A definite integral on the interval $(a,b) =$ the antiderivative at $b -$ the antiderivative at $a$. Am I correct in saying that the antiderivative $=$ the original function? Because the integral is the derivative of the original function, and the answer to the integral is the original function... right?
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Drive slow, homie.
The definite integral of $f(x)$ on $(a,b)$, written $\int_a^b f(x) \,dx$, is a number. This number is the (signed) area under the graph of $f(x)$ along $(a,b)$.
An antiderivative is a function $F(x)$ with the property that $F'(x) = f(x)$.
The Fundamental Theorem of Calculus tells us that we can compute definite integrals using antiderivatives, i.e. $\int_a^b f(x) \, dx = F(b) - F(a)$.
Well, I can't tell what you are referring to by "original" function, but if you are asking if $\int f(x)\,dx = f(x)$, of course that is not true.
Just try one or two of the simplest examples (such as $\int x\,dx$ or $\int x^2\,dx$ or $\int \sin x\,dx$, etc.) you can think of and you will see this is not true.
So the fundamental theorem of calculus says that $$ \int_a^b f'(x) dx = f(b) - f(a).$$
Can you think of any two functions which have the same derivative? If you could, then you would know that the antiderivative isn't necessarily the original function. Suppose $f$ and $g$ have the same derivative. Then by linearity, the function $f-g$ must have a derivative that is identically zero. What functions have derivatives which are identically zero?