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人気のある微分積分 >

integral from 0 to 3 of sqrt(x^2+16)

解

\int_{0}^{3} x^{2} + 16 d x

解

16 (\frac{15}{32} + \frac{1}{2} ln (2))

+1

十進法表記

13.04517 \dots

解答ステップ

\int_{0}^{3} x^{2} + 16 d x

三角関数による置換を適用する

= \int_{0}^{a r c t a n (\frac{3}{4})} 16 sec^{3} (u) d u

定数を除く:

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

= 16 \cdot \int_{0}^{a r c t a n (\frac{3}{4})} sec^{3} (u) d u

積分換算を適用する

= 16 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{3}{4})} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (\frac{3}{4})} sec (u) d u

\int_{0}^{a r c t a n (\frac{3}{4})} sec (u) d u = ln (2)

= 16 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{3}{4})} + \frac{1}{2} ln (2)

簡素化

16 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{3}{4})} + \frac{1}{2} ln (2) : 16 ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{3}{4})} + \frac{1}{2} ln (2))

= 16 ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{3}{4})} + \frac{1}{2} ln (2))

限度を計算する:

\frac{15}{32}

= 16 (\frac{15}{32} + \frac{1}{2} ln (2))

グラフ

人気の例

(dy}{dx}+y/x =\frac{-y^2)/x

\frac{d y}{d x} + \frac{y}{x} = \frac{- y ^{2}}{x}

integral of (4x)/((2x^2+3)^3)

\int \frac{4 x}{( 2 x ^{2} + 3 ) ^{3}} d x

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integral of 1/(sqrt(4x^2+1))

\int \frac{1}{4 x ^{2} + 1} d x

limit as x approaches 1 of 3f(x)-2h(x)

x \to 1 lim (3 f (x) - 2 h (x))