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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?

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  • $\begingroup$ $\frac{f(x)}{8}+C$ because it's indefinite integral. You are almost right $\endgroup$ – Jakobian Apr 23 at 11:06
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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 $$\left (\frac{2x+1}{2} \right )^2 + C$$

Unless you made some typo (highly likely, and I advise you to double check), the answer provided is wrong.

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$\left (\frac{2x+1}{2} \right )^2 + C$ is wrong. Correct is $\frac{f(x)}{8}+C.$

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