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人気のある微分積分 >

integral from 4 to 18 of sqrt(16+3x^2)

解

\int_{4}^{18} 16 + 3 x^{2} d x

解

18247 - 16 + \frac{8 ( ln ( \frac{9 3 + 247}{2} ) - ln ( 2 + 3 ) )}{3}

+1

十進法表記

273.51401 \dots

解答ステップ

\int_{4}^{18} 16 + 3 x^{2} d x

三角関数による置換を適用する

= \int_{\frac{π}{3}}^{a r c t a n (\frac{9 3}{2})} \frac{16}{3} sec^{3} (u) d u

定数を除く:

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

= \frac{16}{3} \cdot \int_{\frac{π}{3}}^{a r c t a n (\frac{9 3}{2})} sec^{3} (u) d u

積分換算を適用する

= \frac{16}{3} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{\frac{π}{3}}^{a r c t a n (\frac{9 3}{2})} + \frac{1}{2} \cdot \int_{\frac{π}{3}}^{a r c t a n (\frac{9 3}{2})} sec (u) d u

\int_{\frac{π}{3}}^{a r c t a n (\frac{9 3}{2})} sec (u) d u = ln (\frac{9 3 + 247}{2}) - ln (3 + 2)

= \frac{16}{3} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{\frac{π}{3}}^{a r c t a n (\frac{9 3}{2})} + \frac{1}{2} (ln (\frac{9 3 + 247}{2}) - ln (3 + 2))

簡素化

\frac{16}{3} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{\frac{π}{3}}^{a r c t a n (\frac{9 3}{2})} + \frac{1}{2} (ln (\frac{9 3 + 247}{2}) - ln (3 + 2)) : \frac{16}{3} [\frac{1}{2} sec (u) tan (u)]_{\frac{π}{3}}^{a r c t a n (\frac{9 3}{2})} + \frac{1}{2} (ln (\frac{9 3 + 247}{2}) - ln (3 + 2))

= \frac{16}{3} [\frac{1}{2} sec (u) tan (u)]_{\frac{π}{3}}^{a r c t a n (\frac{9 3}{2})} + \frac{1}{2} (ln (\frac{9 3 + 247}{2}) - ln (3 + 2))

限度を計算する:

\frac{9 741}{8} - 3

= \frac{16}{3} (\frac{9 741}{8} - 3 + \frac{1}{2} (ln (\frac{9 3 + 247}{2}) - ln (3 + 2)))

簡素化

= 18247 - 16 + \frac{8 ( ln ( \frac{9 3 + 247}{2} ) - ln ( 2 + 3 ) )}{3}

グラフ

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