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人気のある微分積分 >

integral from 0 to 1 of sqrt(1+16x^2)

解

\int_{0}^{1} 1 + 16 x^{2} d x

解

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

+1

十進法表記

2.32339 \dots

解答ステップ

\int_{0}^{1} 1 + 16 x^{2} 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}{4} 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}{4} \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}{4} ([\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}{4} ([\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}{4} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (4)} + \frac{1}{2} ln (4 + 17)) : \frac{1}{4} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (4)} + \frac{1}{2} ln (4 + 17))

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

グラフ

人気の例

integral from 0 to 4 of-2x+3

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

integral from 0 to 2 of x^3+1

\int_{0}^{2} x^{3} + 1 d x

integral from 0 to 2 of (x^2+2x)

\int_{0}^{2} (x^{2} + 2 x) d x

integral from 1 to 4 of 4x

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

integral from 1 to 6 of xsqrt(x+3)

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