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受欢迎的微积分 >

integral from-1 to 1 of 2sqrt(1-y^2)

解答

\int_{- 1}^{1} 2 1 - y^{2} d y

解答

π

+1

十进制

3.14159 \dots

求解步骤

\int_{- 1}^{1} 2 1 - y^{2} d y

提出常数:

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

= 2 \cdot \int_{- 1}^{1} 1 - y^{2} d y

使用三角换元法

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

使用三角恒等式改写

= 2 \cdot \int_{- \frac{π}{2}}^{\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

= 2 \cdot \frac{1}{2} \cdot \int_{- \frac{π}{2}}^{\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

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

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

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

= 2 \cdot \frac{1}{2} (π + 0)

化简

2 \cdot \frac{1}{2} (π + 0) : π

= π

作图

流行的例子

integral from 8 to 20 of 1

\int_{8}^{20} 1 d x

integral from l to u of t*r*e^{-t*r}

\int_{l}^{u} t \cdot r \cdot e^{- t \cdot r} d t

integral from 8 to 0 of (2x^2+x+2)

\int_{8}^{0} (2 x^{2} + x + 2) d x

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\int_{0}^{1} (x - 1 - x^{2}) d x

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\int_{- \infty}^{0} 4 x e^{5 x} d x