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人気のある微分積分 >

integral from 0 to 3 of sqrt(x^2+9)

解

\int_{0}^{3} x^{2} + 9 d x

解

\frac{9 ( 2 + ln ( 1 + 2 ) )}{2}

+1

十進法表記

10.33014 \dots

解答ステップ

\int_{0}^{3} x^{2} + 9 d x

三角関数による置換を適用する

= \int_{0}^{\frac{π}{4}} 9 sec^{3} (u) d u

定数を除く:

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

= 9 \cdot \int_{0}^{\frac{π}{4}} sec^{3} (u) d u

積分換算を適用する

= 9 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} \cdot \int_{0}^{\frac{π}{4}} sec (u) d u)

\int_{0}^{\frac{π}{4}} sec (u) d u = ln (1 + 2)

= 9 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

簡素化

9 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)) : 9 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

= 9 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

限度を計算する:

\frac{1}{2}

= 9 (\frac{1}{2} + \frac{1}{2} ln (1 + 2))

簡素化

= \frac{9 ( 2 + ln ( 1 + 2 ) )}{2}

グラフ

人気の例

tangent y=2x^3+3x^2-12x+2

t an g e n t y = 2 x^{3} + 3 x^{2} - 12 x + 2

integral of x^2-x-6

\int x^{2} - x - 6 d x

derivative of 4sin(4x)

\frac{d}{d x} (4 sin (4 x))

(\partial)/(\partial y)(xy+y^2)

\frac{\partial}{\partial y} (x y + y^{2})

y^{^{'}} = y (1000 - y)