# If f(x) = 5x^4 - 4x - 3/x. F(x) = integral f(x), find F(1)= 0 find f(x)

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f(x) = 5x^4 - 4x - 3/x

Let us integrate f(x) :

F(x) = intg f(x)

= intg (5x^4 - 4x - 3/x) dx

= ing 5x^4 dx - intg 4x dx - intg 3/x dx

= 5x^5/5 - 4x^2/2 - 3 ln x + C

= x^5 - 2x^2 - 3lnx +C

==> F(x) = x^5 - 2x^2 - 3lnx + C

But F(1) = 0

==> F(1) = 1 - 2 - 3ln1 + C = 0

==> -1 + C = 0

==> C = 1

**==> F(x) = x^5 - 2x^2 - 3ln + 1**

f(x) = 5x^4-4x-3/x.

To find F(1) = 0

Solution:

F(x) = Intf(x) dx = Int {5x^4-4x-3/x} dx.

We use Int x^n dx = (x^(n+1))/n. And Int dx/x = lnx

F(x) = (5x^5)/5 -(4x^2)/2 -3lnx + C, where C is constant

F(x) = x^5 -2x^2 -3lnx + 1.

Put x =1

Put F(1) = 1^5 -2*1^2 - 3ln(1) + C

0 = 1-2-3*0 +C, as F(1) = 0 by data . ln(1) = 0.

0 = -1+C

C= 1.

Therefore F(x) = x^5-2x^2-3/nx