Fundamental Theorem of Calculus

1. (4e^x-3)dx plus the number in x (6,1)

what do you mean plus the number

and
do you mean integrate from x = 1 to x = 6 or what?

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 ]

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

To evaluate the definite integral ∫[a,b] (4e^x-3) dx using the Fundamental Theorem of Calculus, we can follow these steps:

Step 1: Find the antiderivative of the function inside the integral.
Considering the function (4e^x-3), the antiderivative of 4e^x is 4e^x, and the antiderivative of -3 is -3x.

Step 2: Apply the Fundamental Theorem of Calculus.
According to the Fundamental Theorem of Calculus, if F(x) is the antiderivative of f(x), then ∫[a,b] f(x) dx = F(b) - F(a).

Step 3: Evaluate the definite integral using the formula from Step 2.
In this case, the initial value is a = 6, and the final value is b = 1.

Using the formula, the definite integral becomes:
∫[6,1] (4e^x-3) dx = (4e^1-3) - (4e^6-3)

Calculating further, we get:
= (4e - 3) - (4e^6 - 3)

Please note that this is the final answer, and the calculation cannot be simplified further without knowing the approximate values of 'e'.