# If lnx is the exponent we raise e to--- how is it the area under the curve of 1/x of the formular of lnx = 1/x?

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The area under the curve 1/x, is the definite integral of f(x) - another curve or line and between the limits x = a and x = b.

Since there are not specified the limits x = a and x = b, also it is not specified the other curve or line, we'll calculate the indefinite integral of 1/x and not the area under the curve.

The indefinite integral of f(x) = 1/x is:

Int f(x) = Int dx/x

Int dx/x = ln x + C

C - family of constants.

To understand the family of constants C, we'll consider the result of the indefinite integral as the function f(x).

f(x) = ln x + C

We'll differentiate f(x):

f'(x) = (ln x + C)'

f'(x) = 1/x + 0

Since C is a constant, the derivative of a constant is cancelling.

So, C could be any constant, for differentiating f(x), the constant will be zero.

Now, we'll calculate the area located between the curve 1/x, x axis, x=a and x=b:

Integral [f(x) - ox]dx, x = a to x = b

Int f(x) dx = F(b) - F(a), using Leibniz-Newton formula

Int dx/x = ln b - ln a

Since the logarithms have matching bases, we'll transform the difference into a product:

Int dx/x = ln |b/a|

**The area located between the curve 1/x, x axis, x=a and x=b is:**

**Int dx/x = ln |b/a|**

ln x is the exponent raised to what power of e.

Answer is lnx = e^lnx.

By definition of logarithms, if e^y = x, then y = lnx. Alternatively, if y = lnx then e^y = x.

Area of the curve under f(x) is Integral f(x) dx.

So Inegral (1/x)dx = lnx.

So area under f(x) = Area under (1/x) = lnx