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

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

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

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

$= 2$

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