Evaluate the following integral.
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F.5 Maths
2011-03-21 7:08 am
回答 (2)
2011-03-21 8:08 am
✔ 最佳答案
Let x = tan u, dx=sec^2 u duS [x/V(x^2+1)] dx
= S (tan u sec^2 u du) / V(tan^2 u + 1)
= S (tan u sec^2 u du) / V(sec^2 u)
= S (tan u sec^2 u du) / sec u
= S (tan u sec u du)
= sec x +C
= V(x^2+1) +C
S (V3, -V3) [x/V(x^2+1)] dx
=V(3^2+1) +C - V[(-3)^2+1] +C
= 0
2011-03-21 7:34 am
Fairly speaking, since f(x) = x/√(x^2 + 1) is an odd function, i.e f(-x) = -f(x). The integral of ∫ (-b -> b) f(x) dx (where b is a constant) should equal to 0 and we don't need to do any calculation further.
收錄日期: 2021-04-26 14:06:35
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