If f'(x) = 8x^3 + 3x^2 + 2 , find f(x) if f(0) = -5.

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We are given that f'(x) = 8x^3 + 3x^2 + 2. To find f(x) we need to find the integral of f'(x).

Int[f'(x)] = Int [ 8x^3 + 3x^2 + 2]

=> 8x^4 / 4 + 3x^3 / 3 + 2x + C

=> 2x^4 + x^3 + 2x +...

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We are given that f'(x) = 8x^3 + 3x^2 + 2. To find f(x) we need to find the integral of f'(x).

Int[f'(x)] = Int [ 8x^3 + 3x^2 + 2]

=> 8x^4 / 4 + 3x^3 / 3 + 2x + C

=> 2x^4 + x^3 + 2x + C

Also, f(0) = -5

2*0^4 + 0^3 + 2*0 + C = -5

=> C = -5

Therefore f(x) = 2x^4 + x^3 + 2x - 5

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f'(x) = 8x^3 + 3x^2 + 2

We know that f(x) = integral of f'(x).

==> f(x) = intg (8x^3 + 3x^2 + 2) dx

              = intg 8x^3 dx + intg 3x^2 dx +  intg 2 dx

              = 8x^4/4  + 3x^3/3 + 2x + C

==> f(x) = 2x^4 + x^3 + 2x + C

But we are given that f(0) = -5

==> f(0) = 0+ C = -5

==> C = -5

==> f(x) = 2x^4 + x^3 + 2x -5

 

Approved by eNotes Editorial Team