(1)计算定积分∫(1/√2~1)√(1-x^2)/x^2 *dx (2)计算定积分∫(-3~0)√(1-x)*dx 求详细过程谢
人气:171 ℃ 时间:2020-05-22 06:14:52
解答
∫(1/√2->1)√(1-x^2)/x^2 *dx
= -[1/x+x](1/√2->1)
= √2 +1/√2 - 2
= (3/2)√2 -2
∫(-3->0)√(1-x)*dx
= - (2/3) [(1-x)^(3/2)] (-3->0)
=-(2/3)( 1-8)
=7/3=-(2/3)( 1-8)=7/3是不是应该等于14/3啊?是14/3
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