`f''(x) = 8x^3 + 5, f(1) = 0, f'(1) = 8` Find `f`.

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`f''(x)=8x^3+5`

`f'(x)=int(8x^3+5)dx`

`f'(x)=8(x^4/4)+5x+c_1`

`f'(x)=2x^4+5x+c_1`

Now let's find constant c_1 , given f'(1)=8

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scisser | Student

Find the anti derivatives and solve for the constants.

`int(8x^3+5)=f'(x)=2x^4+5x+C`

`f'(1)=8` , therefore,

`2(1)^4+5(1)+C=8`

`C=1`

Thus, `f'(x)=2x^4+5x+1`

Do the same thing again

`int(2x^4+5x+1)=f(x)=(2/5)x^5+(5/2)x^2+x+C`

Plug in the point` f(1)=0`

`0=(2/5)(1)^5+(5/2)(1)^2+(1)+C`

`C=-39/10`

Thus,

`f(x)=2/5x^5+5/2x^2+x-39/10`

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