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Just like the integration of the Taylor series of $\,\sin x\,$ gives $\,\cos x +C. \,$ I did the same with the general taylor series. The integration of the taylor series of $f(x)$ around x=a is: $$f(a)x+\frac{(x-a)^2}{2}f^{'}(a)+......$$ which gives $a*f(a)$ at x=a.

I integrated the general Taylor series around $x=a$ of any function and evaluated it at $x=a.$ That gives $a\cdot f(a),$

which is absolutely false because it would mean that the value of the integral of any function $f(x)$ at $x=a$ is $a\cdot f(a)$.

And, it means that the definite integral of any function from $a$ to $b$ is $bf(b)-af(a)$ which is false.

So, what did I do wrong?

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    What exactly did you do? You have to be careful about manipulating infinite series because you can only rearrange them if they're absolutely convergent. – 2017-02-20
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    @lordoftheshadows I didn't understand that. Did you mean that I can use the integral of a taylor series at $x=a$ only to evaluate the integral at x not equal to a? – 2017-02-20
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    Please show you integration of "the general Taylor series" and then evaluating at $x=a$. What you state is not very clear at all. – 2017-02-20
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    @amWhy I've added it. – 2017-02-20
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    Thanks for your responsiveness, @Dove! – 2017-02-20
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    Note that the integration of the series also requires a constant $C$. The primitive of a function does NOT define an unique function. – 2017-02-20
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    @MiguelAtencia What about the definite integral? That would cancel the constant. Doesn't what I've done here mean that the definite integral of any function from a to b is bf(b)-af(a)? – 2017-02-20
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    @Dove Nope, because the parentheses in the next terms do not cancel at $x=b$. The point is that, without additional data, you will never be able to calculate the value of the integral: it is (more or less hidden) part of the statement of the problem. – 2017-02-20

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You forgot the integration constant, which makes

$$F(a)=f(a)a+C$$ which is perfectly valid as it can take any value.

And when performing the definite integration, you assumed that $F(b)=f(b)b$, which is wrong.

The true relation is

$$F(b)-F(a)=\left.f(a)x+f'(a)\frac{(x-a)^2}2+f''(a)\frac{(x-a)^3}3+\cdots\right|_a^b\\ =f(a)(b-a)+f''(a)\frac{(b-a)^2}2+f''(a)\frac{(b-a)^3}3+\cdots$$

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    No. I took the taylor series around x=b to get F(b)=f(b)b+C and subtracted f(a)a+C from it. – 2017-02-20
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    @Dove: don't be bad faith, there is no mention of $C$ in your post. And you may not mix a development around $a$ with one around $b$. – 2017-02-20