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Integrals

lesson · about 3 minutes

Accumulation: areas, averages, and the fundamental theorem.

Pen and paper is fine · no calculator needed why?

Opens at level 41. You're level 1. You can read and practice here now.

the idea

A definite integral, written ∫ₐᵇ f(x) dx, adds up f from x = a to x = b: the area under the graph, with parts below the axis counting as negative.

To work one out, find an antiderivative F, a function whose derivative is f, and take F(b) − F(a). That shortcut is the fundamental theorem of calculus. It also runs the other way: differentiating ∫ₐˣ g(t) dt gives back g(x).

Integrals give averages too. The average value of f over an interval is its integral divided by the interval's length.

techniques

Antiderivative, then top minus bottom

A definite integral of a polynomial, or an average value.

Raise each power by one and divide by the new power: 6x² becomes 2x³.
A plain number c becomes cx.
Answer: F(top limit) − F(bottom limit).
For an average value, divide that by the interval’s length, b − a.

worked example

Example: What is ∫₁² (3x² + 4x) dx?

Antiderivative: F(x) = x³ + 2x².
F(2) = 8 + 8 = 16. F(1) = 1 + 2 = 3.
16 − 3 = 13.

Answer: 13

Differentiate an integral

F(x) = ∫ₐˣ g(t) dt, and you need F′ at a number.

F′(x) = g(x): differentiating undoes the integrating.
Put the number into g. The lower limit a doesn’t matter.

worked example

Example: F(x) = ∫₂ˣ (t² − 3t) dt. What is F′(4)?

F′(x) = x² − 3x, the integrand with t replaced by x.
F′(4) = 16 − 12 = 4.

Answer: 4

Area between curves

Two curves meet at x = p and x = q, and you need the area between them.

Test one x between p and q to see which curve is on top.
Subtract: top minus bottom, as one polynomial.
Integrate that from p to q. The area comes out positive.

worked example

Example: The curves y = x² and y = 2x meet at x = 0 and x = 2. What is the area of the region between them? Enter a fraction or a whole number.

At x = 1, 2x is 2 and x² is 1, so y = 2x is on top.
Top minus bottom: 2x − x². Antiderivative: x² − x³/3.
At 2: 4 − 8/3 = 4/3. At 0 it is 0.

Answer: 4/3

watch out for

Raising the power without dividing by the new power. The antiderivative of 3x² is x³, not 3x³.
Giving the integral as the average value. Divide it by the interval’s length.
Working out F(c) when asked for F′(c). The theorem hands you the integrand at c, with nothing to integrate.
Subtracting the top curve from the bottom one. An area is never negative.

practice

Definite integrals

worked example

What is ∫₂³ (−4x − 4) dx?

Answer: -14

Antiderivative: F(x) = −2x² − 4x.
F(3) = −30 and F(2) = −16.
So the integral is −30 − (−16) = −14.

Average value

worked example

What is the average value of f(x) = x² on the interval [2, 5]?

Answer: 13

Average value = (integral of f over the interval) ÷ (length of the interval).
An antiderivative is F(x) = x³/3, and F(5) − F(2) = 125/3 − 8/3 = 39.
The length is 5 − 2 = 3, so the average value is 39 ÷ 3 = 13.

Fundamental theorem

worked example

F(x) = ∫₃ˣ (−3t + 5) dt. What is F′(−3)?

Answer: 14

The fundamental theorem of calculus: the derivative of ∫₃ˣ g(t) dt is g(x), so F′(x) = −3x + 5.
The lower limit 3 doesn’t matter: F′(−3) = −3(−3) + 5 = 9 + 5 = 14.

Area between curves

worked example

The curves y = x² + 2x − 9 and y = 2x² − 3x − 3 meet at x = 2 and x = 3. What is the area of the region between them? Enter a fraction or a whole number.

Answer: 1/6

Between x = 2 and x = 3, y = x² + 2x − 9 is on top.
Top minus bottom is −x² + 5x − 6; an antiderivative is −x³/3 + 5x²/2 − 6x.
It is −9/2 at x = 3 and −14/3 at x = 2: −9/2 − (−14/3) = 1/6.