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We need to learn to slice flat food first. After it's been de-boned, a fish such as salmon is relatively flat.

If we're planning on making fish sticks (we totally are), we'll need to cut that salmon into thin strips. We slice the fish from top to bottom into nice, even strips. We want them even so the 5-year-olds don't get into a fight at the dinner table.

We do the same thing when we break the area under a curve up into rectangular strips. We can use a left-hand sum to approximate the definite integral of a non-negative function.

If we want to approximate

where "f" stands for fish, we have to know the function *f*(*x*) and the interval [*a*,*b*]:

We slice up the region between the function and the *x*-axis into *n* slices. Each slice has the same width, Δ *x*:

Look at the slice whose left endpoint is at coordinate *x*:

The area of this slice is approximately the same as the area of the rectangle with width Δ *x* and height *f*(*x*).

The area of the rectangle is

*f*(*x*) Δ *x*.

To approximate the area between *f*(*x*) and the *x* axis on [*a*,*b*], we add up the areas of all *n* rectangles.

If we take the limit of the sum as the number of rectangles *n* approaches ∞, the intervals Δ *x* get really small. We get the exact area between *f*(*x*) and the *x*-axis on [*a*,*b*]:

In other words, a definite integral is a sum in the limit that we sum over an infinite number of intervals.

We're slicing the region into infinitely many slices of infinitesimal width.

We use *dx* instead of Δ *x* for the infinitesimal width because the "change in *x*" is too small to measure.

Here are a couple review examples to remind us of how to slice areas into vertical intervals.

Our general plan of attack for these problems is pretty straightforward.

1) Figure out the region whose area you're finding. Catch a fish.

2) Slice up the region into strips and find the approximate area of each strip. Make fish sticks.

3) Add up all the approximate areas and take the limit as the number of strips goes to ∞ to get an integral that gives the exact area of the region. Eat an infinite number of fish sticks.

We need to be careful here. While following this plan of attack, there are a few things we need to keep in mind.

*We should make sure that we know the region whose area we're finding.*

For example, the area of the region between the graph of *y* = *x*^{2} and the line *y* = 1 is this:

Not this:

and not this:

If we don't start with the correct region, we won't find the correct area. It's like trying to find the size of an apple by weighing an elephant. As a first step, it's useful to *draw a picture of the region.*

*We should make sure to know which way we're slicing the region and what our variable of integration means.*

Whichever way we slice it, the variable of integration somehow tells the position of a strip. If we slice horizontally, then *y* tells the position of the strip by giving its distance from the *x*-axis. If we slice vertically, then *x* tells the position of the strip by giving its distance from the *y*-axis. Whenever we finish writing an integral with *dy* or *dx*, we should make sure we know what *y* or *x* is. Not knowing what our variable of integration means is like trying get a glass of water in the dark at night without our glasses on. We *might* find what we're looking for, but it'll be mostly by chance.

*The limits of integration are whatever values make sense for the variable of integration.*

In the last exercise, we found the area of the region between the *y*-axis, the line *y* = 2*x*, and the line *y* = 2:

If we slice vertically and integrate with respect to *x*, the limits of integration are 0 to 1 because those are the values that make sense for *x*. We end up with an integral

If we slice horizontally and integrate with respect to *y*, then the limits of integration are 0 to 2 because those are the values that make sense for *y*. In this case, we end up with an integral

### Assumptions When Finding Area

We're going to be doing

*a lot*of problems making fish sticks or pizza slices or linguine pasta out of different foods. Since mathematicians are lazy, and because we should be focused more on our knife skills–we like our fingers–we're going to make some assumptions to save time and space later.From now on, assume that

- all vertical slices have width Δ
*x*and all horizontal slices have thickness Δ*y*. That way we don't have to keep saying "take a vertical slice of a guava with width Δ*x*."

- when we say "the height of the slice is ..." we mean "the
*approximate*height of the slice is ...". Similarly when we say "the area of the slice is ..." we mean "the*approximate*area of the slice is ..." That way we don't have to keep saying the word "approximately." As much as we like that word, we're going to sound like a Star Wars drone if we keep saying it.

- once we know the area of a slice, we can write down the integral that gives the area of the whole region. We don't have to write "we add up all the approximate regions and take the limit as the number of slices goes to ∞ to get an integral" every single time.

- if a problem asks us to "write an integral expression" for the area of a region, that means we don't need to evaluate the integral. If you want to practice evaluating integrals, then by all means, feel free to do them. We're focusing on how to set up the integrals in this chapter, which is the most difficult part for most people. We can save ourselves the headache of solving them for another day.

- all vertical slices have width Δ
### Triangles

We've made it. We survived years of seemingly endless lectures about trigonometry, proofs, and functions. Our 7

^{th}grade math teacher rattled off formula after formula for the areas and volumes of different polygons and solids. All we could do is sit there, listen, and memorize every formula we were given.We were like power lines transporting the power—our geometry formula—from a power plant that produced the formula to the paper or project, the formula's final home. With calculus, we'll never make a memory error about a length, area, or volume formula again. If we aren't sure, we can produce our own formula on the spot.

With our new found power, we'll begin our production slowly. Triangles are the simplest and most useful polygons we come across, so we'll

*derive*the formula for the area of a triangle first.### Circles

A circular pizza is made of triangular-like slices with toppings like pepperoni, peppers, and everyone's favorite, fried green tomatoes. In this section, we'll use our knowledge of the areas and deliciousness of triangles to build the areas of circles.

When we found the areas of triangles, we used the similar triangle trick. Back in the dark ages of our mathematical youth, we derived the Pythagorean Theorem using triangles. For circles, we're going to use the Pythagorean Theorem as a trick for finding the areas of circles.

### Sample Problem

Write an integral expression for the area of half a circular pepperoni pizza.

Answer.

There are (at least) two ways to do this. We could find the area of a quarter-circle slice and then multiply by 2, or we could take horizontal slices. We're going to take horizontal slices, because it's better to have more slices to share with friends.

We have another choice to make: how to measure the position of the slice. We could measure the depth of the slice below the top of the circle, or the height of the slice from the center of the circle. We'll measure its position from the center of the circle, using

*y*for the height of the slice. Since the distance from the center of this circle to any point on its edge is 4, a beautiful right triangle shows up.Looking at our right triangle, we see the length of the slice at height

*y*is 2*x*, where*x*^{2}+*y*^{2}= 4^{2}.Rearranging this equation, thank you Pythagorus, we get

The thickness of the slice is Δ

*y*since*y*is the variable of integration, which means the area of the slice at height*y*isSince

*y*moves from 0 at the bottom of the half-circle to 4 at the top of the half-circle. The area of the half-circle isIf you evaluate this integral on your calculator, you'll get approximately

which is the area of a circle of radius 4, divided by 2.

If that example seemed harder than getting three people to agree on topping choices for a pizza, no worries. There's another way to slice up a circle to find its area.

One of our favorites is slicing it into concentric rings like a target. When we do this, we use the letter

*r*for the variable of integration. The variable*r*tells how far a ring is from the center of the circle:If we had a circle of radius 6, the innermost ring would be at

*r*= 0 and the outermost ring would be at*r*= 6:To find the area using rings, we have look at one ring at a time. If we snip the ring open and straighten it out, we get something that's more or less a very, very, very skinny rectangle. The ring is like a piece of very thin spaghetti-pizza. We can take this spaghetti-pizza slice, lay it down in a circle with the ends just barely touching, and straighten it out again. Whether we lay the string down in a circle or lay it down straight, the length of the string stays the same. The circle has radius

*r*, and we know the circumference of the circle is 2π*r*. This means the length of the spaghetti-pizza is just 2π*r*.The spaghetti-pizza ring isn't very thick, but it does have a thickness of Δ

*r*. When we straighten it out, it still has a thickness of Δ*r*. So the area of the string, or ring, is 2π*r*Δ*r*.Let's go back to our circle of radius 6.

To find the area of this circle, we need to add up the areas of all the ring-slices. As we let the number of rings approach ∞, we get an integral. The variable

*r*goes from 0 to 6, so these are our limits of integration. This means the area of the circle isWe can evaluate this integral to get

Notice that along the way, we derived the formula for the area of a circle, or a pizza...or a pie. If the circle had radius

*R*instead of 6, we would have ended up withIf you've forgotten about the basics of a circle, no problem. Here's a little video to help refresh your memory.

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