Integration, one idea at a timeMathematics Methods, Unit 3 & 4

Before the method: the idea behind it

Running the derivative backwards

Differentiation takes a function and tells you its rate of change. Integration is the reverse question: you’re handed the rate of change and asked for the original function. That alone would make it useful. What makes it remarkable is that the same operation also solves a completely different-looking problem, finding the area under a curve, and the fact that these two problems are secretly the same is one of the great discoveries in mathematics.

Part one

The reverse question

You’re given a car’s speedometer reading at every instant of a trip. Can you work out how far it travelled?

Speed is the derivative of distance. So this is a derivative-in-reverse question: from the rate, recover the total. The operation is called anti-differentiation, or integration, and it works for any function, not just speeds.

There’s a catch, and it’s an important one. If you know only the speed, you can never tell where the trip started. A trip from 0 km to 50 km and a trip from 200 km to 250 km have identical speedometer readings the whole way. So reversing a derivative always leaves one unknown: a constant, written c, standing for the starting value that the derivative can’t see.

y = x² + c, for different values of c the tangent at x = 1 never changes slope

Every curve here has the same derivative, 2x. Drag the slider to change c and watch the red tangent: it moves up and down with the curve, but its slope stays exactly the same.

Curve: y = x² + 0.00
Gradient at x = 1: 2.00, for every c

That’s why an anti-derivative always comes with a “+ c” attached: it isn’t one function but a whole family, one for each starting value, and the derivative is blind to which member you mean. If a question gives you one extra fact, such as a point the curve passes through, that pins down c and picks out the single curve you want.

\[ \int x^n\,dx = \dfrac{x^{n+1}}{n+1} + c \quad (n \ne -1), \qquad\qquad \int (ax+b)^n\,dx = \dfrac{(ax+b)^{n+1}}{a(n+1)} + c \]

The rule is the power rule run in reverse: add one to the power, then divide by the new power. For a bracket like (ax + b), you also divide by a, because the chain rule would have multiplied by a on the way forward. You can always check an anti-derivative by differentiating it, and you should.

Part two

The area problem

Now a different question entirely. A block of land has one curved boundary. How do you find its area?

Rectangles are easy; curves aren’t. So chop the region into thin vertical strips and treat each one as a rectangle. The answer is slightly wrong, because each rectangle’s top is flat while the curve isn’t. But the thinner the strips, the smaller the error. Push the number of strips towards infinity and the estimate settles on a single exact value. That limit is called the definite integral.

area under y = x² from 0 to 2 0 2

Drag the slider to use more rectangles. The left-hand estimate is always too small and the right-hand one always too big, but both close in on the same number.

Rectangles: 4
Left-hand sum: 1.750
Right-hand sum: 3.750
Exact area: 8/3 ≈ 2.667
\[ \int_a^b f(x)\,dx \ = \ \text{the exact area under } f \text{ from } x=a \text{ to } x=b, \text{ the limit of the rectangle sums} \]

Part three

The theorem that ties them together

Here is the surprise. The area problem in Part two is solved by the reverse-derivative from Part one.

Imagine sweeping a vertical line from a to the right, and tracking the area swept out so far. How fast is that area growing? Each tiny step to the right adds a thin sliver whose height is the curve’s height, f(x). So the rate of change of the accumulated area is f(x) itself. In other words, the area function is an anti-derivative of f. That single observation is the fundamental theorem of calculus, and it turns area, once a limit of ever-thinner rectangles, into two substitutions and a subtraction.

\[ \int_a^b f(x)\,dx = F(b) – F(a), \qquad \text{where } F'(x) = f(x) \]

Read it as a recipe: find any anti-derivative F (you can drop the + c, it cancels in the subtraction), evaluate it at the top limit, evaluate it at the bottom limit, subtract. For y = x² between 0 and 2: F(x) = x³/3, so the area is 8/3 − 0 = 8/3, exactly the number the rectangles were closing in on.

Where this shows up in real life

Your odometer integrates your speedometer

A car’s speedometer reports a rate; the odometer reports the total. The odometer is literally accumulating the area under the speed graph as you drive. Any time you go from a rate to a total, rainfall from a rain gauge’s rate, water from a flow meter, money from a marginal cost, you’re integrating.

Probability is an area

Later this year, the probability that a measurement lands between two values will be the area under a curve between those values, found with a definite integral. The normal distribution, the bell curve, is worked with almost entirely through integration.

Before you start

Calculator-free or calculator-assumed?

From this topic on, your ClassPad becomes a serious tool. It can find anti-derivatives, evaluate definite integrals, and solve for the constant c in one line. WACE exams have a calculator-free section where you must do these by hand, and a calculator-assumed section where using the ClassPad well saves time and prevents arithmetic slips. You need both skills, so every practice question is tagged with which one it’s training, and the ClassPad tab on the next page teaches the exact key sequences.

The question asks for…Reach for
the anti-derivative, ∫ … dx, “find y”, “determine f(x)”Reverse power rule, then + c
a specific function through a given pointAnti-derivative, then substitute the point to find c
an estimate using rectanglesAdd up width × height for each strip
an exact area, a definite integral with limitsFundamental theorem: F(top) − F(bottom)

Next: the working method for each, twenty practice questions, ten harder ones, and the ClassPad guide.