Given information:

The function is given as:

  f''(x)=8x3+5 andconditions f'(1)=8,f(1)=0 .

Formula used:

The function F is antiderivative of f(x) on an interval I if F`(x)=f(x) for all x in I

The function F is known as antiderivative of f(x) on an interval I the most general antiderivative of the function on interval I is F(x)+C , where C is an arbitrary constant.

Calculation:

Consider the function,

  f''(x)=8x3+5

The antiderivative of f(x) is,

  F`(x)=f(x)

So, by reverse power rule Integrating both sides of equation,

  ∫F`(x)=∫f(x)F(x)=∫f(x)

Therefore, integrating the function

  f''(x)=8x3+5∫f''(x)dx=∫8x3+5dxf'(x)=8x44+5x+C=2x4+5x+C

Again, integrate the function,

  ∫f'(x)dx=∫2x4+5x+Cdxf(x)=2x55+5x22+Cx+D

Put f'(1)=8 in the equation f'(x)=2x4+5x+C ,

  f'(1)=2⋅(1)4+5(1)+C8−7=CC=1

Put f(1)=0 in the equation f(x)=2x55+5x22+Cx+D ,

  f(1)=2155+5122+C1+D0=25+52+C+DC+D=2910

Put C=1 in the equation,

  D=1910

Put the values of C and D in equation f(x)=2x55+5x22+Cx+D

  f(x)=2x55+5x22+1x+1910

Thus, the particular solution is f(x)=2x55+5x22+1x+1910 .