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人気のある微分積分 >

integral from 0 to 2 of sqrt(4x^2+1)

解

\int_{0}^{2} 4 x^{2} + 1 d x

解

17 + \frac{1}{4} ln (4 + 17)

+1

十進法表記

4.64678 \dots

解答ステップ

\int_{0}^{2} 4 x^{2} + 1 d x

三角関数による置換を適用する

= \int_{0}^{a r c t a n (4)} \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_{0}^{a r c t a n (4)} sec^{3} (u) d u

積分換算を適用する

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (4)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (4)} sec (u) d u)

\int_{0}^{a r c t a n (4)} sec (u) d u = ln (4 + 17)

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (4)} + \frac{1}{2} ln (4 + 17))

簡素化

\frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (4)} + \frac{1}{2} ln (4 + 17)) : \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (4)} + \frac{1}{2} ln (4 + 17))

= \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (4)} + \frac{1}{2} ln (4 + 17))

限度を計算する:

217

= \frac{1}{2} (217 + \frac{1}{2} ln (4 + 17))

簡素化

= 17 + \frac{1}{4} ln (4 + 17)

グラフ

人気の例

integral from-2 to 0 of 4x-x^2-6x

\int_{- 2}^{0} 4 x - x^{2} - 6 x d x

integral from 1 to 4 of (x-9)/(sqrt(x))

\int_{1}^{4} \frac{x - 9}{x} d x

integral from 0 to 16 of (sqrt(x)-3)

\int_{0}^{16} (x - 3) d x

integral from 0 to 3 of (x^2-4x)

\int_{0}^{3} (x^{2} - 4 x) d x

integral from 1959 to infinity of 1/x

\int_{1959}^{\infty} \frac{1}{x} d x