What Does an Unknown Endpoint Mean?
An accumulation condition can specify a starting point and a total signed change, while leaving the endpoint unknown. For example, an equation such as \(\int_0^x f(t)\,dt=K\) asks: At which endpoint or endpoints has the signed accumulation from \(0\) reached the value \(K\)?
The answer is not always a single number. The accumulated value may reach the same target at more than one endpoint, or it may never reach it. A reliable solution uses the signed area represented by the integral, solves for possible endpoints, and then checks each candidate in the original condition.
How the Endpoint Determines Accumulation
For a continuous integrand, the Fundamental Theorem of Calculus gives the accumulation function \(A(x)=\int_a^x f(t)\,dt\), with \(A'(x)=f(x)\). To solve an endpoint condition, we need the value of \(A(x)\), which can often be found by signed area or by evaluating the definite integral.
The integral also has an orientation. If the endpoint is less than the starting point, reversing the limits changes the sign:
This sign is part of the accumulation condition, not a convention that can be ignored. For example, when \(f\) is positive between \(x\) and \(a\), the integral from \(a\) back to \(x\) is negative.
A Derivation: Splitting at an Intermediate Point
When an integrand changes its formula or sign between the starting point and the unknown endpoint, split the accumulation at the relevant point. For \(a\leq c\leq x\), additivity of the definite integral gives:
This identity follows by writing each definite integral as a difference of values of an accumulation function. If \(F\) is an antiderivative of \(f\), then:
The intermediate value \(F(c)\) cancels, so splitting does not change the total; it simply makes each part easier to calculate or interpret. If \(x<c\), the relevant split and interval order must instead match that endpoint. In every case, the split points and interval conditions matter when the integrand is piecewise defined.
Worked Example: Two Endpoints for One Accumulation Value
Worked Example: Two Endpoints for One Accumulation Value
Find all real \(x\) satisfying \(\int_0^x (t-2)\,dt=-\frac{3}{2}\).
Use the Fundamental Theorem of Calculus. An antiderivative of \(t-2\) is \(\frac{t^2}{2}-2t\), so the condition becomes:
Multiply both sides by \(2\):
Move all terms to the left and factor:
The candidates are \(x=1\) and \(x=3\). Check each in the original integral:
Both endpoints work. In terms of signed area, the integrand is negative from \(0\) to \(2\) and positive after \(2\). By \(x=1\), the accumulated value is \(-\frac{3}{2}\); by \(x=3\), the positive area from \(2\) to \(3\) has brought the total back to that same value.
Piecewise Integrands Require Interval Checks
If the integrand has different formulas on different intervals, first identify which formula applies at the candidate endpoint. An equation derived from one piece is valid only when its endpoint lies in that piece’s interval. Solving the equations without checking those intervals can produce answers that do not satisfy the original condition.
Worked Example: Splitting at a Change in the Integrand
For \(x\geq 0\), let \(f(t)=2\) for \(0\leq t\leq 2\), and \(f(t)=-1\) for \(t>2\). Find all \(x\) satisfying \(\int_0^x f(t)\,dt=3\).
First consider \(0\leq x\leq 2\). Throughout this interval, \(f(t)=2\), so:
Set this equal to \(3\): \(2x=3\), giving \(x=\frac{3}{2}\). This candidate is in the interval \(0\leq x\leq 2\), and checking gives \(\int_0^{3/2}f(t)\,dt=2\left(\frac{3}{2}\right)=3\).
Now consider \(x>2\). Split at \(t=2\):
Set this expression equal to \(3\): \(6-x=3\), so \(x=3\). It satisfies \(x>2\). Check in the original integral:
Thus the solutions for \(x\geq0\) are \(x=\frac{3}{2}\) and \(x=3\). The interval checks establish that each equation was used only where its integrand formula applies.
Reversed Bounds and a Negative Target
An unknown endpoint can lie below the starting point. In that case, the reversal of limits determines the sign. A useful check is to rewrite the integral in the forward direction first, then attach the negative sign. Do not simply treat the region between the endpoints as positive area.
Worked Example: An Endpoint Below the Starting Point
Find the endpoint \(x\) satisfying \(\int_2^x (t-1)\,dt=-\frac{1}{2}\).
An antiderivative of \(t-1\) is \(\frac{t^2}{2}-t\). Evaluate the definite integral with its limits in the given order:
Set this equal to \(-\frac{1}{2}\) and multiply by \(2\):
The candidate is \(x=1\), which is below the starting point \(2\). Check it in the original condition:
The endpoint \(x=1\) satisfies the condition. The minus sign appears because the limits are reversed, even though \(t-1\) is positive between \(1\) and \(2\).
A Reliable Solving Process
Each problem may call for an area calculation or an evaluation of the definite integral. The reasoning is the same: translate the condition into a signed-accumulation equation, solve, and verify.
Write down the fixed lower limit, the unknown endpoint, and the required accumulated value.
Use the graph or evaluate the definite integral, respecting the order of the bounds and any changes in the integrand.
Keep every real solution provisionally; one accumulation value may occur at multiple endpoints.
Reject candidates outside the interval used to derive an equation, then substitute each remaining endpoint into the original integral condition.
Common Pitfall: Treating Signed Area as Positive Area
A frequent error is to use the geometric area between the graph and the axis without accounting for its sign. For instance, in the last example, the region from \(1\) to \(2\) has geometric area \(\frac{1}{2}\), but the requested integral runs from \(2\) to \(1\). Writing \(\int_2^1(t-1)\,dt=+\frac{1}{2}\) would therefore give the wrong sign. The correction is:
A second error is to stop after solving the algebraic equation. A candidate must also satisfy any interval condition used in a piecewise calculation and the original integral equation. Conversely, do not assume the endpoint must be unique: as the first example shows, distinct endpoints can yield the same accumulated value.
Why Endpoint Checks Matter
Accumulation conditions connect a rate function to a total change over an interval. Finding an unknown endpoint can identify when a specified net change has occurred. The signed interpretation is essential: increases and decreases can offset one another, so the total is not necessarily the sum of the magnitudes of all changes.
For AP problems, a complete answer should give the endpoint or endpoints and enough work to justify them. In particular, state how the integral was split or evaluated, respect the direction of the bounds, check interval restrictions, and verify the final candidates in the original condition.
Check Your Understanding
For each question, pay attention to the direction of integration and to whether a candidate fits the interval used in your calculation.
- Find all \(x\) satisfying \(\int_0^x(t-1)\,dt=0\). Check each candidate in the original condition.
- If \(f(t)=-3\) on \(0\leq t\leq 2\), what endpoint \(x\) in that interval satisfies \(\int_0^x f(t)\,dt=-6\)?
- Explain why \(\int_4^1 f(t)\,dt=-\int_1^4 f(t)\,dt\).
- For a piecewise integrand that changes formula at \(t=2\), why should a solution found using the formula for \(t>2\) be checked against \(x>2\)?
- Can an integral condition have two different valid endpoints? Explain using the signed-accumulation idea.
