A definite integral measures the accumulated value of a function over an interval. On a graph of y = f(x), this accumulation can be seen as signed area between the curve and the x-axis from x = a to x = b. Area above the x-axis counts as positive, while area below the x-axis counts as negative.
This idea is important because many quantities in physics and math are found by adding up continuously changing values.
Signed area explains why an integral can be zero even when the graph encloses visible regions. Positive and negative regions can cancel, so the definite integral gives net area, not always total geometric area. To find total area, split the interval at x-axis crossings and add the absolute values of the signed areas.
This distinction is useful for displacement versus distance, net charge versus total charge, and other accumulation problems.
Key Facts
- Definite integral: ∫_a^b f(x) dx gives the signed area from x = a to x = b.
- If f(x) > 0 on an interval, then ∫ f(x) dx represents positive area.
- If f(x) < 0 on an interval, then ∫ f(x) dx represents negative area.
- Net area = positive area + negative area, where regions below the x-axis subtract.
- Total area = ∫_a^b |f(x)| dx.
- If c is between a and b, then ∫_a^b f(x) dx = ∫_a^c f(x) dx + ∫_c^b f(x) dx.
Vocabulary
- Definite integral
- A definite integral is a number that represents the signed accumulation of a function over a specific interval.
- Signed area
- Signed area is area counted as positive above the x-axis and negative below the x-axis.
- Net area
- Net area is the result after positive and negative signed areas are combined.
- Total area
- Total area is the sum of all geometric areas between a curve and the x-axis, with every region counted as positive.
- X-intercept
- An x-intercept is a point where a graph crosses or touches the x-axis, so f(x) = 0.
Common Mistakes to Avoid
- Counting all shaded regions as positive, which is wrong because the definite integral uses signed area and regions below the x-axis subtract.
- Forgetting to split at x-intercepts, which is wrong when finding total area because the sign of f(x) can change across those points.
- Confusing net area with total area, which is wrong because cancellation can make the integral smaller than the actual amount of geometric area shown.
- Reversing the limits without changing the sign, which is wrong because ∫_b^a f(x) dx = -∫_a^b f(x) dx.
Practice Questions
- 1 A graph has 12 square units of area above the x-axis from x = 0 to x = 3 and 5 square units below the x-axis from x = 3 to x = 5. Find ∫_0^5 f(x) dx and the total area.
- 2 For f(x) = x - 2, calculate ∫_0^5 f(x) dx. Then find the total area between the graph and the x-axis on 0 ≤ x ≤ 5.
- 3 A velocity graph is above the time axis for the first part of a trip and below it for the second part. Explain why the definite integral gives displacement rather than total distance traveled.