# Determine the area of the surface between the lines x=0,x=1 and the curve f(x)=(3x^2 + 3)/(x^3 + 3x).

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### 3 Answers

f(x) = (3x^2 + 3)/(x^3 + 3x)

x= 0 and x= 1

We need to calculate the area between f(x) and x= 0 and x= 1

We know that the area under f(x) is integral f(x) = F(x)

F(x) = intg f(x)

= intg (3x^2+ 3)/(x^3 + 3x)

Let t = x^3 + 3x

==> dt = (3x^2 + 3x )dx

==> intg (3x^2 + 3)x / (x^3 + 3x) dx

==> ln (x^3 + 3x)

Now we need to find the area between x=0 and x= 1

==> A = F(1) - F(0)

= ln 4 - ln 0

A = ln 4 - inf.

Then the area is not bounded.

f(x) = (3x^2+3)/(x^3+3x) . To find the area under the curve between the ordinates at x= 0 and x =1.

Solution:

We know that the area A of the under a curve f(x) beween x = a and x = b is guven by:

A = Integral f(x) dx from x= a to x = b.

Given f(x) = (3x^2+3)/(x^3+3x). So,

A = Integral {(3x^2+3)dx)(x^3+3x) ,from x = 0 to x= 1}

We shall have a transformation x^3+3x = t. Then by diffrentiation,( 3x^2+3)dx = dt. When x =0, t = 0^3+3*0 = 0 and when x= 1, t = 1^3+3*1 = 4. So the are under the curve becomes:

A = Integral (dt/t , t= 0 to t = 4}

A={ logt at 4} -{ logt at t= 0}

A = log4 - log (0)

A = log(4/0) Or

A = log4 - (- infinity)

A = unbounded.

So the area unbouded and is not finite quantity.

The area which has to be determined is bounded by the given curve f(x), the lines x = 0 and x = 1, also the x axis.

To calculate the area, we'll use the formula:

S = Integral (f(x) - ox)dx=Int f(x)dx = Int(3x^2 + 3)dx/(x^3 + 3x).

We'll calculate the integral, using substitution technique.

We'll note x^3 + 3x = t.

We'll differentiate x^3 + 3x.

(x^3 + 3x)' = t'

(3x^2+3)dx = dt

We notice that the result of differentiating the function is the numerator of the function.

We'll re-write the integral:

Int dt/t = ln t = ln (x^3 + 3x) + C

Now, we'll calculate the value of the area, using Leibnitz Newton formula::

S = F(1) - F(0), where

F(1) = ln (1^3 + 3*1) = ln 4

F(0) = ln (0^3 + 3*0) = ln 0 impossible!

S = ln 4 - infinite, impossible!