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

Question

4895 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-06T17:22:18

$43.$

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

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

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

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

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

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

$=6+37$

$=43.$

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