Suppose that $A(x)= \int_a^x t^2 \mathrm dt $ and that $A(x)= \frac{x^3}2$. Find $a$.
I think that the only solution is $a = 0$.
Assuming this is correct, is there a general method to find number a?
In general, we are interested in determining the primitive of a function f for which constant C is null. This particular primitive seems more " basic" as others, since the others can be obtained from it adding some constant C.
For example, all primitives of $f(x)=x^2$ will be written in terms of the "basic" primitive $F(x) = \frac{x^3}3$
Not all area functions ("accumulation functions") of a function, say, once again $f(x)=x^2$, will be identical to its "basic" primitive. For example, the area function defined as $x \mapsto \int_5^x t^2 \mathrm dt$ is not identical to the “basic” primitive of $f$, that is, to $F(x)= \frac{x^3}3$. (See image below.)
Apparently, only the function $x \mapsto \int_0^x t^2\mathrm dt$ is identical to the basic primitive $F(x)$ as defined above.
Hence my question: is there a method to determine which particular area function of a function $f$ (with which domain, and in particular which number $a$ as limit of integration) is identical to the "basic/standard" primitive of $f$?
The image below shows that the area function A(x) with a=0 is identical to the " basic" primitive of f(x)=x². this is not the case with a=5.
We require $A(x) = x^3/3$ and we compute $$ A(x) = \int_a^x t^2 \,\mathrm{d}t = \left. \frac{t^3}{3} \right|_{t=a}^x = \frac{x^3}{3} - \frac{a^3}{3} \text{.} $$ Equality of these two expressions for $A(x)$ requires $-a^3/3 = 0$, which forces $a = 0$.
The general technique, demonstrated above, is to compute the definite integral, then solve the resulting equality for the value or values of $a$ that satisfy it.
Warning: The set of accumulation functions is not the full set of antiderivatives and it can be confusing to think otherwise at this time.
An example of this deficiency of accumulation functions is: find $b$ such that $B(x)$, the accumulation function of $2x$ is $B(x) = x^2 + 10$. We proceed by the generic technique: $$ \int_b^x 2t \,\mathrm{d}t = \left. \frac{2t^2}{2} \right|_{b}^x = x^2 - b^2 \text{.} $$ This gives the equation $10 = -b^2$, which has no solution (because the left-hand side is positive and the right-hand side is negative for all choices of $b$). However, $\frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{x^2}{2}+10 \right) = 2x$, so this $B$ is an example of an antiderivative of $2x$ which is not an accumulation of $2x$.
Also, beware this notion of "basic": Any identity that happens to contain a constant offset makes "basic"ness unclear. Which is more "basic": $\sec^2 \theta$ or $\tan^2 \theta + 1$. The two expressions are equal for all $\theta$ and the "${}+1$" is not optional.