integral from 1 to 2 of 1/(sqrt(4-t^2))

解答

\int_{1}^{2} \frac{1}{4 - t ^{2}} d t

\frac{π}{3}

十进制

1.04719 \dots

求解步骤

\int_{1}^{2} \frac{1}{4 - t ^{2}} d t

使用三角换元法

= \int_{\frac{π}{6}}^{\frac{π}{2}} 1 d u

常数积分:

\int a d x = a x

= [1 \cdot u]_{\frac{π}{6}}^{\frac{π}{2}}

化简

= [u]_{\frac{π}{6}}^{\frac{π}{2}}

带入上下限计算后得:

\frac{π}{3}

= \frac{π}{3}

t an g e n t - 3 x^{2} + 2 x y + 2 y^{3} = 125, (3, 4)

\frac{d}{d x} y = ln (10 x^{2} - 8)

\frac{d}{d x} (I n x)

t an g e n t f (x) = x^{3} - 2 x^{2} + 1, at x = 1

\frac{d}{d x} (arcsin (e^{2 x}))