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人気のある微分積分 >

integral from 1 to 4 of sqrt(1+4x^2)

解

\int_{1}^{4} 1 + 4 x^{2} d x

解

\frac{1}{2} (465 - 5 + \frac{1}{2} (ln (8 + 65) - ln (2 + 5)))

+1

十進法表記

15.33969 \dots

解答ステップ

\int_{1}^{4} 1 + 4 x^{2} d x

三角関数による置換を適用する

= \int_{a r c t a n (2)}^{a r c t a n (8)} \frac{1}{2} 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{1}{2} \cdot \int_{a r c t a n (2)}^{a r c t a n (8)} sec^{3} (u) d u

積分換算を適用する

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (2)}^{a r c t a n (8)} + \frac{1}{2} \cdot \int_{a r c t a n (2)}^{a r c t a n (8)} sec (u) d u)

\int_{a r c t a n (2)}^{a r c t a n (8)} sec (u) d u = ln (8 + 65) - ln (2 + 5)

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (2)}^{a r c t a n (8)} + \frac{1}{2} (ln (8 + 65) - ln (2 + 5)))

簡素化

\frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (2)}^{a r c t a n (8)} + \frac{1}{2} (ln (8 + 65) - ln (2 + 5))) : \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (2)}^{a r c t a n (8)} + \frac{1}{2} (ln (8 + 65) - ln (2 + 5)))

= \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (2)}^{a r c t a n (8)} + \frac{1}{2} (ln (8 + 65) - ln (2 + 5)))

限度を計算する:

465 - 5

= \frac{1}{2} (465 - 5 + \frac{1}{2} (ln (8 + 65) - ln (2 + 5)))

グラフ

人気の例

integral from 1 to 3 of (y+4)/(y-4)

\int_{1}^{3} \frac{y + 4}{y - 4} d y

integral from 1 to 6 of x^3+x^2+3

\int_{1}^{6} x^{3} + x^{2} + 3 d x

integral from 1 to infinity of x/(x^2+7)

\int_{1}^{\infty} \frac{x}{x ^{2} + 7} d x

integral from-1 to 3 of (x^3-3x^2-1)

\int_{- 1}^{3} (x^{3} - 3 x^{2} - 1) d x

integral from-2 to 4 of |(x+1)/(x+6)|

\int_{- 2}^{4} \frac{x + 1}{x + 6} d x