The Fundamental Theorem of Calculus and Definite Integrals

3. Visual Connection

5. Concept Development

The Fundamental Theorem connects two central ideas

Differentiation measures instantaneous change, while integration accumulates change. The Fundamental Theorem of Calculus (FTC) explains that these processes reverse one another under appropriate continuity conditions.

An antiderivative reverses differentiation

A function \(F\) is an antiderivative of \(f\) on an interval when \(F'(x)=f(x)\) throughout that interval. Antiderivatives differ only by a constant on a connected interval.

An accumulation function produces an antiderivative

If \(f\) is continuous and

then \(A'(x)=f(x)\). The symbol \(t\) is a dummy variable; \(x\) controls the moving endpoint and makes \(A\) a function.

The evaluation theorem gives an exact integral

If \(f\) is continuous on \([a,b]\) and \(F'=f\), then

This replaces a limiting sum of many contributions with two endpoint evaluations.

Any antiderivative gives the same answer

Using \(F(x)+C\) instead of \(F(x)\) does not alter a definite integral because

For that reason, a separate \(+C\) is unnecessary when evaluating a definite integral.

Evaluation-bar notation packages endpoint subtraction

The upper endpoint is substituted first, and the entire lower-endpoint value is subtracted.

Simplify before finding an antiderivative

Rewrite radicals as rational exponents, separate compatible terms, and simplify algebraic expressions before integrating. A clean integrand makes the correct antiderivative pattern easier to recognize.

Keep exact values through endpoint evaluation

Preserve fractions, radicals, logarithms, and multiples of \(\pi\) unless a decimal approximation is requested. Early rounding can amplify endpoint-subtraction error.

A definite integral is net signed accumulation

The FTC evaluates the signed integral. Contributions below the horizontal axis are negative, so a correct answer may be negative even though every geometric area is nonnegative.

Total area requires splitting at sign changes

To find total area between a graph and the axis, locate every zero in the interval and integrate the absolute value piecewise:

Piecewise functions are evaluated piece by piece

Split at every formula boundary, use an appropriate antiderivative on each subinterval, and add the resulting definite integrals. A single antiderivative formula cannot be applied across unrelated pieces.

The FTC recovers a quantity from its rate

If \(Q'(t)=r(t)\), then

The integral is the net change; the initial value must be added when the final amount is requested.

Units provide an error check

If \(f\) has output units per input unit, then \(\int_a^b f(x)\,dx\) has output units. For example, integrating liters per minute over minutes produces liters.

Differentiate to verify the antiderivative

Before evaluating endpoints, mentally differentiate the proposed \(F\). This catches missing coefficients, incorrect exponent changes, and sign errors.

Check the theorem's hypotheses

Continuity on the closed interval guarantees the standard FTC evaluation theorem applies. If the integrand is unbounded or the interval is improper, the problem instead requires limits; do not substitute across a vertical singularity.

A reliable FTC workflow

Inspect the interval and continuity, simplify the integrand, find and verify an antiderivative, write the evaluation bar, compute upper minus lower with parentheses, and interpret sign and units.

Common errors

Frequent errors include using lower minus upper, omitting parentheses around \(F(a)\), adding \(+C\) to the final number, confusing net change with final amount, and treating a signed integral as total area.

6. Detailed Worked Example and Error Check

Example 1: Evaluate a polynomial integral.

Differentiating \(x^3-x^2\) returns \(3x^2-2x\), and the lower value is subtracted as a complete group.

Example 2: Rewrite radicals first.

Example 3: Preserve exact trigonometric values.

Example 4: Distinguish net integral from total area. For \(f(x)=x^2-1\) on \([-2,2]\),

The zeros are \(-1\) and \(1\). Splitting there and making each contribution positive gives total area \(4\), not \(4/3\).

Example 5: Integrate a piecewise function. Let \(f(x)=x+2\) on \([-1,0]\) and \(f(x)=x^2+2\) on \((0,2]\). Then

Example 6: Recover an amount from a rate. If \(Q'(t)=4t-1\) and \(Q(0)=10\), then

The integral \(15\) is the change, while \(25\) is the final amount.

Example 7: Connect both parts of the FTC. Define \(A(x)=\int_2^x(3t^2-4)\,dt\). Evaluation gives

because the lower-endpoint expression is zero. Therefore \(A'(x)=3x^2-4\) and \(A(2)=0\), exactly as the accumulation definition requires.