The curve is y = 1/x.

To find the area under the curve.

Solution:

The cuve is in 1st and 3rd quadrants.

y = 1/x does not exist at x = 0.

So we have to integrate between a and b. We consider only for positive values of x between the ordinates at x= a and x =b. b>a > 0.

y = 1/x

Area under y (from x = a to x = b) = Int y dx.

Let F(x) = Int (1/x) = logx.

F(x) = log(x) +C.

Area under y (from x = a to x = ) = F(b) -F(a)

F(b) -F(a) = log b - log a. The constant C gets cancelled.

F(b) - F(a) = log (b/a) , as log x - log y = log(x/y)

Therefore area under the curve y = 1/x from x= a to x = b is log(b/a).

The area under the curve 1/x, is the definite integral of f(x) minus INtegral of 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

We'll apply Leibniz-Newton formula:

**Int f(x) dx = F(b) - F(a)**

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|**