Given the function f(x)=a*lnx+b, what are a and b if f'(1)=2 and definite integral of f(x)=7, if the limits of integration are 1 and e?

First, we'll differentiate the function to get
f'(x):

f'(x) = a/x

Now, we'll
substitute x by 1:

f'(1) = a/1

But,
from enunciation, f'(1) = 2 => a = 2.

We'll evaluate the
definite integral of f(x).

Int f(x)dx = Int (a*ln x + b)dx = Int
a*ln xdx + Int b dx

Int ln xdx = x*ln x - Int
dx

Int ln xdx = x*ln x - x + C

Int ln
xdx = x(ln x - 1) + C

We'll apply Leibniz Newton to determine the
values of definite integral:

Int ln x dx = F(e) -
F(1)

F(e) = e(ln e - 1) = 0

F(1) = 1(ln
1 - 1) = -1

F(e) - F(1) = - (-1) =
1

Int a*ln xdx = a

Int bdx = bx +
C

F(e) - F(1) = b*e - b = b(e-1)

But,
from enunciation, Int (a*ln x + b)dx = 7

a + b(e-1) =
7

2 +b(e-1) = 7

b(e-1) =
5

b = 5/(e-1)

We'll substitute a and b
and we'll get:

f(x) = 2ln x +
5/(e-1)

f(x) = ln (x^2) +
5/(e-1)

The requested function is f(x) = ln (x^2) +
5/(e-1).