Integration 1 — Foundations & Trapezoidal Rule

Section 1 of 11

Anti-derivative — undoing differentiation

Differentiation goes one way. Anti-differentiation goes the other. If you can differentiate $y = x^{2}$ to get $\dfrac{dy}{dx} = 2x$, you should be able to start with $\dfrac{dy}{dx} = 2x$ and get back to $y$.

$y = x^{2} \;\;\Rightarrow\;\; \dfrac{dy}{dx} = 2x$ (differentiate)

$\dfrac{dy}{dx} = 2x \;\;\Rightarrow\;\; y = \;?$ (anti-differentiate)

Two opposite moves

Differentiate = Multiply by the power, then reduce the power by 1
Anti-differentiate = Increase the power by 1, then divide by the new power

Apply that to $\dfrac{dy}{dx} = 2x$:

Increase power: $2x \;\to\; 2x^{2}$

Divide by new power: $\dfrac{2x^{2}}{2} = x^{2}$

So: $y = x^{2}$ … almost. There's one thing missing.

Section 2 of 11

The constant $+c$ — never forget it

Here's the catch. If $y = x^{2}$ then $\dfrac{dy}{dx} = 2x$. But if $y = x^{2} + 1000$, then also $\dfrac{dy}{dx} = 2x$ — because the derivative of $1000$ is $0$.

$y = x^{2} \;\;\Rightarrow\;\; \dfrac{dy}{dx} = 2x$

$y = x^{2} + 1000 \;\;\Rightarrow\;\; \dfrac{dy}{dx} = 2x$ (same!)

$y = x^{2} - 7 \;\;\Rightarrow\;\; \dfrac{dy}{dx} = 2x$ (still same!)

When we anti-differentiate, we can't tell which constant was there. So we add an unknown constant $c$ to cover all possibilities.

Must learn

$\dfrac{dy}{dx} = 2x \;\;\Rightarrow\;\; y = x^{2} + c$

$c$ is a constant and must NOT be forgotten.

Try one more — $\dfrac{dy}{dx} = 2x + 3$:

$\dfrac{2x^{2}}{2} + \dfrac{3x^{1}}{1} + c$

$y = x^{2} + 3x + c$

Section 3 of 11

Find $f(x)$ when you know $f'(x)$

Same idea, but with function notation. Given $f'(x)$, find $f(x)$ by anti-differentiating each term.

(i) $f'(x) = x^{2} + 5x - 7$ — find $f(x)$

$f(x) = \dfrac{x^{3}}{3} + \dfrac{5x^{2}}{2} - 7x + c$

Increase each power by 1, divide by the new power. The constant $-7$ becomes $-7x$ (anti-derivative of a constant).

(ii) $f'(x) = 6x^{2} - 4x + 1$ — find $f(x)$

$f(x) = \dfrac{6x^{3}}{3} - \dfrac{4x^{2}}{2} + x + c$

$f(x) = 2x^{3} - 2x^{2} + x + c$

You try

Given $f'(x) = 3x^{2} + 8x - 5$, find $f(x)$.

Anti-differentiate each term. Don't forget $+c$.

$f(x) = \dfrac{3x^{3}}{3} + \dfrac{8x^{2}}{2} - 5x + c$

$f(x) = x^{3} + 4x^{2} - 5x + c$

You try — one more

$f'(x) = 10x^{4} - 9x^{2} + 6x - 1$. Find $f(x)$.

Term by term — increase power, divide by new power. $-1$ becomes $-x$.

$f(x) = \dfrac{10x^{5}}{5} - \dfrac{9x^{3}}{3} + \dfrac{6x^{2}}{2} - x + c$

$f(x) = 2x^{5} - 3x^{3} + 3x^{2} - x + c$

Section 4 of 11

Finding $c$ — when you know one point

If the question gives you a point on the curve — like $f(1) = 11$ — you can pin down $c$ exactly.

Three-step recipe

(1) Anti-differentiate to get $f(x)$ with $+c$
(2) Sub in the given point to find $c$
(3) Write $f(x)$ with the actual $c$

Worked — $f'(x) = 4x + 5$, $f(1) = 11$. Find $f(x)$.

(1) $f(x) = \dfrac{4x^{2}}{2} + 5x + c$

$\phantom{(1)} \;\;\; f(x) = 2x^{2} + 5x + c$

(2) $f(1) = 2(1)^{2} + 5(1) + c = 11$

$\phantom{(2)} \;\;\; 2 + 5 + c = 11$

$\phantom{(2)} \;\;\; c = 4$

(3) $f(x) = 2x^{2} + 5x + 4$

You try

$\dfrac{dy}{dx} = 6x - 2$ and the curve passes through $(2, 7)$. Find $y$.

Anti-differentiate, then sub $x=2, y=7$ to find $c$.

$y = \dfrac{6x^{2}}{2} - 2x + c = 3x^{2} - 2x + c$

Sub $(2,7)$: $7 = 3(4) - 2(2) + c$

$7 = 12 - 4 + c \;\;\Rightarrow\;\; c = -1$

$y = 3x^{2} - 2x - 1$

You try — one more

The slope of the tangent to a curve at $(x, y)$ is $\dfrac{dy}{dx} = 2x - 4$, and the curve passes through $(3, 0)$. Find the equation of the curve.

Anti-differentiate, then sub $x=3,\,y=0$ to find $c$. Watch the signs.

$y = \dfrac{2x^{2}}{2} - 4x + c = x^{2} - 4x + c$

Sub $(3,0)$: $0 = 9 - 12 + c$

$c = 3$

$y = x^{2} - 4x + 3$

Section 5 of 11

Area — a quick refresher

Before we tackle areas under curves, let's make sure we can do areas of straight-sided shapes. Three you must know:

Rectangle — length $\times$ breadth

$A = \ell \times b = 10 \times 5 = 50 \text{ m}^{2}$

Triangle — $\tfrac{1}{2}$ base $\times$ height

$A = \tfrac{1}{2} \, b \, h$

Trapezium — two parallel sides

A trapezium has two parallel sides. Easiest way to see why the formula works: split it into two triangles by drawing a diagonal.

$T_{1} = \tfrac{1}{2}(12)(4) = 24$

$T_{2} = \tfrac{1}{2}(8)(4) = 16$

Total: $24 + 16 = 40 \text{ m}^{2}$

Same answer from the trapezium formula:

Trapezium formula

$A = \left(\dfrac{a + b}{2}\right) h$

where $a$, $b$ are the parallel sides and $h$ is the perpendicular distance between them — the height.

$A = \left(\dfrac{12 + 8}{2}\right)(4) = (10)(4) = 40 \text{ m}^{2}$ ✓

You try

A trapezium has parallel sides of $14$ m and $10$ m, separated by a perpendicular distance of $3$ m. Find its area — both ways: as two triangles, and using the trapezium formula.

Two triangles share the diagonal as base — one has base $14$, the other $10$. Both have height $3$.

As two triangles:

$T_{1} = \tfrac{1}{2}(14)(3) = 21$

$T_{2} = \tfrac{1}{2}(10)(3) = 15$

Total $= 21 + 15 = 36 \text{ m}^{2}$

Trapezium formula:

$A = \left(\dfrac{14 + 10}{2}\right)(3) = (12)(3) = 36 \text{ m}^{2}$ ✓

$A = 36 \text{ m}^{2}$

Hold onto the trapezium formula. It's the key to estimating areas under curves.

Section 6 of 11

Splitting a curved shape into strips

Now suppose the top is curved — like a hill, or the cross-section of a river bank. There's no neat formula. But you can chop it into strips of equal width $h$, and each strip is almost a trapezium.

Each strip is approximately a trapezium. The two parallel sides are the ordinates on the left and right; the width $h$ is the height between them.

$A_{1} = \left(\dfrac{y_{1} + y_{2}}{2}\right) h$

$A_{2} = \left(\dfrac{y_{2} + y_{3}}{2}\right) h$

$A_{3} = \left(\dfrac{y_{3} + y_{4}}{2}\right) h$

$A_{4} = \left(\dfrac{y_{4} + y_{5}}{2}\right) h$

Add them all up:

$A = \dfrac{h}{2}\left[\, y_{1} + y_{2} + y_{2} + y_{3} + y_{3} + y_{4} + y_{4} + y_{5} \,\right]$

The middle ones each appear twice. Group them:

$A = \dfrac{h}{2}\left[\, y_{1} + y_{5} + 2(y_{2} + y_{3} + y_{4}) \,\right]$

That's the pattern. First + last, plus twice the middles.

Section 7 of 11

The Trapezoidal Rule

That same pattern works for any number of strips. Replace $y_{5}$ with $y_{n}$ — the last ordinate — and you have the rule.

Trapezoidal Rule — must learn

$A \approx \dfrac{h}{2}\Big[\, y_{1} + y_{n} + 2\big(y_{2} + y_{3} + \ldots + y_{n-1}\big) \Big]$

$y_{1}$ = first ordinate · $y_{n}$ = last ordinate
The ones in the middle get doubled. $h$ = strip width.

It's an estimate. The curved top of each strip isn't exactly a straight line, so the answer is approximate — but very close, and good enough for the exam.

Counting strips and ordinates. If you have $n$ ordinates (vertical lines), you have $n-1$ strips. Easy to mix up — count carefully.

5 ordinates → 4 strips

7 ordinates → 6 strips

If the total width is $W$ and there are $n-1$ strips, then $h = \dfrac{W}{n-1}$.

Section 8 of 11

Trapezoidal Rule — worked example

The cross-section of a piece of land is measured with seven vertical heights, each $6$ m apart. Find the area.

Seven ordinates, six strips, $h = 6$ m. First and last are $y_{1} = 5.5$ and $y_{7} = 5.4$. The middles are $y_{2}, y_{3}, y_{4}, y_{5}, y_{6}$.

$A \approx \dfrac{h}{2}\Big[\, y_{1} + y_{7} + 2(y_{2} + y_{3} + y_{4} + y_{5} + y_{6}) \,\Big]$

$A \approx \dfrac{6}{2}\Big[\, 5.5 + 5.4 + 2(7.3 + 9.1 + 10.2 + 10.1 + 8.7) \,\Big]$

$A \approx 3 \Big[\, 10.9 + 2(45.4) \,\Big]$

$A \approx 3 \big[\, 10.9 + 90.8 \,\big]$

$A \approx 3 \times 101.7$

$A \approx 305.1 \text{ m}^{2}$

You try

A piece of land has six measured heights — $6, 7, 5, 5, 4, 0$ metres — taken at equal intervals. The total width is $15$ m. Use the Trapezoidal Rule to estimate the area.

6 ordinates → 5 strips. So $h = \dfrac{15}{5} = 3$ m. First = $6$, last = $0$, middles = $7, 5, 5, 4$.

$h = \dfrac{15}{5} = 3$

$A \approx \dfrac{3}{2}\big[\, 6 + 0 + 2(7+5+5+4) \,\big]$

$A \approx \dfrac{3}{2}\big[\, 6 + 42 \,\big]$

$A \approx \dfrac{3}{2}(48)$

$A \approx 72 \text{ m}^{2}$

$72 \text{ m}^{2}$

You try — one more

The values of a function $y = f(x)$ are tabulated at five equally spaced values of $x$ from $x = 1$ to $x = 9$:

$x$	$1$	$3$	$5$	$7$	$9$
$y$	$2$	$6$	$8$	$7$	$3$

Use the Trapezoidal Rule to estimate the area between the curve and the $x$-axis from $x = 1$ to $x = 9$.

5 ordinates, 4 strips. $h = 2$. First = $2$, last = $3$, middles = $6, 8, 7$.

$h = 2$ (the $x$-values jump by $2$ each time)

$A \approx \dfrac{2}{2}\big[\, 2 + 3 + 2(6 + 8 + 7) \,\big]$

$A \approx 1 \big[\, 5 + 2(21) \,\big]$

$A \approx 5 + 42$

$A \approx 47 \text{ sq units}$

Section 9 of 11

The bridge — anti-derivative finds areas

Here's the beautiful link. Take two shapes whose areas we already know — and find them again using anti-derivatives. If both methods agree, we're onto something.

Triangle — area between $y = x$ and the $x$-axis from $x = 0$ to $x = 2$

Way 1 — triangle formula.

$A = \tfrac{1}{2} \, b \, h = \tfrac{1}{2}(2)(2) = 2$ sq units

Way 2 — anti-derivative of $f(x) = x$, evaluated from $0$ to $2$.

Anti-derivative: $\dfrac{x^{2}}{2}$

At $x = 2$: $\dfrac{2^{2}}{2} = \dfrac{4}{2} = 2$

At $x = 0$: $\dfrac{0^{2}}{2} = 0$

$A = 2 - 0 = 2$ sq units ✓

Both methods agree. Now check it works for a rectangle too.

Rectangle — area between $y = 5$ and the $x$-axis from $x = 0$ to $x = 3$

Way 1 — rectangle.

$A = 5 \times 3 = 15$ sq units

Way 2 — anti-derivative of $f(x) = 5$.

Anti-derivative: $5x$

At $x = 3$: $5(3) = 15$

At $x = 0$: $5(0) = 0$

$A = 15 - 0 = 15$ sq units ✓

The discovery

Area under a curve = Anti-derivative at the right end − Anti-derivative at the left end.

This is the link between two things that seemed unrelated — differentiation and area. They turn out to be opposite sides of the same coin.

You try

Use the anti-derivative method to find the area between $y = 4$ and the $x$-axis from $x = 0$ to $x = 8$. Then check it with the rectangle formula.

Anti-derivative of $4$ is $4x$. Evaluate at $8$ and at $0$, then subtract.

Anti-derivative of $f(x) = 4$ is $4x$.

At $x = 8$: $4(8) = 32$

At $x = 0$: $4(0) = 0$

$A = 32 - 0 = 32$ sq units

Check: rectangle $= 4 \times 8 = 32$ ✓

$A = 32$ sq units

Section 10 of 11

The integral sign — $\int$ — and what it really means

For straight-sided shapes we used the trapezium formula. For irregular shapes we approximated with strips. The big idea behind integration is to imagine the strips getting thinner and thinner — until they're infinitely thin.

Each thin strip has width $dx$ (infinitely small) and height $f(x)$. So one strip's area is $f(x)\,dx$.

Area of one slice: $f(x) \, dx$

Total area: $f(x_{1})\,dx + f(x_{2})\,dx + f(x_{3})\,dx + \ldots$

Add up infinitely many slices. The Greek $\sum$ we use for finite sums becomes the long $\int$ for infinite sums:

The integral — must learn

Area $= \displaystyle\int f(x) \, dx$

$\displaystyle\int$ = "sum of" · an elongated S
$f(x) \, dx$ = one tiny slice · height $\times$ width
The whole symbol reads: "sum all the tiny slices $f(x)\,dx$"

And the punchline:

Integration = Anti-derivative

The way we compute $\displaystyle\int f(x)\,dx$ is by anti-differentiating $f(x)$.
That's the link Section 9 showed us: anti-derivatives produce areas.

Section 11 of 11

What's next

You now have the three foundations of integration on Paper 1:

• The anti-derivative — undoing differentiation, always with $+c$
• The Trapezoidal Rule — for estimating areas from a list of heights
• The integral $\displaystyle\int$ — anti-derivative as a sum of infinitely thin slices

Coming next. In Integration — Part 2 we work through the 8 indefinite-integration methods (Multiply out, Tables, Factors, Split, Indices, $e^{x}$, Trig with Products-to-Sums, and the Special Case). Then in Part 3 we use definite integrals to find areas under curves — the long-question topic.

That's Integration — Part 1.

Foundations and the Trapezoidal Rule. Next: the methods for working out $\displaystyle\int f(x)\,dx$ for every kind of $f$.