# Double Antiderivation problem

I have to find $f(x)$ given $f''(x)$ and certain initial conditions.

$$f"(x) = 8x^3 + 5$$ and $f(1) = 0$ and $f'(1) = 8$

$$f'(x) = 8 \cdot \frac{x^4}{4} + 5x + C = 2x^4 + 5x + C$$

Since $f'(1) = 8 \Rightarrow 2 + 5 + C = 8$ so $C = 1$

$$f'(x) = 2x^4 + 5x + 1$$

$$f(x) = 2 \cdot \frac{x^5}{5} + 5 \cdot \frac{x^2}{2} + X = \frac{2}{5} x^5 + \frac{5}{2} x^2 + X + D$$

Since $f(1) = 0$ then:

$$\frac{2}{5} + \frac{5}{2} + 1 + D = 0$$ $$\frac{29}{10} + 1 + D = 0$$

So $D = \dfrac{-39}{10}$

So $$f(x) = \frac{2}{5} \cdot x^5 + \frac{5}{2} \cdot x^2 + x - \frac{39}{10}$$

Does that look right?

• Yes it's correct. An alternative way to do it is to work with definite integrals and write $f'(x) - f'(1) = \int_1^x (8t^3+5)\,{\rm d}t$ and $f(x)-f(1) = \int_1^x f'(t)\,{\rm d}t$. This avoids solving for the integration constants, but it's pretty much the same thing. Mar 12, 2018 at 17:32
• Yes, you can also check it by yourself Mar 12, 2018 at 17:33

$$f(x) = \frac{2}{5} \cdot x^5 + \frac{5}{2} \cdot x^2 + x - \frac{39}{10}\implies f'(x)= 2x^4+5x+1\implies f''(x)=8x^3+5$$
$$f(1)=\frac{2}{5}+ \frac{5}{2} + 1 - \frac{39}{10}=\frac{4+25+10-39}{10}=0$$
$$f'(1)=2+5+1=8$$