From Accumulation to an Antiderivative
A definite integral records the net accumulation of a function over an interval. An antiderivative, in contrast, is a function whose derivative gives the original function. The Fundamental Theorem of Calculus connects these two ideas: the accumulated value of a continuous function over an interval is the change in any of its antiderivatives across that interval.
This connection is useful because a definite integral defined through a limit of sums can often be evaluated by finding a function with a known derivative. The key is not merely a convenient calculation rule. The rule works because the accumulation function and every antiderivative of the integrand have the same derivative, so they can differ only by a constant.
The expression \(F(b)-F(a)\) is called the net change in \(F\) from \(a\) to \(b\). It is the value of the antiderivative at the upper endpoint minus its value at the lower endpoint. The integrand is \(f\); the antiderivative is \(F\). Keeping those roles distinct is important.
Why the Formula Works
To see why an antiderivative difference gives the integral, consider the accumulation function with base point \(a\): $$ A(x)=\int_a^x f(t)\,dt. $$ $$
Here \(t\) is a variable of integration; \(x\) is the input of the function \(A\). By the Fundamental Theorem of Calculus, Part 1, if \(f\) is continuous, then \(A\) is differentiable and \(A'(x)=f(x)\) for interior points of the interval.
We are also given that \(F'(x)=f(x)\). Thus \(A'(x)=F'(x)\), so the derivative of \(A-F\) is zero. The Mean Value Theorem implies that a differentiable function with derivative zero throughout an interval is constant there. Therefore \(A-F\) has the same value at every point in the interval.
At \(x=a\), the accumulation function is \(A(a)=\int_a^a f(t)\,dt=0\). Consequently, \(A(a)-F(a)=-F(a)\). Since \(A(x)-F(x)\) is constant, for \(x=b\) we have \(A(b)-F(b)=-F(a)\). Rearranging yields the definite-integral formula.
Let \(A(x)=\int_a^x f(t)\,dt\), the accumulated value from the fixed starting point \(a\) to \(x\).
Continuity of \(f\) gives \(A'(x)=f(x)\), and the antiderivative condition gives \(F'(x)=f(x)\).
\((A-F)'=A'-F'=f-f=0\), so the Mean Value Theorem shows \(A-F\) is constant.
Since \(A(a)=0\), the constant value is \(A(a)-F(a)=-F(a)\). Hence \(A(b)-F(b)=-F(a)\).
Because \(A(b)=\int_a^b f(t)\,dt\), the equality gives \(\int_a^b f(t)\,dt=F(b)-F(a)\).
This argument also explains why the result does not depend on which antiderivative is chosen. If \(F\) and \(G\) are both antiderivatives of \(f\), then \((F-G)'=f-f=0\), so \(F-G\) is constant. Adding the same constant to both endpoint values does not change their difference.
Using an Antiderivative to Find an Integral
Once the theorem's conditions are met, the evaluation has three essential parts: identify an antiderivative \(F\), find its value at each bound, and subtract the lower-bound value from the upper-bound value. The bounds belong to the variable in the integral. If the integral uses \(x\), evaluate \(F(x)\) at \(x=b\) and \(x=a\).
Worked Example: A Polynomial with a Positive Integrand
Evaluate \(\int_1^3 2x\,dx\). The function \(f(x)=2x\) is continuous on \([1,3]\). An antiderivative is \(F(x)=x^2\), since \(F'(x)=2x\). Therefore, $$ \int_1^3 2x\,dx=F(3)-F(1)=3^2-1^2=9-1=8. $$
Check: The region under \(y=2x\) from \(x=1\) to \(x=3\) is a trapezoid with parallel sides of lengths \(2\) and \(6\) and width \(2\). Its area is \(\frac{1}{2}(2+6)(2)=8\), agreeing with the integral. Also, differentiating \(F(x)=x^2\) recovers the integrand \(2x\).
Worked Example: A Polynomial That Changes Sign
Evaluate \(\int_0^2 (3x^2-4x+1)\,dx\). The integrand is continuous, and an antiderivative is $$ F(x)=x^3-2x^2+x, $$ because differentiating term by term gives \(F'(x)=3x^2-4x+1\). Evaluate at the bounds: $$ F(2)=2^3-2(2^2)+2=8-8+2=2, $$ $$ F(0)=0^3-2(0^2)+0=0. $$ Thus $$ \int_0^2 (3x^2-4x+1)\,dx=F(2)-F(0)=2-0=2. $$
Check: The integrand factors as \((3x-1)(x-1)\), so it is positive on \(0\le x<\frac{1}{3}\), negative on \(\frac{1}{3}<x<1\), and positive on \(1<x\le2\). The result is a signed net accumulation, not the total geometric area; contributions below the axis count negatively. Differentiating the stated \(F\) again gives the original integrand.
Worked Example: A Trigonometric Integrand
Evaluate \(\int_0^{\pi/2}\cos x\,dx\). The cosine function is continuous on the interval, and \(F(x)=\sin x\) is an antiderivative because \(F'(x)=\cos x\). Hence $$ \int_0^{\pi/2}\cos x\,dx =F\left(\frac{\pi}{2}\right)-F(0) =\sin\left(\frac{\pi}{2}\right)-\sin(0) =1-0=1. $$
Check: The integrand is nonnegative on the interval, so its integral is positive, consistent with the value \(1\). The derivative of the proposed antiderivative \(\sin x\) is \(\cos x\), as required.
Why the Constant of Integration Does Not Matter
When finding an antiderivative, it is common to write a family such as \(F(x)+C\), where \(C\) is a constant. For a definite integral, any one member of that family gives the same result: $$ \bigl(F(b)+C\bigr)-\bigl(F(a)+C\bigr) =F(b)+C-F(a)-C =F(b)-F(a). $$
The cancellation is not an extra trick. It follows from the fact that antiderivatives of the same function differ only by a constant. A definite integral measures the change in the antiderivative, and a constant shift changes neither endpoint-to-endpoint change nor the integral.
The theorem also clarifies why the integral is a signed quantity. If \(f\) is positive on some portion of the interval, that portion contributes positively; where \(f\) is negative, it contributes negatively. The endpoint difference records the net result. It does not, in general, equal the total geometric area between the graph and the horizontal axis.
A Common Subtraction Error
The endpoint order matters: subtract the value at the lower bound from the value at the upper bound. For example, with \(F(x)=x^2\) on \([1,3]\), an incorrect calculation would be $$ F(1)-F(3)=1-9=-8. $$
This reverses the required order. The correct calculation is $$ F(3)-F(1)=9-1=8. $$ Since \(2x\) is positive on \([1,3]\), its definite integral must be positive, which immediately signals that \(-8\) cannot be correct. A reliable check is to read the formula aloud: upper endpoint value minus lower endpoint value.
Another error is to subtract values of the integrand rather than values of an antiderivative. In general, \(f(b)-f(a)\) is not the definite integral. The integral is computed from \(F(b)-F(a)\), where \(F'=f\). For \(f(x)=2x\) on \([1,3]\), subtracting integrand values gives \(f(3)-f(1)=6-2=4\), whereas the integral is \(8\). The calculation uses the wrong function.
When to Apply the Result
For the formula to apply in the stated form, the integrand must be continuous on the interval and the function used must be an antiderivative of that integrand there. Continuity guarantees the accumulation function has derivative \(f\), and the antiderivative condition identifies the function whose endpoint change matches that accumulation.
Before evaluating, check these points:
- Does the proposed \(F\) differentiate to the integrand \(f\)?
- Is \(f\) continuous throughout the interval of integration?
- Have the bounds been applied to \(F\), not to \(f\)?
- Is the subtraction in the order \(F(b)-F(a)\)?
The main conceptual payoff is that a limit of sums can be evaluated by a change in a function. The definite integral measures accumulation; the antiderivative tracks that accumulation through its rate of change. Their connection is exactly what makes \(F(b)-F(a)\) a method for computing \(\int_a^b f(x)\,dx\).
Check Your Understanding
Use the connection between an integrand, its antiderivative, and endpoint change to answer these questions.
- State the conditions under which \(\int_a^b f(x)\,dx=F(b)-F(a)\).
- Why do an accumulation function and an antiderivative of the same continuous function differ by a constant?
- Evaluate \(\int_0^2 3x^2\,dx\) using an antiderivative, and verify the antiderivative by differentiation.
- If two antiderivatives of the same function differ by \(7\), why do they produce the same definite integral?
- For a function that is negative on part of an interval, does the definite integral represent net signed accumulation or total geometric area? Explain.
