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integral from 0 to 0.6 of (x^2)/(sqrt(9-25x^2))

解答

\int_{0}^{0.6} \frac{x ^{2}}{9 - 25 x ^{2}} d x

解答

\frac{9 π}{500}

+1

十进制

0.05654 \dots

求解步骤

\int_{0}^{0.6} \frac{x ^{2}}{9 - 25 x ^{2}} d x

使用三角换元法

= \int_{0}^{\frac{π}{2}} \frac{9 sin ^{2} ( u )}{125} d u

提出常数:

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

= \frac{9}{125} \cdot \int_{0}^{\frac{π}{2}} sin^{2} (u) d u

使用三角恒等式改写

= \frac{9}{125} \cdot \int_{0}^{\frac{π}{2}} \frac{1 - cos ( 2 u )}{2} d u

提出常数:

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

= \frac{9}{125} \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} 1 - cos (2 u) d u

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{9}{125} \cdot \frac{1}{2} (\int_{0}^{\frac{π}{2}} 1 d u - \int_{0}^{\frac{π}{2}} cos (2 u) d u)

\int_{0}^{\frac{π}{2}} 1 d u = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (2 u) d u = 0

= \frac{9}{125} \cdot \frac{1}{2} (\frac{π}{2} - 0)

化简

\frac{9}{125} \cdot \frac{1}{2} (\frac{π}{2} - 0) : \frac{9 π}{500}

= \frac{9 π}{500}

作图

流行的例子

tangent 2x^4-9x^2

t an g e n t 2 x^{4} - 9 x^{2}

x y^{^{'}} + y = 3

ty^'+(t+1)y=t,y(ln(3))=1

t y^{^{'}} + (t + 1) y = t, y (ln (3)) = 1

limit as x approaches 0 of x^4(cos(2/x))

x \to 0 lim (x^{4} (cos (\frac{2}{x})))

integral of 4/5

\int \frac{4}{5} d x