IndietroApplications of Integration: Displacement, Distance, and Area Between Curves
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Applications of Integration
Displacement and Distance from Velocity
When an object's velocity as a function of time is known, integration can be used to determine both its displacement and the total distance traveled over a given interval.
Displacement: The net change in position of an object over a time interval [a, b] is given by the definite integral of its velocity function:
Distance Traveled: The total distance traveled (regardless of direction) is the integral of the absolute value of velocity:
Example: Given a velocity graph, the displacement from t = 0 to t = 7 is computed by summing the signed areas under the curve, while the distance is the sum of the absolute values of these areas.
Initial Position: If the initial position s(0) is known, the position at time t is:
Application: If s(0) = 40 and the displacement from t = 0 to t = 7 is 3.5, then s(7) = 43.5.
Area Between Curves
Integration can be used to find the area of a region bounded by curves. The area is typically set up as a definite integral, either with respect to x or y, depending on the orientation and boundaries of the region.
Area with Respect to x
General Formula: For curves y = f(x) (upper) and y = g(x) (lower) over a ≤ x ≤ b:
Example: For the region bounded by y = (x-2)^2, y = 9, and x = 3 for x ≥ 3:
Here, the limits of integration are determined by solving (x-2)^2 = 9, giving x = -1 and x = 5. Since x ≥ 3, the relevant interval is x = 3 to x = 5.
Area with Respect to y
General Formula: For curves x = f(y) (right) and x = g(y) (left) over c ≤ y ≤ d:
Example: For the same region as above, solve y = (x-2)^2 for x to get x = 2 + \sqrt{y}. The area is:
The lower limit y = 1 comes from substituting x = 3 into y = (x-2)^2.
Area of a Region Bounded by Multiple Curves
When a region is bounded by more than two curves, it may be necessary to split the area into parts and sum the corresponding integrals.
Example: For the region bounded by y = 2 + \sqrt{x-3}, y = -2x + 18, and y = -\frac{1}{3}x + 3:
To express the area with respect to y, solve each equation for x in terms of y:
y = 2 + \sqrt{x-3} → x = 3 + (y-2)^2
y = -2x + 18 → x = 9 - \frac{1}{2}y
y = -\frac{1}{3}x + 3 → x = 9 - 3y
The area is then:
Summary Table: Area Setups
Variable of Integration | Upper/Right Function | Lower/Left Function | Limits | Integral Expression |
|---|---|---|---|---|
x | y = f(x) | y = g(x) | x = a \to x = b | |
y | x = f(y) | x = g(y) | y = c \to y = d |
Key Points
Always determine intersection points to set correct limits of integration.
For distance, use the absolute value of velocity; for displacement, use the signed value.
When regions are bounded by multiple curves, split the area into subregions as needed.
Additional info: The notes above expand on the brief calculations in the original file, providing general formulas, context, and a summary table for area computations.
