Introduction
The Fundamental Theorem of Calculus connects accumulated change with rates of change. Its formulas are compact, but small errors can change a correct result into an incorrect one: reversing bounds changes the sign, substituting endpoints carelessly changes the value, and a variable endpoint requires the chain rule.
A reliable solution begins by identifying what the expression represents. A definite integral measures signed accumulation across an interval. The integrand describes the rate or quantity being accumulated, while an accumulation function records the integral up to a specified endpoint. These related quantities are not generally equal.
Two Forms of the Fundamental Theorem
For evaluating a definite integral, the Fundamental Theorem of Calculus says that if \(f\) is continuous on \([a,b]\) and \(F'(x)=f(x)\) throughout that interval, then the integral is the upper-endpoint value of \(F\) minus its lower-endpoint value.
For an accumulation function with a variable upper endpoint, the Fundamental Theorem gives a derivative. If \(f\) is continuous on an interval containing the fixed endpoint \(a\) and the relevant values of \(g(x)\), and \(g\) is differentiable, then
The factor \(g'(x)\) is essential whenever the endpoint is not simply \(x\). If the variable endpoint is the lower limit instead, the bounds are reversed, so the derivative has a negative sign:
Why the Variable-Endpoint Formula Has a Chain-Rule Factor
The reason for the factor can be seen by defining a new function \(F(u)=\int_a^u f(t)\,dt\). The Fundamental Theorem says \(F'(u)=f(u)\) where \(f\) is continuous. The accumulation with endpoint \(g(x)\) is \(F(g(x))\); differentiating this composite function requires the chain rule.
Let \(F(u)=\int_a^u f(t)\,dt\). By the Fundamental Theorem, \(F'(u)=f(u)\).
Then \(H(x)=\int_a^{g(x)}f(t)\,dt=F(g(x))\).
\(H'(x)=F'(g(x))g'(x)=f(g(x))g'(x)\).
This argument requires \(f\) to be continuous on an interval that contains the fixed endpoint \(a\) and the values of \(g(x)\) being considered. It also requires \(g\) to be differentiable at the input where the derivative is taken. Continuity matters because it allows the Fundamental Theorem to identify \(F'\) with \(f\).
Worked Example: Reversing the Bounds
Worked Example: Reversing the Bounds
Evaluate \(\int_5^1(2t-3)\,dt\), and check the sign by reversing the bounds.
An antiderivative of \(2t-3\) is \(F(t)=t^2-3t\), because \(F'(t)=2t-3\). Apply upper endpoint minus lower endpoint, even though the upper endpoint is numerically smaller:
For a check, reversing the bounds must reverse the sign. Indeed,
The computed value agrees with the bound-reversal check. Writing \(F(5)-F(1)\) for the original integral would give the wrong sign.
Worked Example: Substituting Both Endpoints Correctly
Worked Example: Substituting Both Endpoints Correctly
Evaluate \(\int_{-1}^2(x^2+1)\,dx\). Pay particular attention to the value at the negative lower endpoint.
An antiderivative is \(F(x)=\frac{x^3}{3}+x\), since \(F'(x)=x^2+1\). Evaluate it separately at the upper and lower endpoints:
Now subtract the lower-endpoint value, keeping its negative sign inside parentheses:
Check: differentiating \(F(x)=\frac{x^3}{3}+x\) gives \(x^2+1\), the integrand. Also, \(x^2+1\) is positive throughout the interval, so a positive integral is consistent with the signed accumulation. The value is \(6\).
Worked Example: A Variable Endpoint and the Missing Factor
Worked Example: A Variable Endpoint and the Missing Factor
Let \(A(x)=\int_2^{x^2+1}(3t+1)\,dt\). Find \(A'(x)\), and verify the result by evaluating the integral first.
The integrand \(f(t)=3t+1\) is continuous for all real \(t\), and the endpoint \(g(x)=x^2+1\) is differentiable. The variable-endpoint formula gives
For an independent check, an antiderivative of \(3t+1\) is \(G(t)=\frac{3}{2}t^2+t\), since \(G'(t)=3t+1\). Thus
Differentiate this expression using the chain rule:
The two methods agree. As a numerical check at \(x=1\), the endpoint is \(2\), so the formula gives \(A'(1)=(3(2)+1)(2)=7\cdot2=14\). The derived expression also gives \(6+8=14\).
Worked Example: An Integrand Value Is Not an Accumulated Value
Worked Example: An Integrand Value Is Not an Accumulated Value
Let \(B(x)=\int_0^x(t+1)\,dt\). Compare \(B(2)\), the integrand value at \(2\), and the rate of change \(B'(2)\).
First calculate the accumulated value. An antiderivative of \(t+1\) is \(Q(t)=\frac{t^2}{2}+t\), because \(Q'(t)=t+1\). Therefore,
By the variable-endpoint form of the Fundamental Theorem, \(B'(x)=x+1\), so \(B'(2)=3\). The integrand at \(2\) is also \(2+1=3\). Thus the accumulated value is \(B(2)=4\), while the instantaneous rate and integrand value at \(2\) are both \(3\).
Check by differentiating the evaluated expression: \(B(x)=\frac{x^2}{2}+x\), so \(B'(x)=x+1\), and \(B'(2)=3\), as above. In general, \(B(x)\) and \(B'(x)\) answer different questions: the first is accumulated change from the starting point, and the second is the current rate.
Common Errors and Their Corrections
Several tempting shortcuts cause recurring errors. For example, a student evaluating the integral in the endpoint-substitution example might write
The error is that \(F(-1)=-\frac{4}{3}\), not \(+\frac{4}{3}\). The correction is to preserve the sign and subtract the entire lower-endpoint value:
Another common error is to differentiate \(\int_2^{x^2+1}(3t+1)\,dt\) as \(3(x^2+1)+1\), omitting the factor \(2x\). That expression accounts for the integrand at the moving endpoint but not for how quickly the endpoint itself moves. The correct derivative is \(\bigl(3(x^2+1)+1\bigr)(2x)\).
Finally, do not replace an integral by the integrand evaluated at one endpoint. In the last example, \(t+1\) evaluated at \(t=2\) is \(3\), but the accumulated value from \(0\) to \(2\) is \(4\). The integral combines contributions across the whole interval; the integrand value describes the rate at a single input.
A Practical Checklist
Before finalizing an application of the Fundamental Theorem, check each of these points:
- For a definite integral, did you evaluate \(F\) at the upper endpoint and subtract \(F\) at the lower endpoint?
- Did you keep parentheses around the lower-endpoint value, especially when it is negative?
- For a variable endpoint, did you multiply by the derivative of that endpoint?
- Is the integrand continuous on an interval containing the fixed endpoint and the relevant values of the variable endpoint when using the derivative formula?
- Are you reporting an accumulated integral, an integrand value, or the derivative of an accumulation function?
- Does the sign, unit, and size of the result make sense for the situation?
These checks are useful because the formulas often produce similar-looking expressions for different quantities. Naming what each expression represents helps prevent confusing an instantaneous rate with accumulated change.
Check Your Understanding
For each question, identify the relevant bound, endpoint, or quantity before calculating.
- Evaluate \(\int_4^1(2t+1)\,dt\). What sign should the answer have compared with \(\int_1^4(2t+1)\,dt\)?
- For \(C(x)=\int_0^{x^3}(t+2)\,dt\), what factor must multiply the integrand value at the variable endpoint when finding \(C'(x)\)?
- In evaluating \(\int_{-2}^1 f(x)\,dx\) from an antiderivative \(F\), which endpoint value is subtracted, and why are parentheses useful?
- If \(D(x)=\int_1^x f(t)\,dt\), how is \(D(3)\) different in meaning from \(D'(3)\)?
- On what interval must \(f\) be continuous, and which endpoints must that interval contain, for \(\frac{d}{dx}\int_a^{g(x)}f(t)\,dt\) to be justified by the Fundamental Theorem?
