Question 819251
<pre>
*[tex \large f''(x)\ =\ 8x^3\ +\ 5]
*[tex \large f'(x)\ =\ \int\,\left(8x^3\ -\ 5\right)\,dx\ =\ 2x^4\ -\ 5x\ +\ C_1]
But
*[tex \large f'(0)\ =\ 4\ \Rightarrow\ C_1\ =\ 4]
Then
*[tex \large f(x)\ =\ \int\,\left(2x^4\ -\ 5x\ +\ 4\right)\,dx\ =\ \frac{2}{5}x^5\ -\ \frac{5}{2}x^2\ +\ 4x\ +\ C]
But
*[tex \large f(1)\ =\ 0\ \Right\ C\ =\ -\frac{2}{5}\ +\ \frac{5}{2}\ -\ 4]
I should think you can do your own arithmetic from here.