How do you evaluate the definite integral #int 3x^3-x^2+x-1# from #[-1,1]#?

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Answer 1 · 2016-09-23T10:14:07

#int_(-1)^1(3x^3-x^2+x-1)dx=-8/3#

#int(3x^3-x^2+x-1)dx#

= #3xx x^4/4-x^3/3+x^2/2-x+c#

= #3x^4/4-x^3/3+x^2/2-x+c#

Hence the value of definite integral from #[-1,1]# is

#int_(-1)^1(3x^3-x^2+x-1)dx#

= #[3x^4/4-x^3/3+x^2/2-x+c}_(-1)^1#

= #(3xx1^4/4-1^3/3+1^2/2-1+c)-(3(-1)^4/4-(-1)^3/3+(-1)^2/2-(-1)+c)#

= #(3/4-1/3+1/2-1+c)-(3/4+1/3+1/2+1+c)#

= #3/4-1/3+1/2-1+c-3/4-1/3-1/2-1-c#

= #-2/3-2=-8/3#