Question: Given a = 0 . Letthepolynomialbe f ( t ) = 3 t + 2 and b = 2 , thenfromthelimitdefinitionofadefiniteintegral,wehave a b f (

Given

a=0.

Letthepolynomialbef(t)=3t+2andb=2,

thenfromthelimitdefinitionofadefiniteintegral,wehave

abf(t)=limn(nba)i=1nf(a+nbai)

02(3t+2)=limn(n20)i=1nf(0+n20i)

=limn(n2)i=1nf(n2i)

=limn(n2)i=1n(3(n2i)+2)

=limn(n2)i=1n(n6i+2)

=limn(n2)[n6i=1ni+i=1n2]

=limn(n2)[n6(2n(n+1))+2n]

=limn(n2)[3(n+1)+2n]

=limn(n2)[5n+3]

=limn(10+n6)

=limn10+limnn6

=10+0

=10

The answer to this is incorrect. Can someone show me the right result? Thanks

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