From an Integrand to an Endpoint Calculation
A definite integral can represent accumulated change across an interval. The integrand describes the rate of accumulation, while the integral gives the total accumulated amount. The second part of the Fundamental Theorem of Calculus provides a useful way to calculate that amount: instead of accumulating infinitely many small contributions directly, find an antiderivative and compare its values at the interval's endpoints.
The central computation is short, but its conditions and order matter. The integrand must be continuous on the interval, and the function used in the calculation must genuinely have that integrand as its derivative. Once those requirements are met, evaluating a definite integral becomes a pair of function evaluations followed by a subtraction.
The Evaluation Rule
The value \(F(b)\) is the antiderivative's value at the upper endpoint, and \(F(a)\) is its value at the lower endpoint. The order is upper minus lower. This difference gives the definite integral; it is not necessary to add a constant to the antiderivative before evaluating. Adding a constant to both endpoint values would cancel in their difference.
In practice, the rule has three main tasks:
Confirm that the integrand is continuous on the entire closed interval and that the function you plan to use differentiates to that integrand.
Choose \(F\) so that \(F'(x)=f(x)\) throughout the interval.
Calculate \(F(b)\) and \(F(a)\), then compute \(F(b)-F(a)\).
The method turns integration into differentiation run in reverse, followed by endpoint arithmetic. Differentiating the proposed \(F\) is an important check: if it does not return \(f\), the theorem does not justify using it.
Why the Rule Makes Sense
Let \(A(x)=\int_a^x f(t)\,dt\) be the accumulated amount from a fixed starting point \(a\) to a variable point \(x\). When \(f\) is continuous, the first part of the Fundamental Theorem of Calculus says that \(A'(x)=f(x)\). Thus the derivative of the accumulation function is the original integrand: both \(A\) and an antiderivative \(F\) describe functions whose rate of change is \(f\).
This explains the idea behind the endpoint calculation. The integral from \(a\) to \(b\) records the accumulated change as the endpoint moves from \(a\) to \(b\); the theorem says that this accumulated change is measured by the corresponding change in an antiderivative. In using the rule below, each antiderivative is checked by differentiation before its endpoint values are used.
Worked Example: A Polynomial Integrand
Evaluate \(\int_0^2 (3x^2-4x+1)\,dx\).
The integrand is a polynomial, so it is continuous on \([0,2]\). An antiderivative is found term by term:
The derivative check confirms that \(F\) is an antiderivative of the integrand. Evaluate it at the upper endpoint:
Evaluate it at the lower endpoint:
Subtract the lower-endpoint value from the upper-endpoint value:
Check: differentiating \(x^3-2x^2+x\) gives \(3x^2-4x+1\), and the endpoint values produce \(2-0=2\). Therefore the evaluation is consistent with the theorem and the arithmetic.
Trigonometric and Logarithmic Integrands
The same rule applies when an antiderivative involves a trigonometric or logarithmic function. The key is still to verify the derivative and then evaluate at the two specified endpoints. In each example, the integrand is continuous throughout the interval of integration.
Worked Example: A Trigonometric Integrand
Evaluate \(\int_0^{\pi/2}(2\cos x+\sin x)\,dx\).
The sine and cosine functions are continuous on \([0,\pi/2]\). An antiderivative is \(F(x)=2\sin x-\cos x\), because
At the upper endpoint, \(\sin(\pi/2)=1\) and \(\cos(\pi/2)=0\), so
At the lower endpoint, \(\sin(0)=0\) and \(\cos(0)=1\), so
Now subtract the lower value:
Check: \(F'(x)=2\cos x+\sin x\), the given integrand, and the endpoint subtraction is \(2-(-1)=3\).
Worked Example: A Logarithmic Antiderivative
Evaluate \(\int_1^e \left(\frac{1}{x}+2x\right)\,dx\).
The integrand is continuous on \([1,e]\), since this interval does not contain \(x=0\). An antiderivative is \(F(x)=\ln x+x^2\). Check it by differentiating:
At the upper endpoint, \(\ln e=1\), so
At the lower endpoint, \(\ln 1=0\), so
Apply the theorem and simplify:
Check: differentiating \(F(x)=\ln x+x^2\) recovers the integrand, and subtracting the endpoint values cancels the two \(1\)'s, leaving \(e^2\).
Endpoint Arithmetic and a Common Mistake
The subtraction order is part of the theorem, not a matter of preference. Always write the upper endpoint's function value first and subtract the lower endpoint's function value second. Keeping the two values visible before simplifying helps prevent sign errors, especially when the lower value is negative.
Worked Example: Avoiding a Sign Error
Consider \(\int_0^1 (2x-3)\,dx\). The integrand is continuous on \([0,1]\), and an antiderivative is \(F(x)=x^2-3x\), since \(F'(x)=2x-3\).
The endpoint values are
The correct evaluation is upper minus lower:
A common incorrect calculation reverses the subtraction and writes \(F(0)-F(1)=0-(-2)=2\). That is not the formula in the theorem. The correction is to keep \(F(1)-F(0)\), which gives \(-2\). Check: differentiating \(x^2-3x\) returns \(2x-3\), and the endpoint values confirm the result \(-2\).
Conditions and Checks
Before applying the rule, make two separate checks. First, is \(f\) continuous on the whole closed interval? Second, does the proposed \(F\) satisfy \(F'(x)=f(x)\) on the interval? Continuity is a hypothesis of the stated version of the theorem; finding an expression that looks like an antiderivative does not remove that requirement.
For example, \(1/x\) is not continuous on any interval containing \(0\). Therefore, the theorem stated here cannot be applied directly to an integral over an interval that includes \(0\), using \(F(x)=\ln|x|\). The integrand fails the continuity condition at that point. On an interval such as \([1,e]\), however, it is continuous, and the usual endpoint calculation applies.
These checks also help catch algebraic errors. If a proposed antiderivative differentiates to only part of the integrand, then it is not suitable. If the antiderivative is correct but its endpoint values are substituted incorrectly, the final calculation is still wrong. A reliable evaluation shows the antiderivative, verifies its derivative, computes both endpoint values, and displays the subtraction.
Check Your Understanding
For each question, identify the continuity and antiderivative checks that justify the calculation.
- Find an antiderivative of \(4x^3-2\), then state which endpoint values are subtracted to evaluate its integral from \(0\) to \(1\).
- Evaluate \(\int_0^\pi \sin x\,dx\), showing the antiderivative and both endpoint values.
- Why must the proposed antiderivative be differentiated before using the endpoint formula?
- Can the stated version of the theorem be applied directly to \(\int_{-1}^1 \frac{1}{x}\,dx\)? Explain which condition is relevant.
- If \(F(3)=7\) and \(F(1)=-2\), what is the endpoint difference for an integral from \(1\) to \(3\)?
