what do you mean plus the number
and
do you mean integrate from x = 1 to x = 6 or what?
Fundamental Theorem of Calculus
1. (4e^x-3)dx plus the number in x (6,1)
4 answers
yes it is mean integrate the x = 1 and x = 6
ok
first split it
4 e^x dx - 3 dx
4 e^x - 3 x
evaluate at 6 and subtract the value at x = 1
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or perhaps you mean
4 e^(x-3) dx
then you have
4 e^(x-3) = 4 e^-3 e^x
so integrate
(4 e^-3) integral e^x dx
which is just
4 e^-3 e^x
or
4 e^-3 [ e^6 - e1 ]
first split it
4 e^x dx - 3 dx
4 e^x - 3 x
evaluate at 6 and subtract the value at x = 1
-------------------------
or perhaps you mean
4 e^(x-3) dx
then you have
4 e^(x-3) = 4 e^-3 e^x
so integrate
(4 e^-3) integral e^x dx
which is just
4 e^-3 e^x
or
4 e^-3 [ e^6 - e1 ]
The FTC says that the value is F(6)-F(1)
where dF/dx = 4e^x-3
F(x) = 4e^x - 3x + c
F(6) = 4e^6 - 18 + c
F(1) = 4e^1 - 3 + c
So, ∫[1,6]4e^x-3 dx = 4e^6-4e-15
where dF/dx = 4e^x-3
F(x) = 4e^x - 3x + c
F(6) = 4e^6 - 18 + c
F(1) = 4e^1 - 3 + c
So, ∫[1,6]4e^x-3 dx = 4e^6-4e-15