log problem - how do you switch fractions into log? (1 Viewer)

[Fiction]

For the questions:
c. Given y = 2xe^(2x)

i) Find dy/dx 2

ii) Hence evaluate int 4 xe^(2x).dx 3
(give your answer in exact form)

in ii), the answers go as the following:




I can follow until the lines indicated by the question mark. I don't understand how the answer got from int 2e^(2x) to e^(2x) (right hand side on transition from 4line to 5th line)

In my answer, I did int 23^(2x) + int 4xe^(2x) = 2xe^(2x) + c
= 4e ^(2x) +D + int 4xe^(2x) = 2xe^(2x)+C

2
Int 4xe^(2x) =2xe^(2x) - 4e^(2x)
0
= [2xe^(2x) - 4e^(2x) ] with limits 2 and 0
= 4e^4 - 4e^4 -0-4e^0
= -4

 

[InteGrand]

For the questions:
c. Given y = 2xe^(2x)

i) Find dy/dx 2

ii) Hence evaluate int 4 xe^(2x).dx 3
(give your answer in exact form)

in ii), the answers go as the following:


View attachment 31962

I can follow until the lines indicated by the question mark. I don't understand how the answer got from int 2e^(2x) to e^(2x) (right hand side on transition from 4line to 5th line)

In my answer, I did int 23^(2x) + int 4xe^(2x) = 2xe^(2x) + c
= 4e ^(2x) +D + int 4xe^(2x) = 2xe^(2x)+C

2
Int 4xe^(2x) =2xe^(2x) - 4e^(2x)
0
= [2xe^(2x) - 4e^(2x) ] with limits 2 and 0
= 4e^4 - 4e^4 -0-4e^0
= -4

To go from line 4 to line 5 in the solutions, they just computed the integral in line 4: (the +C unimportant because a definite integral was being evaluated).

And in your answer, you accidentally wrote the derivative of 2e^2x rather than its primitive.