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integral from 7 to 14 of sqrt(x^2-16)

解答

\int_{7}^{14} x^{2} - 16 d x

解答

16 (- \frac{1}{2} ln (- \frac{( 3 5 + 7 ) ( 33 - 7 )}{8}) + \frac{21 5}{8} - \frac{7 33}{32})

+1

十进制

67.68059 \dots

求解步骤

\int_{7}^{14} x^{2} - 16 d x

使用三角换元法

= \int_{arcsec (\frac{7}{4})}^{arcsec (\frac{7}{2})} 16 tan^{2} (u) sec (u) d u

提出常数:

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

= 16 \cdot \int_{arcsec (\frac{7}{4})}^{arcsec (\frac{7}{2})} tan^{2} (u) sec (u) d u

使用三角恒等式改写

= 16 \cdot \int_{arcsec (\frac{7}{4})}^{arcsec (\frac{7}{2})} (- 1 + sec^{2} (u)) sec (u) d u

乘开

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

= 16 \cdot \int_{arcsec (\frac{7}{4})}^{arcsec (\frac{7}{2})} - sec (u) + sec^{3} (u) d u

使用积分加法定则:

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

= 16 (- \int_{arcsec (\frac{7}{4})}^{arcsec (\frac{7}{2})} sec (u) d u + \int_{arcsec (\frac{7}{4})}^{arcsec (\frac{7}{2})} sec^{3} (u) d u)

\int_{arcsec (\frac{7}{4})}^{arcsec (\frac{7}{2})} sec (u) d u = ln (- \frac{( 3 5 + 7 ) ( 33 - 7 )}{8})

\int_{arcsec (\frac{7}{4})}^{arcsec (\frac{7}{2})} sec^{3} (u) d u = \frac{21 5}{8} - \frac{7 33}{32} + \frac{1}{2} ln (- \frac{( 3 5 + 7 ) ( 33 - 7 )}{8})

= 16 (- ln (- \frac{( 3 5 + 7 ) ( 33 - 7 )}{8}) + \frac{21 5}{8} - \frac{7 33}{32} + \frac{1}{2} ln (- \frac{( 3 5 + 7 ) ( 33 - 7 )}{8}))

化简

- ln (- \frac{( 3 5 + 7 ) ( 33 - 7 )}{8}) + \frac{21 5}{8} - \frac{7 33}{32} + \frac{1}{2} ln (- \frac{( 3 5 + 7 ) ( 33 - 7 )}{8}) : - \frac{1}{2} ln (- \frac{( 3 5 + 7 ) ( 33 - 7 )}{8}) + \frac{21 5}{8} - \frac{7 33}{32}

= 16 (- \frac{1}{2} ln (- \frac{( 3 5 + 7 ) ( 33 - 7 )}{8}) + \frac{21 5}{8} - \frac{7 33}{32})

作图

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