Applications of Integration: Displacement, Distance, and Area Between Curves

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:

$s(b) - s(a) = \int_a^b v(t)\,dt$

• Distance Traveled: The total distance traveled (regardless of direction) is the integral of the absolute value of velocity:

$\text{Distance} = \int_a^b |v(t)|\,dt$

• 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:

$s(t) = s(0) + \int_0^t v(u)\,du$

• 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:

$\text{Area} = \int_a^b [f(x) - g(x)]\,dx$

• Example: For the region bounded by y = (x-2)^2, y = 9, and x = 3 for x ≥ 3:

$\text{Area} = \int_3^5 [9 - (x-2)^2] \,dx$

• 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:

$\text{Area} = \int_c^d [f(y) - g(y)]\,dy$

• Example: For the same region as above, solve y = (x-2)^2 for x to get x = 2 + \sqrt{y}. The area is:

$\text{Area} = \int_1^9 [(2 + \sqrt{y}) - 3] \,dy$

• 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:

$\text{Area} = \int_3^7 \left[2 + \sqrt{x-3} - \left(-\frac{1}{3}x + 3\right)\right] dx + \int_7^9 \left[ -2x + 18 - \left(-\frac{1}{3}x + 3\right) \right] dx$

• 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:

$\text{Area} = \int_0^2 \left[ (9 - \frac{1}{2}y) - (9 - 3y) \right] dy + \int_2^4 \left[ (9 - \frac{1}{2}y) - (3 + (y-2)^2) \right] dy$

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	$\int_a^b [f(x) - g(x)] dx$
y	x = f(y)	x = g(y)	y = c \to y = d	$\int_c^d [f(y) - g(y)] dy$

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.