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人気のある微分積分 >

integral from 4 to 8 of sqrt(x^2-16)

解

\int_{4}^{8} x^{2} - 16 d x

解

16 (- \frac{1}{2} ln (2 + 3) + 3)

+1

十進法表記

17.17714 \dots

解答ステップ

\int_{4}^{8} x^{2} - 16 d x

三角関数による置換を適用する

= \int_{0}^{\frac{π}{3}} 16 tan^{2} (u) sec (u) d u

定数を除く:

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

= 16 \cdot \int_{0}^{\frac{π}{3}} tan^{2} (u) sec (u) d u

三角関数の公式を使用して書き換える

= 16 \cdot \int_{0}^{\frac{π}{3}} (- 1 + sec^{2} (u)) sec (u) d u

拡張

(- 1 + sec^{2} (u)) sec (u) : - sec (u) + sec^{3} (u)

= 16 \cdot \int_{0}^{\frac{π}{3}} - sec (u) + sec^{3} (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

= 16 (- \int_{0}^{\frac{π}{3}} sec (u) d u + \int_{0}^{\frac{π}{3}} sec^{3} (u) d u)

\int_{0}^{\frac{π}{3}} sec (u) d u = ln (2 + 3)

\int_{0}^{\frac{π}{3}} sec^{3} (u) d u = 3 + \frac{1}{2} ln (2 + 3)

= 16 (- ln (2 + 3) + 3 + \frac{1}{2} ln (2 + 3))

簡素化

- ln (2 + 3) + 3 + \frac{1}{2} ln (2 + 3) : - \frac{1}{2} ln (2 + 3) + 3

= 16 (- \frac{1}{2} ln (2 + 3) + 3)

グラフ

人気の例

integral from-1 to 1 of 7t^6

\int_{- 1}^{1} 7 t^{6} d t

integral from 0 to 1 of t^2+2

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

integral from y to sqrt(y of)1

\int_{y}^{y} 1 d x

integral from 0 to 2 of 1/6 x

\int_{0}^{2} \frac{1}{6} x d x

integral from 1 to 4 of 1/6 (x-1)

\int_{1}^{4} \frac{1}{6} (x - 1) d x