 \dots

解答ステップ

\int_{0}^{2} - x^{2} + 4 d x

三角関数による置換を適用する

= \int_{0}^{\frac{π}{2}} 4 cos^{2} (u) d u

定数を除く:

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

= 4 \cdot \int_{0}^{\frac{π}{2}} cos^{2} (u) d u

三角関数の公式を使用して書き換える

= 4 \cdot \int_{0}^{\frac{π}{2}} \frac{1 + cos ( 2 u )}{2} d u

定数を除く:

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

= 4 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} 1 + cos (2 u) d u

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

= 4 \cdot \frac{1}{2} (\int_{0}^{\frac{π}{2}} 1 d u + \int_{0}^{\frac{π}{2}} cos (2 u) d u)

\int_{0}^{\frac{π}{2}} 1 d u = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (2 u) d u = 0

= 4 \cdot \frac{1}{2} (\frac{π}{2} + 0)

簡素化

4 \cdot \frac{1}{2} (\frac{π}{2} + 0) : π

= π

グラフ

人気の例

integral from 0 to 7 of 3sqrt(x+9)

\int_{0}^{7} 3 x + 9 d x

integral from-2 to 14 of ln(t)sin(t)

\int_{- 2}^{14} ln (t) sin (t) d t

integral from 0 to x of (sin(t))/(pi-t)

\int_{0}^{x} \frac{sin ( t )}{π - t} d t

integral from 0 to 9 of [2(37)+3(16)]

\int_{0}^{9} [2 (37) + 3 (16)] d x

integral from 1 to e of (ln^2(x))/(x^2)

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