This set of notes covers the applied side of integration from the same homework. The problems here use integration to solve real-world-style questions: how much oil a well produces, how high a helicopter climbs, how fast and in what direction a storm moves. These connect what you learned about antiderivatives and the FTC to physical quantities like position, velocity, and accumulated totals.
You should be comfortable with definite and indefinite integrals (Parts 1 and 2) and with basic differentiation (the quotient rule appears in one problem). These applications are the reason integration exists in the first place: measuring things that accumulate over time.
Integration turns rates into totals and derivatives into original functions. To find total oil produced, integrate the production rate. To find height from velocity, integrate velocity. To recover a curve from its slope and a point, integrate and solve for the constant. The FTC Part 2 lets you differentiate an integral with a variable upper limit to recover the integrand.
Rate function
A function f(t) that gives the rate of change of a quantity at time t (barrels per month, feet per second, etc.). Integrating a rate over an interval gives the total change in the quantity.
Initial value problem (IVP)
A problem that gives you a derivative (dy/dx = ...) and an initial condition (a point the curve passes through), and asks you to find the original function. You integrate, then use the initial condition to solve for the constant C.
Velocity and position
Velocity v(t) is the derivative of position s(t). To recover position from velocity, integrate: s(t) = ∫ v(t) dt. Speed is |v(t)|, always non-negative.
FTC Part 2 (differentiation form)
If F(x) = ∫₀ˣ f(t) dt, then F'(x) = f(x). In simple terms, differentiating an integral whose upper limit is x gives back the function inside the integral, evaluated at x.
When a problem gives you a rate of change and asks for the total amount over an interval, you integrate the rate.
Example:
Oil flows from a well at f(t) = 264/(t + 5)² barrels per month (in multiples of 100). Find the total oil produced in the second three months (months 3 to 6).
Set up the integral: T = ∫₃⁶ 264/(t + 5)² dt.
Recognise that 264/(t + 5)² = 264·(t + 5)⁻². The antiderivative of (t + 5)⁻² is −(t + 5)⁻¹, so the antiderivative of the full expression is −264/(t + 5).
Evaluate: [−264/(t + 5)]₃⁶ = −264/11 − (−264/8) = −24 + 33 = 9.
Since the rate is given in multiples of 100: total = 9 × 100 = 900 barrels.
Integrating velocity over a time interval gives displacement (change in position).
Example:
A helicopter has velocity v(t) = 12t + 54/(t + 3)² ft/s. Find its height after 3 seconds (starting from the ground).
h = ∫₀³ [12t + 54/(t + 3)²] dt.
The antiderivative of 12t is 6t². The antiderivative of 54/(t + 3)² is −54/(t + 3).
h = [6t² − 54/(t + 3)]₀³ = (54 − 54/6) − (0 − 54/3) = (54 − 9) − (−18) = 45 + 18 = 63 feet.
Given the slope dy/dx and a point on the curve, integrate to find y, then use the point to solve for C.
Example:
Find the y-intercept of the curve through (1, 4) with slope dy/dx = 6x − 2.
Integrate: y = 3x² − 2x + C.
Use the initial condition y(1) = 4: 3(1) − 2(1) + C = 4, so 1 + C = 4, giving C = 3.
The curve is y = 3x² − 2x + 3. The y-intercept is y(0) = 3.
These problems test whether you can interpret the sign of a derivative as a direction of motion.
Example (storm system):
A storm's distance from Austin is s(t) = 30 + t − t² miles, where t is hours after noon. Find its speed and direction at 5:00 pm.
Velocity: s'(t) = 1 − 2t. At t = 5: s'(5) = 1 − 10 = −9.
The negative sign means the distance is decreasing, so the storm is moving toward Austin. Speed is the absolute value: 9 mph.
Key point: speed is always non-negative. A negative velocity means the object is moving in the negative direction (toward the reference point), not that speed is negative.
Example (reading speed from a graph):
A student's distance from a building is given by a graph. On the interval [2, 4], the graph is a straight line from (2, 200) to (4, 280).
Speed = slope = (280 − 200)/(4 − 2) = 40 yds/min. Since the distance is increasing, the student is moving away from the building.
Sometimes the function is given in terms of parameters rather than numbers.
Example:
If f is linear with slope m and y-intercept b, evaluate ∫₀² f(x) dx.
Since f(x) = mx + b, the integral is:
∫₀² (mx + b) dx = [(1/2)mx² + bx]₀² = (1/2)m(4) + b(2) = 2m + 2b.
This is a good template problem: it confirms that you can work with symbolic parameters just as well as with specific numbers.
The second part of the Fundamental Theorem says that if you define a function as an integral with a variable upper limit, differentiating it gives back the integrand.
Example:
Given ∫₀ˣ f(t) dt = 4x/(x² + 2), find f(1).
By FTC Part 2, f(x) = d/dx [4x/(x² + 2)].
Apply the quotient rule: f(x) = [4(x² + 2) − 4x(2x)] / (x² + 2)² = (4x² + 8 − 8x²) / (x² + 2)² = (8 − 4x²) / (x² + 2)².
At x = 1: f(1) = (8 − 4) / (1 + 2)² = 4/9.
The key recognition: you do not need to figure out what f is by "un-integrating." The FTC tells you that f(x) is simply the derivative of the right-hand side.
Students confuse velocity with speed. Velocity can be negative (indicating direction); speed is always |velocity|. A negative velocity does not mean the object is slowing down.
In initial value problems, students sometimes forget to use the given point to solve for C. Finding the antiderivative is only half the problem.
When integrating a rate function like 264/(t + 5)², students sometimes try to use the power rule on the whole fraction without rewriting it as 264·(t + 5)⁻² first.
On FTC Part 2 problems, students sometimes try to evaluate the integral on the left side rather than just differentiating the right side. The whole point is that you do not need to integrate at all.
⚠️ Application problems are where exams test understanding rather than just computation. Expect at least one question that gives a velocity and asks for position, or gives a rate and asks for total accumulation.
⚠️ Initial value problems test the constant of integration in a meaningful way. C is not just a formality here; it determines which specific curve passes through the given point.
⚠️ The FTC Part 2 (differentiating an integral) often appears as a standalone problem. Recognise the form ∫₀ˣ f(t) dt = [expression in x] and know to differentiate the right side.
⚠️ Direction-of-motion questions require you to interpret the sign of the derivative and state whether the object is approaching or receding. The exam will penalise "speed = −9" because speed is non-negative by definition.
True or false: If v(t) = −5, the object is moving backward and its speed is −5. (False. Speed is |v(t)| = 5. Speed is never negative.)
Fill in the blank: To find the total oil produced between months 3 and 6 from a rate function f(t), you compute ______. (∫₃⁶ f(t) dt)
True or false: If dy/dx = 6x − 2 and y(1) = 4, the constant of integration is C = 3. (True. Integrating gives y = 3x² − 2x + C; plugging in (1, 4) gives 1 + C = 4.)
Fill in the blank: By FTC Part 2, if ∫₀ˣ f(t) dt = x³ + 1, then f(x) = ______. (3x². Differentiate the right side.)
True or false: Integrating velocity from t = 0 to t = 3 gives the object's position at t = 3. (True only if the object starts at position 0. Otherwise it gives displacement, not absolute position.)
Q: Oil flows at f(t) = 264/(t + 5)² barrels per month (×100). Find the total produced in months 3 to 6.
A: ∫₃⁶ 264/(t + 5)² dt = [−264/(t + 5)]₃⁶ = −264/11 + 264/8 = −24 + 33 = 9. Multiply by 100: 900 barrels.
Q: A helicopter rises with velocity v(t) = 12t + 54/(t + 3)². How high is it after 3 seconds?
A: h = ∫₀³ v(t) dt = [6t² − 54/(t + 3)]₀³ = (54 − 9) − (0 − 18) = 45 + 18 = 63 feet.
Q: Find the y-intercept of the curve through (1, 4) with slope dy/dx = 6x − 2.
A: Integrate: y = 3x² − 2x + C. Use y(1) = 4: C = 3. So y = 3x² − 2x + 3 and y(0) = 3.
Q: A storm is s(t) = 30 + t − t² miles from Austin (t = hours after noon). Find its speed and direction at 5 pm.
A: s'(t) = 1 − 2t. At t = 5: s'(5) = −9. The storm has not yet reached Austin (s(5) = 0, s(6) = 0, so it arrives at t = 6), so negative velocity means it is moving toward Austin at 9 mph.
Q: If f is linear with slope m and y-intercept b, what is ∫₀² f(x) dx?
A: f(x) = mx + b. ∫₀² (mx + b) dx = [(1/2)mx² + bx]₀² = 2m + 2b.
Q: Given ∫₀ˣ f(t) dt = 4x/(x² + 2), find f(1).
A: By FTC Part 2, f(x) = d/dx [4x/(x² + 2)] = (8 − 4x²)/(x² + 2)². At x = 1: f(1) = 4/9.
The velocity-to-position relationship here is the same one that underpins kinematics problems later in the course and in physics. If you go on to study differential equations, the initial value problems in this set are the simplest case of what that entire subject is about: finding a function from its derivative plus a starting condition.
The FTC Part 2 problem (Q20) connects to accumulation functions, which appear in AP-style and university exam questions that ask you to analyse the behaviour of F(x) = ∫₀ˣ f(t) dt using properties of f.
applications of integration, integrating a rate function, total from rate, velocity to position, displacement integral, helicopter height, oil production integral, initial value problem, IVP, finding the curve, slope to equation, y-intercept from derivative, position velocity direction, speed vs velocity, storm problem, FTC Part 2, differentiating an integral, quotient rule with FTC, accumulation function, integral calculus, UT Austin, HW 02