# 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?

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### 1 Answer

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).**