How do you evaluate the definite integral #int (4t^3-2t)dt# from [-1,1]?

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Answer 1 · 2017-01-30T17:40:22

#0.#

At a glance, we can answer #0#, if we use the following Result :

If,#f:[-a,a] to RR" is continuous odd function, "int_(-a)^a f(x)dx=0.#

Otherwise,

#int_(-1)^1 (4t^3-2t)dt=2[(2t^4)/4-t^2/2]_-1^1#

#=2[(2/4-1/2)-(2/4-1/2)]#

#=0#.

Enjoy Maths.!