The Fundamental Theorem of Calculus Part 2 connects two big ideas in calculus: accumulation and change. It says that a definite integral can be evaluated by finding any antiderivative of the function. Instead of adding infinitely many thin rectangles, you can compute one difference, F(b) - F(a).
This is important because it turns area, displacement, and total change problems into algebraic calculations.
If F'(x) = f(x), then the net signed area under f(x) from x = a to x = b is integral_a^b f(x) dx = F(b) - F(a). The function F measures accumulated change, while f gives the rate of change at each input. Positive parts of f add to the integral, and negative parts subtract from it.
In applications, this theorem lets you find total distance from velocity, total mass from density, or total cost from marginal cost when the rate function is known.
Understanding Calculus: The Fundamental Theorem Part 2
The calculation has a strict order. First, identify a function whose derivative is the given rate. Next, substitute the upper endpoint into that function.
Then substitute the lower endpoint. Subtract the second result from the first. Parentheses matter when the lower endpoint is negative or when its output is negative.
For example, if the lower output is negative five, subtracting it produces a positive five. Many incorrect answers come from dropping these parentheses, not from using the wrong calculus rule. Keep the endpoints attached to the original integral until the final substitution.
The power rule gives a fast way to build many antiderivatives. For a power of x, raise the exponent by one, then divide by the new exponent. Thus an antiderivative of three x squared is x cubed, because the derivative of x cubed is three x squared.
Constants stay as factors. An antiderivative of seven x to the fourth power is seven fifths x to the fifth power. There are important exceptions.
The antiderivative of one over x involves the natural logarithm of the absolute value of x. Expressions containing sine, cosine, or exponential functions use their own derivative patterns. A quick derivative check is useful.
Differentiate your proposed answer before evaluating it at the endpoints. If the original function does not return, the antiderivative needs correction.
Units help explain what the final number means. If a velocity is measured in meters per second and time is measured in seconds, integrating velocity over time produces meters. That result is displacement, which tracks final position change.
A person can walk east, turn around, then return to the start. The displacement is zero even though the person traveled a positive distance. To find total distance, split the time interval wherever velocity changes sign.
Evaluate each piece, change negative accumulated amounts to positive amounts, then add them. This same idea appears in profit and loss, electric charge flow, and water entering or leaving a tank. The integral records the net effect unless the situation specifically calls for a total amount.
Graphs provide a strong reality check. A rate graph above the horizontal axis should contribute a positive result over that interval. A graph below the axis should contribute a negative result.
A result near zero can be reasonable when positive and negative regions nearly balance. It may be unreasonable if the graph stays far above the axis. The theorem is easiest to use when the rate function is continuous across the full interval.
If a formula has a break, a vertical asymptote, or a jump, examine the interval carefully. Some cases require splitting the interval into separate pieces.
Others create an improper integral that needs a limit before any numerical value can be trusted. A calculator can evaluate expressions quickly, but it cannot decide whether the model, bounds, or sign interpretation make sense.
Key Facts
- If F'(x) = f(x), then integral_a^b f(x) dx = F(b) - F(a).
- The definite integral integral_a^b f(x) dx gives net signed area, not always ordinary geometric area.
- Any antiderivative of f can be used because constants cancel: (F(b) + C) - (F(a) + C) = F(b) - F(a).
- If f(x) is continuous on [a, b], then the theorem applies directly.
- Reversing the bounds changes the sign: integral_b^a f(x) dx = - integral_a^b f(x) dx.
- For a constant multiple, integral_a^b c f(x) dx = c integral_a^b f(x) dx.
Vocabulary
- Definite integral
- A definite integral represents the net signed accumulation of a function over an interval.
- Antiderivative
- An antiderivative of f is a function F whose derivative is f.
- Net signed area
- Net signed area counts area above the x-axis as positive and area below the x-axis as negative.
- Bounds of integration
- The bounds of integration are the lower and upper input values that define the interval of accumulation.
- Accumulation function
- An accumulation function gives the total change collected from a starting value up to a variable endpoint.
Common Mistakes to Avoid
- Forgetting to evaluate both endpoints is wrong because a definite integral requires F(b) - F(a), not just F(b).
- Adding the constant of integration in a definite integral is unnecessary because the constant cancels when subtracting endpoint values.
- Treating all shaded area as positive is wrong because the definite integral measures net signed area, so regions below the x-axis subtract.
- Using f(b) - f(a) instead of F(b) - F(a) is wrong because the theorem requires an antiderivative F, not the original integrand f.
Practice Questions
- 1 Evaluate integral_1^4 3x^2 dx using an antiderivative.
- 2 Evaluate integral_0^2 (5x - 1) dx and state whether the net accumulation is positive or negative.
- 3 A velocity graph is above the x-axis from t = 0 to t = 3 and below the x-axis from t = 3 to t = 5. Explain why the definite integral of velocity gives displacement rather than total distance.