From an Antiderivative to a Definite Integral
A definite integral records the net accumulation of a function across an interval. Once an antiderivative is known, the Fundamental Theorem of Calculus lets us calculate that accumulation using only the two endpoints. The essential operation is simple: evaluate the antiderivative at the upper bound, then subtract its value at the lower bound.
The order matters. Reversing the subtraction changes the sign, just as reversing the bounds of an integral does. Before using the rule, also check that the integrand is continuous on the interval and that the function you have found really is an antiderivative there.
The Evaluation Rule
The notation \([F(x)]_a^b\) is shorthand for \(F(b)-F(a)\). The upper bound is substituted first; the lower-bound value is subtracted from it. Thus, the evaluation rule is often written in either of these equivalent forms:
The definite integral has a value, not a family of values. By contrast, an indefinite integral represents a family of antiderivatives, which differ by constants. That constant does not affect the evaluation rule: if \(F(x)+C\) is used in place of \(F(x)\), then
This is why you do not need to append \(+C\) to the result of a definite integral. The constant cancels when the endpoint values are subtracted.
Why the Rule Works
The evaluation rule connects the accumulation of \(f\) from \(a\) to \(b\) to the change in any antiderivative of \(f\). The following argument explains that connection using the Fundamental Theorem’s accumulation function.
Let \(H(x)=\int_a^x f(t)\,dt\), where \(f\) is continuous on \([a,b]\). By the Fundamental Theorem of Calculus, \(H'(x)=f(x)\) on the interval.
Since \(F'(x)=f(x)\) and \(H'(x)=f(x)\), the derivative of their difference is \((F-H)'(x)=F'(x)-H'(x)=f(x)-f(x)=0\).
By the Mean Value Theorem, a function with derivative zero throughout an interval is constant there. Therefore \(F(x)-H(x)\) is constant. At \(x=a\), \(H(a)=\int_a^a f(t)\,dt=0\), so this constant is \(F(a)\).
At \(x=b\), the relation \(F(b)-H(b)=F(a)\) gives \(H(b)=F(b)-F(a)\). Because \(H(b)=\int_a^b f(t)\,dt\), the evaluation rule follows.
The proof also shows why any antiderivative works: any two antiderivatives on the interval differ by a constant, and that constant disappears in the endpoint difference.
How to Evaluate an Integral
A reliable calculation follows three steps: find or identify an antiderivative, substitute the upper bound, and subtract the value at the lower bound. Then check the antiderivative by differentiating it. Keep the endpoint expressions in parentheses until after the subtraction, especially when they contain negative terms.
Worked Example: Evaluate a Polynomial Integral
Evaluate \(\int_1^3 (2x-1)\,dx\).
Step 1: Find an antiderivative. Integrating each term gives \(F(x)=x^2-x\), because the antiderivative of \(2x\) is \(x^2\), and the antiderivative of \(-1\) is \(-x\).
Step 2: Check it. Differentiate: \(F'(x)=2x-1\), which is the integrand.
Step 3: Evaluate at the upper bound. \(F(3)=3^2-3=9-3=6\).
Step 4: Evaluate at the lower bound and subtract. \(F(1)=1^2-1=1-1=0\), so \(F(3)-F(1)=6-0=6\).
Check: Differentiating \(x^2-x\) returns \(2x-1\), and the endpoint calculation gives \(6\). Thus the definite integral is \(6\).
Worked Example: Evaluate an Integral with a Logarithmic Antiderivative
Evaluate \(\int_0^2 \frac{1}{x+1}\,dx\).
Step 1: Identify an antiderivative. On \([0,2]\), \(x+1\) is positive, so an antiderivative is \(F(x)=\ln(x+1)\).
Step 2: Check it. \(F'(x)=\frac{1}{x+1}\), the given integrand. The denominator is never zero on the interval, so this antiderivative is defined throughout it.
Step 3: Evaluate at the bounds. At the upper bound, \(F(2)=\ln(2+1)=\ln 3\). At the lower bound, \(F(0)=\ln(0+1)=\ln 1=0\).
Check: The derivative of \(\ln(x+1)\) is \(\frac{1}{x+1}\), and the lower endpoint contributes zero. The exact value is \(\ln 3\).
Worked Example: Evaluate a Trigonometric Integral
Evaluate \(\int_0^{\pi/2}\sin x\,dx\).
Step 1: Find an antiderivative. Since the derivative of \(-\cos x\) is \(\sin x\), take \(F(x)=-\cos x\).
Step 2: Check it. \(F'(x)=-(-\sin x)=\sin x\), as required.
Step 3: Substitute the upper bound. \(F(\pi/2)=-\cos(\pi/2)=-0=0\).
Step 4: Substitute the lower bound and subtract. \(F(0)=-\cos(0)=-1\), so \(F(\pi/2)-F(0)=0-(-1)=1\).
Check: Differentiating \(-\cos x\) gives \(\sin x\), and the endpoint difference is \(1\).
Bounds, Signs, and Net Accumulation
The formula always uses upper-bound value minus lower-bound value. It does not promise that the result is positive. A definite integral measures signed, or net, accumulation: values of the integrand below the horizontal axis contribute negatively. The endpoint order also matters. Reversing the bounds reverses the sign because \(F(a)-F(b)=-\bigl(F(b)-F(a)\bigr)\).
Worked Example: Reverse the Bounds Correctly
Evaluate \(\int_3^1 2x\,dx\).
Step 1: Find and check an antiderivative. Let \(F(x)=x^2\). Its derivative is \(F'(x)=2x\), the integrand.
Step 2: Use the stated upper bound first. The upper bound is \(1\), so \(F(1)=1^2=1\).
Step 3: Subtract the value at the lower bound. The lower bound is \(3\), so \(F(3)=3^2=9\). Therefore, \(F(1)-F(3)=1-9=-8\).
Check: Reversing the bounds of \(\int_1^3 2x\,dx=9-1=8\) gives \(-8\), matching the evaluation. The negative answer is consistent with the reversed orientation of the interval.
Common Mistake: Subtracting in the Wrong Order
A frequent error is to calculate the lower-bound value minus the upper-bound value. For example, consider \(\int_1^3 2x\,dx\), with antiderivative \(F(x)=x^2\). A wrong calculation is \(F(1)-F(3)=1-9=-8\). This uses the bounds in the opposite order.
The correction is to evaluate the upper bound first and subtract the lower-bound value: \(F(3)-F(1)=9-1=8\). Differentiating \(x^2\) confirms that it is an antiderivative of \(2x\). The result is positive because the integrand is positive on \([1,3]\). If the bounds had actually been \(3\) and \(1\), then \(-8\) would be correct.
When the Evaluation Rule Applies
Before using the rule, confirm that the integrand is continuous over the entire closed interval and that the proposed antiderivative differentiates to the integrand there. These conditions make the Fundamental Theorem’s evaluation rule applicable. In practical problems, also check that substitutions into logarithms, denominators, and other expressions are defined at both endpoints and throughout the interval.
The rule does not require \(F(a)=0\), nor does it require the integral to be positive. It requires only a suitable antiderivative on the interval and the correct endpoint subtraction. If the upper and lower bounds are equal, the endpoint difference is \(F(a)-F(a)=0\), consistent with an integral over an interval of zero length.
Evaluating definite integrals this way is useful because it replaces an accumulation calculation with endpoint arithmetic. The integral still represents net change over the interval; the antiderivative packages that change so that only its values at the endpoints are needed.
Check Your Understanding
For each question, use the upper-bound value minus the lower-bound value, and check any proposed antiderivative by differentiating it.
- Evaluate \(\int_0^2 (3x^2+1)\,dx\).
- Evaluate \(\int_1^e \frac{1}{x}\,dx\).
- If \(F'(x)=f(x)\), write the evaluation rule for \(\int_a^b f(x)\,dx\) using \(F\).
- Explain why adding a constant to an antiderivative does not change the value of a definite integral.
- Evaluate \(\int_2^0 4x\,dx\), and explain why its sign differs from the integral with the bounds in the opposite order.
