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

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Answer 1 · 2018-03-12T01:36:29

#2#

Given: #int(2x-3)dx# from #[1,3]#

#int_1^3(2x-3)dx =int_1^3(2x)dx - int_1^3 3 dx #

Use: #int(kx^n)dx = k*1/(n+1)x^(n+1) + C#

#int_1^3(2x)dx - int_1^3 3 dx = 2*1/2 x^2 - 3x |_1^3#

# = x^2 - 3x |_1^3 = (3^2 - 3*3) - (1^2 - 3*1)#

#= (9-9) - (1-3) = 0 -(-2)#

#= 2#