From an Antiderivative to a Definite Integral
A definite integral has a numerical value, while an antiderivative is a function. The Fundamental Theorem of Calculus connects these two ideas: once an antiderivative of the integrand is known, the integral over an interval can be evaluated using the antiderivative at the interval’s endpoints.
The endpoint order matters. For bounds \(a\) and \(b\) with \(a<b\), the required calculation is the value at \(b\) minus the value at \(a\). A reliable evaluation therefore involves finding an antiderivative, evaluating it at each endpoint, and subtracting in that order.
For the evaluation rule below, assume that \(f\) is continuous on the closed interval \([a,b]\), and that \(F\) is an antiderivative of \(f\) on that interval. In particular, \(F'(x)=f(x)\) for interior points of the interval. Continuity ensures that the definite integral exists; the derivative condition identifies a suitable antiderivative.
The Evaluation Form of the Fundamental Theorem
The theorem asks for an antiderivative, not a unique one. If \(F\) is an antiderivative, then \(F+C\) is also an antiderivative for any constant \(C\). The constant does not affect the evaluation:
For that reason, we usually choose one convenient antiderivative and do not add \(+C\) when evaluating a definite integral. The \(+C\) is needed to describe the full family of antiderivatives, but it cancels in the endpoint difference.
Why Endpoint Evaluation Works
The theorem follows from the relationship between the accumulation of \(f\) and the derivative of an antiderivative. Here is a derivation of the evaluation rule using the first part of the Fundamental Theorem of Calculus: for continuous \(f\), the function \(G(x)=\int_a^x f(t)\,dt\) satisfies \(G'(x)=f(x)\).
For \(x\in[a,b]\), define \(G(x)=\int_a^x f(t)\,dt\). Since \(f\) is continuous, the first part of the Fundamental Theorem gives \(G'(x)=f(x)\) in the interior of the interval.
Because \(F'(x)=f(x)\) and \(G'(x)=f(x)\), the function \(H(x)=F(x)-G(x)\) satisfies \(H'(x)=F'(x)-G'(x)=0\).
For any \(x,y\in[a,b]\), the Mean Value Theorem applied to \(H\) gives \(H(y)-H(x)=H'(c)(y-x)=0\) for some \(c\) between \(x\) and \(y\). Thus \(H(y)=H(x)\), so \(H\) is constant on the interval.
By its definition, \(G(a)=\int_a^a f(t)\,dt=0\), while \(G(b)=\int_a^b f(t)\,dt\). Since \(F-G\) is constant, \(F(b)-G(b)=F(a)-G(a)\). Rearranging and using \(G(a)=0\) gives \(\int_a^b f(t)\,dt=F(b)-F(a)\).
This explains why the endpoint difference measures the integral: both \(F\) and the accumulated function \(G\) have the same derivative, so they differ only by a constant. Taking their difference between the endpoints removes that constant.
A Reliable Evaluation Procedure
Use the same order of operations in each problem. First find an antiderivative and verify its derivative. Then substitute the upper bound and lower bound separately. Finally, subtract the lower-endpoint value from the upper-endpoint value.
If a substitution was used to find an antiderivative, make sure the expression is written in the original variable before substituting the original bounds. This keeps the antiderivative and the endpoint values in the same variable.
Worked Example: A Polynomial Integrand
Evaluate \(\int_1^3 (3x^2-4x+1)\,dx\).
Integrate each term to obtain an antiderivative:
Check it by differentiating: \(F'(x)=3x^2-4x+1\), which is the integrand. Apply the theorem by evaluating at the upper endpoint \(3\) and lower endpoint \(1\):
Subtract the lower value from the upper value:
As a final check, the endpoint difference is \(12\), consistent with the evaluated values \(F(3)=12\) and \(F(1)=0\), and differentiating \(F\) recovers the original integrand.
Exponential and Trigonometric Integrands
The same theorem applies to different function types as long as the integrand is continuous on the interval and an antiderivative is available. The endpoint procedure does not change when the antiderivative contains an exponential or trigonometric function.
Worked Example: An Exponential Integrand
Evaluate \(\int_0^{\ln 3} e^x\,dx\). The function \(e^x\) is continuous on this interval, and an antiderivative is \(F(x)=e^x\), since \(F'(x)=e^x\).
Evaluate at the upper endpoint. Because \(e^{\ln 3}=3\),
Evaluate at the lower endpoint:
Now subtract the lower value from the upper value:
The derivative check is \(F'(x)=(e^x)'=e^x\), as required. The endpoint values \(3\) and \(1\) also confirm the final difference \(2\).
Worked Example: A Trigonometric Integrand
Evaluate \(\int_0^{\pi/2}(\sin x+2\cos x)\,dx\), with angles measured in radians. Both trigonometric functions are continuous on the interval. An antiderivative is \(F(x)=-\cos x+2\sin x\), because
At the upper endpoint, use \(\cos(\pi/2)=0\) and \(\sin(\pi/2)=1\):
At the lower endpoint, use \(\cos(0)=1\) and \(\sin(0)=0\):
Subtract the lower-endpoint value, including its negative sign:
The derivative of \(F(x)=-\cos x+2\sin x\) is \(\sin x+2\cos x\), so it matches the integrand. The evaluated endpoint values also give \(2-(-1)=3\).
A Common Endpoint-Order Mistake
A frequent error is to calculate \(F(a)-F(b)\) instead of \(F(b)-F(a)\). The theorem specifies upper endpoint minus lower endpoint. Reversing the subtraction changes the sign, even if the antiderivative and both endpoint evaluations are correct.
Another source of mistakes is evaluating an expression without parentheses. For example, if \(F(a)=-1\), subtracting \(F(a)\) means subtracting the entire value: \(-F(a)=-(-1)=1\). In the trigonometric example, this is why \(2-(-1)=3\), not \(2-1=1\).
Why This Method Matters
The evaluation theorem turns a definite integral into an endpoint calculation once an antiderivative is known. It also provides a useful check on integration work: differentiate the proposed antiderivative to confirm that it returns the integrand, and then evaluate each endpoint independently before performing the subtraction.
Keep the assumptions visible. The evaluation rule applies when the integrand is continuous on the closed interval and the chosen function is an antiderivative there. When those conditions hold, the calculation is always the same: \(F(b)-F(a)\). Writing out both endpoint values makes the process easier to audit and helps catch sign and substitution errors.
Check Your Understanding
For each question, write down the antiderivative and show both endpoint evaluations before giving the definite integral.
- Evaluate \(\int_0^2 (2x+3)\,dx\).
- Evaluate \(\int_0^{\ln 2} e^x\,dx\).
- Evaluate \(\int_0^{\pi/2}\cos x\,dx\).
- Why does adding a constant to an antiderivative leave the endpoint difference unchanged?
- What is the sign error in replacing \(F(b)-F(a)\) with \(F(a)-F(b)\)?
