# Given that $f(x)=(2x+1)^3$, find $\int (\lim_{h \to 0} \frac{f(x+h)-f(x)}{8h})\,dx$

I thought this was as simple as: $$\int \left (\lim_{h \to 0} \frac{f(x+h)-f(x)}{8h}\right)\,dx = \frac{1}{8}\int f'(x)\, dx=\frac{f(x)}{8} + C$$

But the answer is supposed to be:

$$\left (\frac{2x+1}{2} \right )^2 + C$$

How?

• $\frac{f(x)}{8}+C$ because it's indefinite integral. You are almost right – Jakobian Apr 23 at 11:06

For a continuously differentiable function $$f$$, we have
$$\int f'(x)dx = f(x) + C$$ i.e., you need to write the constant. The indefinite integral is always a family of functions, not just a single function. So the correct answer would be $$\frac{f(x)}{8} + C.$$
The answer to the question you wrote is certainly not $$\left (\frac{2x+1}{2} \right )^2 + C$$
$$\left (\frac{2x+1}{2} \right )^2 + C$$ is wrong. Correct is $$\frac{f(x)}{8}+C.$$