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

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Answer 1 · 2018-03-31T18:30:45

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

#int_-1^1(t^(1/3)-2)dt=int_-1^1t^(1/3)dt-int_-1^1 2dt#

We can split up sums or differences when dealing with integrals.

Now, integrate each term individually and evaluate from #[-1, 1].#

#3/4t^(4/3)|_-1^1-2dt|_-1^1=(3/4*1^(4/3)-3/4*-1^(4/3))-(2(1)-2(-1))#

#1^(4/3)=root(3)(1^4)=1#
#(-1)^(4/3)=root(3)((-1)^4)=root(3)1=1#

So, we then have

#(3/4-3/4)-(2-(-2))=4#

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