How do you find the definite integral of #int (1+3x)dx# from #[ -1,5]#?

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Answer 1 · 2016-10-06T17:22:18

#43.#

We know that, if #intf(x)dx=F(x)+C,# then,

#int_a^bf(x)dx=[F(x)]_a^b=F(b)-F(a)#.

#int_-1^5 (1+3x)dx=[x+3x^2/2]_-1^5#

#=[{5+(3/2)5^2}-{-1+(3/2)(-1)^2}]#

#=5+75/2+1-3/2#

#=6+(75-1)/2#

#=6+37#

#=43.#