Question

试用定积分的几何意义计算∫(上2下0)根号下(4-x²)dx的值

Answer 1

被积函数是√(4-x²),即曲线为y=√(4-x²)
圆的方程为x²+y²=4,半径为2,圆心为(0,0)
定积分下限为0,上限为2,x截距和y截距都是2,所求是1/4圆的面积
整个圆的面积为πr²=4π
而1/4圆的面积为4π/4=π
直接解定积分亦可:
∫<0,2>√(4-x²)dx
设x=2siny,dx=cosy
当x=0,y=0,当x=2,y=π/2
=∫<0,π/2>cosy√(4-4sin²y)dy
=2∫<0,π/2>2cos²ydy
=4∫<0,π/2>[(1+cos2y)/2]dy
=2∫<0,π/2>(1+cos2y)dy
=2∫<0,π/2>dy+2∫<0,π/2>cos2ydy
=2y<0,π/2>+2*(1/2)∫<0,π/2>cos2yd(2y)
=[2*π/2]-[2*0]+sin2y<0,π/2>
=π-[sin(2*π/2)]-[sin(2*0)]
=π-0-0
=π

Answer 2

被积函数是√(4-x²),即曲线为y=√(4-x²)
圆的方程为x²+y²=4,半径为2,圆心为(0,0)
定积分下限为0,上限为2,x截距和y截距都是2,所求是1/4圆的面积
整个圆的面积为πr²=4π
而1/4圆的面积为4π/4=π
直接解定积分亦可:
∫<0,2>√(4-x²)dx
设x=2siny,dx=cosy
当x=0,y=0,当x=2,y=π/2
=∫<0,π/2>cosy√(4-4sin²y)dy
=2∫<0,π/2>2cos²ydy
=4∫<0,π/2>[(1+cos2y)/2]dy
=2∫<0,π/2>(1+cos2y)dy
=2∫<0,π/2>dy+2∫<0,π/2>cos2ydy
=2y<0,π/2>+2*(1/2)∫<0,π/2>cos2yd(2y)
=[2*π/2]-[2*0]+sin2y<0,π/2>
=π-[sin(2*π/2)]-[sin(2*0)]
=π-0-0
=π