integral from pi/2 to pi of sin(2x)

解答

\int_{\frac{π}{2}}^{π} sin (2 x) d x

- 1

求解步骤

\int_{\frac{π}{2}}^{π} sin (2 x) d x

使用换元积分法

= \int_{π}^{2 π} sin (u) \frac{1}{2} d u

提出常数:

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

= \frac{1}{2} \cdot \int_{π}^{2 π} sin (u) d u

使用积分常见定则:

\int sin (u) d u = - cos (u)

= \frac{1}{2} [- cos (u)]_{π}^{2 π}

带入上下限计算后得:

- 2

= \frac{1}{2} (- 2)

化简

= - 1

t a y l or \frac{sin ( x ) - x}{x ^{2}}

x \to 0 lim (sin (\frac{1}{∣ x ∣}))

d er i v a t i v e \frac{3 x ^{2}}{4}

\frac{d}{d x} (2 + \frac{1}{x ^{2}})

\frac{\partial}{\partial x} (3 x^{2} + 2 y)