# Which is the value of area bounded by the graph of the function f(x)=2x/x^2,the ox axis and lines x=1, x=3.

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To find the area bounded by the function f(x), x=1 and x =2. we need to fing intg f(x) and then subrtact intg f(1) from intg f(2):

f(x) = 2x/x^2 = 2/x.

==> intg f(x) dx = intg (2/x) dx = 2ln x

Now the area is

intg f(3) - intg f(1)

= 2ln3-2ln1 = 2ln3 - 2(0) = 2ln3

To find out the value of the area bounded by the graph of the function f(x), 0x axis and the lines x=1 and x=3, we just have to calculate the definite integral of f(x).

Integral f(x)dx=Integral [(2x)/x^2]dx=

= Integral (2xdx/x^2)

To calculate Integral (2xdx/x^2), we'll apply the substitution method.

We'll substitute x^2=t. Differentiating both sides, we'll get 2xdx=dt.

**Int(2xdx/x^2)=Int dt/t=ln t=ln x^2=ln(3^2) - ln1=2 ln 3= ln9**

To find the area bounded by f(x) = 2x/x^2, x axis and x=1 and x =2.

Solution:

f(x) = 2x/x^2 = 2/x.

The area is given by the inyegral f(x) dx for x = 1 to x =3.

=Int(2/x)dx for x = 1 to x =3.

= [(2lnx)at x=3]- [(2lnx) at x=1]

=2ln3-2ln1

= 2ln3, as ln1 =0

=ln3^2

=ln9