AreaFind the the area between x axis and the curve x^-1.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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The first step is to re-write the given curve y = x^-1, using the property of negative power.

x^-1 = 1/x

The area under the curve 1/x, is the definite integral of y 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, we'll calculate the indefinite integral of 1/x and not the area under the curve.

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

Int f(x) dx = 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 and we'll consider the limit lines 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|

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