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人気のある微分積分 >

integral from 12 to 16 of (x^2-6)^{1/2}

解

\int_{12}^{16} (x^{2} - 6)^{\frac{1}{2}} d x

解

6 (- \frac{1}{2} (ln (5 \cdot 5^{\frac{1}{2}} + 8 \cdot 2^{\frac{1}{2}}) - \frac{1}{2} ln (3) - ln (2 3^{\frac{1}{2}} + 2 \cdot 6^{\frac{1}{2}})) + \frac{20 \cdot 1 0 ^{\frac{1}{2}}}{3} - 13 8^{\frac{1}{2}})

+1

十進法表記

55.13000 \dots

解答ステップ

\int_{12}^{16} (x^{2} - 6)^{\frac{1}{2}} d x

三角関数による置換を適用する

= \int_{arcsec (2 \cdot 6^{\frac{1}{2}})}^{arcsec (\frac{8 \cdot 2 ^{\frac{1}{2}}}{3 ^{\frac{1}{2}}})} 6 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

= 6 \cdot \int_{arcsec (2 \cdot 6^{\frac{1}{2}})}^{arcsec (\frac{8 \cdot 2 ^{\frac{1}{2}}}{3 ^{\frac{1}{2}}})} tan^{2} (u) sec (u) d u

三角関数の公式を使用して書き換える

= 6 \cdot \int_{arcsec (2 \cdot 6^{\frac{1}{2}})}^{arcsec (\frac{8 \cdot 2 ^{\frac{1}{2}}}{3 ^{\frac{1}{2}}})} (- 1 + sec^{2} (u)) sec (u) d u

拡張

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

= 6 \cdot \int_{arcsec (2 \cdot 6^{\frac{1}{2}})}^{arcsec (\frac{8 \cdot 2 ^{\frac{1}{2}}}{3 ^{\frac{1}{2}}})} - 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

= 6 - \int_{arcsec (2 \cdot 6^{\frac{1}{2}})}^{arcsec (\frac{8 \cdot 2 ^{\frac{1}{2}}}{3 ^{\frac{1}{2}}})} sec (u) d u + \int_{arcsec (2 \cdot 6^{\frac{1}{2}})}^{arcsec (\frac{8 \cdot 2 ^{\frac{1}{2}}}{3 ^{\frac{1}{2}}})} sec^{3} (u) d u

\int_{arcsec (2 \cdot 6^{\frac{1}{2}})}^{arcsec (\frac{8 \cdot 2 ^{\frac{1}{2}}}{3 ^{\frac{1}{2}}})} sec (u) d u = ln (5 \cdot 5^{\frac{1}{2}} + 8 \cdot 2^{\frac{1}{2}}) - \frac{1}{2} ln (3) - ln (2 3^{\frac{1}{2}} + 2 \cdot 6^{\frac{1}{2}})

\int_{arcsec (2 \cdot 6^{\frac{1}{2}})}^{arcsec (\frac{8 \cdot 2 ^{\frac{1}{2}}}{3 ^{\frac{1}{2}}})} sec^{3} (u) d u = \frac{20 \cdot 1 0 ^{\frac{1}{2}}}{3} - 13 8^{\frac{1}{2}} + \frac{1}{2} (ln (5 \cdot 5^{\frac{1}{2}} + 8 \cdot 2^{\frac{1}{2}}) - \frac{1}{2} ln (3) - ln (2 3^{\frac{1}{2}} + 2 \cdot 6^{\frac{1}{2}}))

= 6 (- (ln (5 \cdot 5^{\frac{1}{2}} + 8 \cdot 2^{\frac{1}{2}}) - \frac{1}{2} ln (3) - ln (2 3^{\frac{1}{2}} + 2 \cdot 6^{\frac{1}{2}})) + \frac{20 \cdot 1 0 ^{\frac{1}{2}}}{3} - 13 8^{\frac{1}{2}} + \frac{1}{2} (ln (5 \cdot 5^{\frac{1}{2}} + 8 \cdot 2^{\frac{1}{2}}) - \frac{1}{2} ln (3) - ln (2 3^{\frac{1}{2}} + 2 \cdot 6^{\frac{1}{2}})))

類似した元を足す:

- (ln (5 \cdot 5^{\frac{1}{2}} + 8 \cdot 2^{\frac{1}{2}}) - \frac{1}{2} ln (3) - ln (2 \cdot 6^{\frac{1}{2}} + 2 3^{\frac{1}{2}})) + \frac{1}{2} (ln (5 \cdot 5^{\frac{1}{2}} + 8 \cdot 2^{\frac{1}{2}}) - \frac{1}{2} ln (3) - ln (2 \cdot 6^{\frac{1}{2}} + 2 3^{\frac{1}{2}})) = - \frac{1}{2} (ln (5 \cdot 5^{\frac{1}{2}} + 8 \cdot 2^{\frac{1}{2}}) - \frac{1}{2} ln (3) - ln (2 \cdot 6^{\frac{1}{2}} + 2 3^{\frac{1}{2}}))

= 6 (- \frac{1}{2} (ln (5 \cdot 5^{\frac{1}{2}} + 8 \cdot 2^{\frac{1}{2}}) - \frac{1}{2} ln (3) - ln (2 3^{\frac{1}{2}} + 2 \cdot 6^{\frac{1}{2}})) + \frac{20 \cdot 1 0 ^{\frac{1}{2}}}{3} - 13 8^{\frac{1}{2}})

グラフ

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