# Calculate the definite integral of f(x)=(3x^2-1)/x^3, ox axis, lines x=1, x=2?

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intg f(x)=Intg[(3x^2-1)/x^3]dx=

==> Intg (3x^2/x^3)dx - Intg (1/x^3)dx

==> Intg (3x^2/x^3) dx - Intg [x^(-3)]dx

Now assume that u= x^3 ==> du = 3x^2du

==> Intg (3x^2/x^3)dx= intg du/u

but, u = x^3

==> ln u= ln x^3

now for the lines x=1 and x=2, we subtract functions value when x=1 from the function value when x =2

==> ln(2^2) - ln1= 2ln2-0 = ln4

Now for Intg [x^-3)dx ,

==> Integral [x^(-3)]dx = [x^(-3+1)]/(-3+1) = x^(-2)/-2

for x=2 and x=1,

==> -1/2(2)^2 + 1/2(1)^2= -1/8 + 1/2 = 3/8

Then intg f(x) = ln4 - 3/8

To calculate the area bouded by f(x) = (3x^2-1)/x^2 , x axis and the lines x =1 and x =2.

Solution: The area between bounded by f(x), x axis and the lines x=1 and x = 2 is given by:

Area = Int f(x) dx for x =1 to x =2

=Int [(3x^2-1)/x^3]dx for x =1 to x =2.

= {Int (3x^2/x^3 -1/x^3) dx for x =1 to x=2}

= (int 3 *dx/x - int dx/x^3) for x =1 to x=2

={[3lnx -(- 1/(2x^2))] x =2} - ={[3lnx -(- 1/(2x^2)] x =1}

= {[3ln2+1/4]- {[3*0+1/2} = 3ln2 - 3/2 = **ln8-1/4**

Integral f(x)dx=Integral [(3x^2-1)/x^3]dx=

= Integral (3x^2dx/x^3) - Integral (dx/x^3)=

= Integral (3x^2dx/x^3) - Integral [x^(-3)]dx

To calculate Integral (3x^2dx/x^2), we'll apply the substitution method.

we'll note x^3=t. Differentiating both sides, we'll get 3x^2dx=dt.

Int(3x^2dx/x^3)=Int dt/t=ln t=ln x^3=ln(2^2) - ln1=2ln2=ln4

Integral [x^(-3)]dx = [x^(-3+1)]/(-3+1) = x^(-2)/-2 = -1/2x^2= -1/8 + 1/2 = 3/8

**Integral [(3x^2-1)/x^3]dx = ln4 - 3/8 **