Definite integration

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2010-08-08 5:23 am

(a) If f(-x) = -f(x) for all values of x, show that ∫[a~-a] f(x) dx = 0 for any constant a.
(b) Hence evaluation ∫[3~-3] │x│sin x^3 dx.

更新1:

(b) Hence evaluate ∫[3~-3] │x│sin x^3 dx

回答 (1)

螞蟻雄兵

2010-08-08 5:51 am

✔ 最佳答案

(a) If f(－x)=－f(x) for allvalues of x, show that ∫[a~－a]f(x)dx=0 for any constant a.
Sol
∫[a~0]f(x)dx
Set y=－x
dx=－dy
∫[a~0] f(x)－dx
=∫[－a~0] f(－y)－dy
=∫[－a~0][－f(y)]－dy
=∫[－a~0]f(y)dy
=－∫[0~－a] f(y) dy
=－∫[0~－a] f(x) dx
So
∫[a~－a] f(x) dx
= ∫[a~0] f(x) dx+ ∫[0~－a] f(x) dx
=-∫[0~－a] f(x) dx+∫[0~－a] f(x) dx
=0

(b) Hence evaluation ∫[3~－3] │x│Sin x^3 dx.
Sin^3 (－x)=(－Sinx)^3=－Sin^3 x
Set g(x)=｜x｜Sin^3 x
g(－x)=｜－x｜Sin^3 (－x)=－｜x｜Sin^3 (x)=－g(x)
∫[3~－3] ｜x｜Sin x^3 dx.=0

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