integration of (0.6x+1.35)dx, the value of x varies from 0 to 6

Question

It is a integration problem

Answer 1

integral(0.6x+1.35dx)=0.3x^2 +1.35x... evaluate at x=0 giv 0... at x=6, yu get 0.3*36+6*1.35... =10.8 +8.1... =18.9

Answer 2

integral(.6x+1.35)dx=0.3x^2+1.35x+c

integral from 0 to 6

x=0 is lower limit x=6 upper limt

=[0..3(6^2)+1.35(6)-0.3(0)^2+1.35(0)]

=[18.9-0]

=18.9

Answer 3

Given; ∫(0.6x + 1.35)dx

from the definition of integration

∫x^n dx = (1/{n+1}x^{n+1})

therefore

∫(0.6x + 1.35)dx = 0.3(x^2) + 1.35x

introducing the upper limit                                                                             introducing the lower limit

= 0.3(6)^2 + 1.35(6)                                                                                    =0.6(0) + 1.35(0)

= 10.8 + 8.1                                                                                             = 0

=18.9

therefore

∫(0.6x + 1.35)dx = 11.7 - 0 = 18.9