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🌌 Calculus Kingdom · Calculus

Applications of Integrals

Put the definite integral to work: the area between two curves, the average value of a function, the derivative of an accumulation function, displacement against distance, substitution for a composite integrand, and total change from a rate in context.

In short

  • Area between curves is the integral of (top - bottom). Decide which curve is on top before integrating, and split the region wherever they cross.
  • The average value of f on [a, b] is the integral divided by the width b - a: the height of the rectangle with the same area. The Mean Value Theorem for Integrals says the curve actually reaches that height somewhere in the interval.
  • The first part of the fundamental theorem says d/dx of the integral from a to x of f(t) dt is f(x). A variable upper limit other than x brings the chain rule with it; x in the lower limit brings a minus sign.
  • An integral of a rate is a total change, so it is signed and it is not a level. Displacement needs the starting position added, total distance needs the interval split where the velocity changes sign.

The area between two curves

A definite integral of a positive function measures the area trapped between the graph and the x-axis. Trade the x-axis for a second curve and the same idea measures the area trapped between the two.

area = integral from a to b of (top curve - bottom curve) dx

The subtraction is the whole trick. At each x, the height of the region is the top value minus the bottom value, and the integral adds up those heights across the interval.

Two decisions come before any integrating:

  • Which curve is on top? Test one x-value inside the interval and compare. If the two curves cross inside the interval, the region has to be split at the crossing, because the top curve changes.
  • Where does the region start and end? Often the limits are the points where the curves meet. Set the two expressions equal, bring everything to one side and solve.

An example. y = x + 6 and y = x2 meet where x2 = x + 6, that is x2 - x - 6 = 0, so x = -2 and x = 3. Testing x = 0 gives 6 for the line and 0 for the parabola, so the line is on top:

area = integral from -2 to 3 of (x + 6 - x2) dx = 125/6

The order matters and the answer tells you so. Subtracting the wrong way round gives -125/6. An area is never negative, so a minus sign in an area answer is a signal to swap the two functions round, not to write a minus sign down.

The average value of a function

The average of five numbers is their sum divided by five. A function has infinitely many values on an interval, so the sum becomes an integral and the count becomes the width:

average value of f on [a, b] = (1/(b - a)) x integral from a to b of f(x) dx

Read it as a picture. The integral is the area under the curve. Divide that area by the width and you get the height of the rectangle with the same area — the level the curve would sit at if it were flattened out.

The division is the step that gets forgotten. An integral has the units of f times the units of x; an average value has the units of f alone. Water flowing for 6 minutes might accumulate 90 litres, but the average rate is 90/6 = 15 litres per minute, not 90.

The Mean Value Theorem for Integrals says something stronger: if f is continuous on [a, b] then there is some c in [a, b] with

f(c) = the average value of f on [a, b]

The curve really does pass through its own average height somewhere. To find that c, work out the average value first, then solve f(c) = that value, keeping only the solution inside the interval.

For a straight line, c is always the midpoint (a + b)/2 — a useful check, and a useful warning: for anything that bends, the midpoint is the wrong answer. For f(x) = 4/x2 on [1, 9], the integral is 4/1 - 4/9 = 32/9, the average value is (32/9)/8 = 4/9, and solving 4/c2 = 4/9 gives c = 3, not 5.

The fundamental theorem, part one

Fix a lower limit and let the upper limit run. What you get is an accumulation function:

F(x) = integral from a to x of f(t) dt

F(x) is the amount of signed area f has piled up between a and x. The letter t inside is a dummy: it is there only so that the variable being integrated is not confused with the variable x that marks where to stop.

The first part of the fundamental theorem says that differentiating this undoes the accumulating:

d/dx [ integral from a to x of f(t) dt ] = f(x)

No integrating is needed at all. Copy the integrand out and write x where t stood. The lower limit a moves the value of F up or down but never its rate of change, so it plays no part in the answer.

Two adjustments cover almost everything else.

A variable upper limit that is not just x brings in the chain rule. With u on top instead of x, substitute u into the integrand and multiply by du/dx:

d/dx [ integral from 1 to x2 of (t2 + 1) dt ] = ((x2)2 + 1) x 2x = 2x(x4 + 1)

x underneath instead of on top costs a minus sign, because swapping the limits of a definite integral flips its sign:

d/dx [ integral from x to 5 of f(t) dt ] = d/dx [ -integral from 5 to x of f(t) dt ] = -f(x)

Because F'(x) = f(x), everything already known about derivatives transfers: F is increasing exactly where f is positive, and F has a maximum where f changes from positive to negative.

Motion: displacement is not distance

Velocity is the rate of change of position, so integrating velocity recovers the change in position:

displacement from a to b = integral from a to b of v(t) dt

Displacement is signed. Time spent moving backwards subtracts, and it is entirely possible for a cart to travel for ten seconds and have a displacement of zero.

Total distance ignores direction, so it is the integral of the speed:

total distance from a to b = integral from a to b of the size of v(t) dt

In practice that means: find where v changes sign, split the interval there, integrate each piece separately, make each piece positive, then add.

Take v(t) = t - 2 metres per second on 0 to 5 seconds. The velocity is negative before t = 2 and positive after.

from 0 to 2: integral = -2 from 2 to 5: integral = 9/2 displacement = -2 + 9/2 = 5/2 metres total distance = 2 + 9/2 = 13/2 metres

Two different questions with two different answers, from the same integral.

A position is not a displacement either. An integral gives a change, and a change has to be added to where the object already was:

s(b) = s(a) + integral from a to b of v(t) dt

Substitution, and accumulation in context

Substitution is the chain rule run backwards. When the integrand contains an inside function whose derivative is also present, name the inside u and the integral collapses:

integral of 2x(x2 + 5)3 dx let u = x2 + 5, du = 2x dx = integral of u3 du = u4/4 = (x2 + 5)4/4 + C

When the derivative of the inside is only a constant multiple away, the constant is fixed up on the way through:

integral of (3x + 2)4 dx let u = 3x + 2, du = 3 dx, so dx = du/3 = (3x + 2)5/15 + C integral of e2x dx = e2x/2 + C integral of cos(4x) dx = sin(4x)/4 + C, with x in radians integral of x sqrt(x^2 + 9) dx let u = x2 + 9, x dx = du/2 = (x2 + 9)3/2/3 + C

Every one of these can be checked in seconds by differentiating it back, and that check catches the missing constant every time.

For a definite integral there are two honest routes: substitute back into x and use the original limits, or change the limits into u-values and never return to x. What is not allowed is to change the variable and keep the old limits.

Accumulation in context is the same mathematics wearing work clothes. If r(t) is a rate — litres per minute, people per hour, centimetres per year — then

integral from a to b of r(t) dt = the total change between t = a and t = b

Check it with units: litres per minute, multiplied by minutes, leaves litres. The integral of a rate is always an amount, never a rate.

And an amount is not a level. If a cistern holds 40 litres at t = 2 and water flows in at r(t) = 3t2 + 4t + 5 litres per minute, then the amount at t = 5 is

40 + integral from 2 to 5 of (3t2 + 4t + 5) dt = 40 + 174 = 214 litres

Leave the 40 out and you have answered a question nobody asked.

Worked examples

Example 1

Find the area of the region enclosed by y = 2x + 3 and y = x2.

  1. Find where they meet: x2 = 2x + 3, so x2 - 2x - 3 = 0, giving (x - 3)(x + 1) = 0 and x = -1, x = 3.
  2. Decide which is on top inside the interval. At x = 0 the line gives 3 and the parabola gives 0, so the line is above.
  3. Set up the integral: area = integral from -1 to 3 of (2x + 3 - x2) dx.
  4. An antiderivative is x2 + 3x - x3/3.
  5. At x = 3 that is 9 + 9 - 9 = 9. At x = -1 it is 1 - 3 + 1/3 = -5/3.
  6. Area = 9 - (-5/3) = 32/3. It is positive, which confirms the two curves were subtracted the right way round.

Example 2

A cart has velocity v(t) = t2 - 4t metres per second for t between 0 and 5 seconds, and it is at s(0) = 10 metres. Find (a) the displacement, (b) the total distance travelled, (c) its position at t = 5.

  1. Find where the velocity changes sign: t2 - 4t = t(t - 4) = 0 at t = 0 and t = 4, so t = 4 is the only sign change inside the interval.
  2. An antiderivative is s(t) = t3/3 - 2t2.
  3. (a) Displacement = s(5) - s(0) = (125/3 - 50) - 0 = -25/3 metres. Negative: the cart finishes behind where it started.
  4. (b) Split at t = 4. From 0 to 4 the integral is (64/3 - 32) - 0 = -32/3. From 4 to 5 it is (125/3 - 50) - (64/3 - 32) = 7/3.
  5. Make each piece positive and add: total distance = 32/3 + 7/3 = 13 metres. Notice it is larger than the size of the displacement, as it must be.
  6. (c) Position = starting position + displacement = 10 + (-25/3) = 5/3 metres.

Example 3

F(x) is the integral from 2 to x2 of sqrt(t^2 + 5) dt. Find F prime of x.

  1. The first part of the fundamental theorem applies to a variable upper limit, and here the upper limit is x2 rather than x.
  2. Write u = x2, so F is the accumulation up to u and du/dx = 2x.
  3. Evaluate the integrand at the upper limit: sqrt(u^2 + 5) = sqrt(x^4 + 5).
  4. Multiply by du/dx, which the chain rule demands: F prime of x = 2x sqrt(x^4 + 5).
  5. Check the shape of the answer: it is still a function of x, and the lower limit 2 has vanished, as it always does.

Practice problems, with solutions

Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.

Problem 1

Difficulty 1 of 5

Find the area of the region between y = 2x2 + 2x - 1 and y = 3, from x = -2 to x = 1. Give an exact answer (a fraction in lowest terms if it is not a whole number).

Answer: 9

  1. Set them equal: 3 = 2x2 + 2x - 1, so -2x2 - 2x + 4 = 0, which happens at x = -2 and x = 1.
  2. At x = -0.5 the two curves are worth 3 and -1.5, so y = 3 is the top curve on that stretch.
  3. Top - bottom = -2x2 - 2x + 4.
  4. An antiderivative is -(2/3)x3 - x2 + 4x.
  5. Area = value at 1 minus value at -2 = 9.

Problem 2

Difficulty 3 of 5

Find the average value of f(x) = 2x2 + 5 on the interval [1, 3]. Give an exact answer (a fraction in lowest terms if it is not a whole number).

Answer: 13 2/3

  1. An antiderivative is F(x) = (2/3)x3 + 5x.
  2. Integral = F(3) - F(1) = 82/3.
  3. Width of the interval = 3 - (1) = 2.
  4. Average value = (82/3) / 2 = 41/3.

Problem 3

Difficulty 4 of 5

F(x) is the integral from 0 to x of f(t) dt, where f(t) = t2 - 5t + 4. f is zero at t = 1 and t = 4, negative between them and positive outside them. On which x-values is F increasing?

  1. x > 4
  2. 1 < x < 4
  3. x < 1
  4. x < 1 or x > 4

Answer: D. x < 1 or x > 4

  1. By the fundamental theorem, F'(x) = f(x) = x2 - 5x + 4.
  2. f is positive for x < 1 and for x > 4, and negative in between.
  3. So F is increasing on x < 1 or x > 4.

Common mistakes

  • Subtracting the curves the wrong way round, and writing down a negative area. Bottom minus top gives the right size with the wrong sign — swap them rather than keeping the minus.
  • Finding the integral and calling it the average value. An average value divides that integral by the width of the interval, b - a, and dividing by the right endpoint b instead is the same error wearing a disguise.
  • Forgetting the chain factor in the fundamental theorem. With x2 on top, the answer is f(x2) multiplied by 2x; substituting into the integrand and stopping there loses the whole outer factor.
  • Reporting displacement when the question asked for total distance. Integrating straight across lets the backwards stretch cancel the forwards one, so the interval has to be split where the velocity changes sign and each piece made positive before adding.
  • Substituting u without adjusting for du. The integral of e2x is e2x/2, not e2x: differentiate any antiderivative back and the spare factor shows up immediately.
  • Answering a net-change question with the change alone. If the tank already held 40 litres, the amount at the end is 40 plus the integral, not the integral by itself.

What you should be able to do

  • Find the area between two curves, and the area against the axis where a curve dips below.
  • Find the average value of a function on an interval.
  • Differentiate an accumulation function with the first part of the fundamental theorem.
  • Find displacement, distance and net change with an integral, using substitution where the integrand asks for it.

Where this fits in the curriculum

Common Core

  • CHA-5.A

    AP Calculus AB, Unit 8 — Calculate areas in the plane using the definite integral.

    The Common Core has no calculus standards; this is the AP Calculus AB learning objective.

  • CHA-4.B

    AP Calculus AB, Unit 8 — Determine the average value of a function using definite integrals.

    The Common Core has no calculus standards; this is the AP Calculus AB learning objective.

  • CHA-4.C

    AP Calculus AB, Unit 8 — Determine values for positions and rates of change using definite integrals in problems involving rectilinear motion.

    The Common Core has no calculus standards; this is the AP Calculus AB learning objective.

  • CHA-4.E

    AP Calculus AB, Unit 8 — Determine net change using definite integrals in applied contexts.

    The Common Core has no calculus standards; this is the AP Calculus AB learning objective.

  • FUN-5.A

    AP Calculus AB, Unit 6 — Represent accumulation functions using definite integrals, and differentiate them.

    The Common Core has no calculus standards; this is the AP Calculus AB learning objective.

  • FUN-6.D

    AP Calculus AB, Unit 6 — For integrands requiring substitution or rearrangement into equivalent forms, determine indefinite integrals and evaluate definite integrals.

    The Common Core has no calculus standards; this is the AP Calculus AB learning objective.

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