# Application of Integration in Real Life

September 8, 2017 | Author: Izyan Mohd Subri | Category: Integral, Trigonometric Functions, Sine, Derivative, Mathematical Analysis

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Introduction to Integration Integration is a way of adding slices to find the whole. Integration can be used to find areas, volumes, central points and many useful things. But it is easiest to start with finding the area under the curve of a function like this:

What is the area under y = f(x) ?

Slices We could calculate the function at a few points and add up slices of width Δx like this (but the answer won't be very accurate):

We can make Δx a lot smaller and add up many small slices(answer is getting better):

And as the slices approach zero in width, the answer approaches thetrue answer. We now write dx to mean the Δx slices are approaching zero in width.

That is a lot of adding up! But we don't have to add them up, as there is a "shortcut". Because ...

... finding an Integral is the reverse of finding a Derivative. (So you should really know about Derivatives before reading more!) Like here:

Example: What is an integral of 2x?

We know that the derivative of x2 is 2x ...

... so an integral of 2x is x2

Notation The symbol for "Integral" is a stylish "S" (for "Sum", the idea of summing slices):

After the Integral Symbol we put the function we want to find the integral of (called the Integrand), and then finish with dx to mean the slices go in the x direction (and approach zero in width). And here is how we write the answer:

Plus C We wrote the answer as x2 but why + C ? It is the "Constant of Integration". It is there because of all the functions whose derivative is 2x:

The derivative of x2+4 is 2x, and the derivative of x2+99 is also 2x, and so on! Because the derivative of a constant is zero. So when we reverse the operation (to find the integral) we only know 2x, but there could have been a constant of any value. So we wrap up the idea by just writing + C at the end.

Tap and Tank

Integration is like filling a tank from a tap. The input (before integration) is the flow rate from the tap. Integrating the flow (adding up all the little bits of water) gives us thevolume of water in the tank.

Imagine the flow starts at 0 and gradually increases (maybe a motor is slowly opening the tap).

As the flow rate increases, the tank fills up faster and faster. With a flow rate of 2x, the tank fills up at x2. We have integrated the flow to get the volume.

Example: (assuming the flow is in liters per minute) after 3 minutes (x=3):  the flow rate has reached 2x = 2×3 = 6 liters/min,  and the volume has reached x2 = 32 = 9 liters.

We can do the reverse, too:

Imagine you don't know the flow rate. You only know the volume is increasing by x2. We can go in reverse (using the derivative, which gives us the slope) and find that the flow rate is 2x. Example: at 2 minutes the slope of the volume is 4, meaning it is increasing at 4 liters/minute, which is the flow rate. Likewise at 3 minutes the slope is 6, etc.

So Integral and Derivative are opposites.

We can write that down this way:

The integral of the flow rate 2x tells us the

∫2x dx =

volume of water: x2 + C

And the slope of the volume increase x2+C gives

(x2 + C) us back the flow rate: = 2x

And hey, we even get a nice explanation of that "C" value ... maybe the tank already has water in it! 

The flow still increases the volume by the same amount

And the increase in volume can give us back the flow rate.

Which teaches us to always add "+ C".

Other functions Well, we have played with y=2x enough now, so how do we integrate other functions?

If we are lucky enough to find the function on the result side of a derivative, then (knowing that derivatives and integrals are opposites) we have an answer. But remember to add C.

Example: what is

∫cos(x) dx ?

From the Rules of Derivatives table we see the derivative of sin(x) is cos(x) so:

∫cos(x) dx = sin(x) + C But a lot of this "reversing" has already been done (see Rules of Integration).

Example: What is

∫x3 dx ?

On Rules of Integration there is a "Power Rule" that says:

∫x

n

dx = xn+1/(n+1) + C

We can use that rule with n=3:

∫x

3

dx = x4 /4 + C

Knowing how to use those rules is the key to being good at Integration.

So get to know those rules and get lots of practice. Learn the Rules of Integration and Practice! Practice! Practice!

Definite vs Indefinite Integrals We have been doing Indefinite Integrals so far. A Definite Integral has actual values to calculate between (they are put at the bottom and top of the "S"):

Indefinite Integral

Definite Integral

Definite Integrals You might like to read Introduction to Integration first!

Integration Integration can be used to find areas, volumes, central points and many useful things. But it is often used to find the area under the graph of a function like this:

The area can be found by adding slices that approach zero in width: And there are Rules of Integration that help us get the answer.

Notation The symbol for "Integral" is a stylish "S" (for "Sum", the idea of summing slices):

After the Integral Symbol we put the function we want to find the integral of (called the Integrand), and then finish with dx to mean the slices go in the x direction (and approach zero in width).

Definite Integral A Definite Integral has start and end values: in other words there is an interval (a to b). The values are put at the bottom and top of the "S", like this:

Indefinite Integral (no specific values)

Definite Integral (from a to b)

We can find the Definite Integral by calculating the Indefinite Integral at points a and b, then subtracting:

Example: The Definite Integral, from 1 to 2, of 2x dx:

The Indefinite Integral is:

∫2x dx = x

 At x=1:

∫2x dx = 1

2

+C

 At x=2:

∫2x dx = 2

2

+C

2

+C

Subtract:

(22 + C) − (12 + C) 22 + C − 12 − C 4−1+C−C=3 And "C" gets cancelled out ... so with Definite Integrals we can ignore C. In fact we can give the answer directly like this:

We can check that, by calculating the area of the shape: Yes, it has an area of 3. Let's try another example:

Example: The Definite Integral, from 0.5 to 1.0, of cos(x) dx:

(Note: x must be in radians)

The Indefinite Integral is:

∫cos(x) dx = sin(x) + C

We can ignore C when we do the subtraction (as we saw above): = sin(1) − sin(0.5) = 0.841... − 0.479... = 0.362... And another example to make an important point:

Example: The Definite Integral, from 0 to 1, of sin(x) dx:

The Indefinite Integral is:

∫sin(x) dx = −cos(x) + C

Since we are going from 0, can we just calculate the area at x=1? −cos(1) = −0.540... What? The Area at x=1 is negative? No, we need to subtract the integral at x=0. We shouldn't assume that it is zero. So let us do it properly, subtracting one from the other (and C gets cancelled so we don't need to show it):

= −cos(1) − (−cos(0)) = −0.540... − (−1) = 0.460... That's better! But we can have negative areas, when the curve is below the axis:

Example: The Definite Integral, from 1 to 3, of cos(x) dx:

Notice that some of it is positive, and some negative. The definite integral will work out the net area.

The Indefinite Integral is:∫cos(x) dx = sin(x) + C So let us do the calculations: = sin(3) − sin(1)

= 0.141... − 0.841... = −0.700... Try integrating cos(x) with different start and end values to see for yourself how positive and negative areas work.

Continuous Oh yes, the function we are integrating must be Continuous between a and b: no holes, jumps or vertical asymptotes (where the function heads up/down towards infinity).

Example: A vertical asymptote between a and b affects the definite integral.

Properties Reversing the interval

Reversing the direction of the interval gives the negative of the original direction.

Interval of zero length

When the interval starts and ends at the same place, the result is zero:

Integration Rules

Integration Integration can be used to find areas, volumes, central points and many useful things. But it is often used to find the area underneath the graph of a function like this:

The integral of many functions are well known, and there are useful rules to work out the integral of more complicated functions, many of which are shown here.

Common Functions Constant

Function

Integral

∫a dx

ax + C

Variable

∫x dx

x2/2 + C

Square

∫x

x3/3 + C

2

dx

∫(1/x) dx

Reciprocal Exponential

∫e

x

dx

ex + C

∫a

x

dx

ax/ln(a) + C

∫ln(x) dx

x ln(x) − x + C

∫cos(x) dx

sin(x) + C

∫sin(x) dx

-cos(x) + C

∫sec (x) dx

tan(x) + C

Function

Integral

∫cf(x) dx

c∫f(x) dx

2

Rules Multiplication by constant Power Rule (n≠-1)

ln|x| + C

∫x

n

dx

xn+1/(n+1) + C

Sum Rule

∫(f + g) dx

∫f dx + ∫g dx

Difference Rule

∫(f - g) dx

∫f dx - ∫g dx

Examples Example: what is the integral of sin(x) ? From the table above it is listed as being −cos(x) + C It is written as:

∫sin(x) dx = −cos(x) + C Power Rule Example: What is

∫x3 dx ?

The question is asking "what is the integral of x3 ?" We can use the Power Rule, where n=3:

∫x

n

dx = xn+1/(n+1) + C

∫x Example: What is

3

dx = x4/4 + C

∫√x dx ?

√x is also x0.5 We can use the Power Rule, where n=½:

∫x

n

∫x

dx = xn+1/(n+1) + C 0.5

dx = x1.5/1.5 + C

Multiplication by constant

Example: What is

∫6x2 dx ?

We can move the 6 outside the integral:

∫6x

2

dx = 6∫x2 dx

And now use the Power Rule on x2:

= 6 x3/3 + C Simplify:

= 2x3 + C

Sum Rule Example: What is

∫cos x + x dx ?

Use the Sum Rule:

∫cos x + x dx = ∫cos x dx + ∫x dx Work out the integral of each (using table above):

= sin x + x2/2 + C

Difference Rule Example: What is

∫ew − 3 dw ?

Use the Difference Rule:

∫e

w

− 3 dw =∫ew dw −

∫3 dw

Then work out the integral of each (using table above):

= ew − 3w + C

Sum, Difference, Constant Multiplication And Power Rules Example: What is

∫8z + 4z

3

− 6z2 dz ?

Use the Sum and Difference Rule:

∫8z + 4z

3

− 6z2 dz =∫8z dz +

∫4z

3

dz −

∫6z

2

dz

Constant Multiplication:

= 8∫z dz + 4∫z3 dz − 6∫z2 dz Power Rule:

= 8z2/2 + 4z4/4 − 6z3/3 + C Simplify:

= 4z2 + z4 − 2z3 + C

Integration by Parts Integration by Parts is a special method of integration that is often useful when two functions are multiplied together, but is also helpful in other ways.

You will see plenty of examples soon, but first let us see the rule:

∫u v dx = u∫v dx −∫u' (∫v dx) dx 

u is the function u(x)

v is the function v(x)

As a diagram:

And let us get straight into an example:

Example: What is

∫x cos(x) dx ?

First choose u and v:  u=x  v = cos(x) Differentiate u: u' = x' = 1 Integrate v: together:

∫v dx = ∫cos(x) dx = sin(x)

Now put it

Simplify and solve:

x sin(x) −

∫sin(x) dx

x sin(x) + cos(x) + C

So we followed these steps: 

Choose u and v

Differentiate u: u'

Integrate v:

Put u, u' and

Simplify and solve

∫v dx ∫v dx here: u∫v dx −∫u' (∫v dx) dx

(u integral v) minus integral of (derivative u, integral v)

Let's try some more examples:

∫ln(x)/x

Example: What is

2

dx ?

First choose u and v:  u = ln(x)  v = 1/x2 Differentiate u: ln(x)' = 1/x Integrate v:

∫1/x

2

dx =

∫x

-2

dx = −x-1 = -1/x (by

the power rule) Now put it together:

Simplify:

−ln(x)/x −

∫−1/x

2

dx = −ln(x)/x − 1/x + C

−(ln(x) + 1)/x + C

Example: What is

∫ln(x) dx ?

But there is only one function! How do we choose u and v ?

Hey! We can just choose v as being "1":  u = ln(x)  v=1 Differentiate u: ln(x)' = 1/x Integrate v:

∫1 dx = x

Now put it together:

Simplify:

x ln(x) −

∫1 dx = x ln(x) − x + C

Example: What is Choose u and v:  u = ex  v=x

∫ex x dx ?

Differentiate u: (ex)' = ex

∫x dx = x /2 2

Integrate v:

Now put it together:

Well, that was a spectacular disaster! It just got more complicated. Maybe we could choose a different u and v?

Example:

∫ex x dx (continued)

Choose u and v differently:  u=x  v = ex Differentiate u: (x)' = 1 Integrate v:

∫e

x

dx = ex

Now put it together:

Simplify:

x ex − ex + C ex(x−1) + C The moral of the story: Choose u and v carefully! Choose a u that gets simpler when you differentiate it and a v that doesn't get any more complicated when you integrate it. A helpful rule of thumb is

I LATE. Choose u based on which of

these comes first: 

I: Inverse trigonometric functions such as sin-1(x), cos-1(x), tan-1(x)

L: Logarithmic functions such as ln(x), log(x)

A: Algebraic functions such as x2, x3

T: Trigonometric functions such as sin(x), cos(x), tan (x)

E: Exponential functions such as ex, 3x

And here is one last (and tricky) example:

Example:

∫ex sin(x) dx

Choose u and v:  u = sin(x)  v = ex Differentiate u: sin(x)' = cos(x) Integrate v:

∫e

x

dx = ex

Now put it together:

∫e

x

sin(x) dx = sin(x) ex -∫cos(x) ex dx

Looks worse, but let us persist! We can use integration by parts again: Choose u and v:  u = cos(x)  v = ex Differentiate u: cos(x)' = -sin(x) Integrate v:

∫e

x

dx = ex

Now put it together:

∫e

sin(x) dx = sin(x) ex - (cos(x) ex −∫−sin(x) ex dx) x

Simplify:

∫e

x

sin(x) dx = ex sin(x) - ex cos(x) −∫ ex sin(x)dx

Now we have the same integral on both sides (except one is subtracted) ... ... so bring the right hand one over to the left and we get:

2∫ex sin(x) dx = ex sin(x) − ex cos(x) Simplify:

∫e

x

sin(x) dx = ex (sin(x) - cos(x)) / 2 + C

Where Did "Integration by Parts" Come From? It is based on the Product Rule for Derivatives:

(uv)' = uv' + u'v Integrate both sides and rearrange:

∫(uv)' dx = ∫uv' dx + ∫u'v dx uv =

∫uv' dx + ∫u'v dx

∫uv' dx = uv − ∫u'v dx

Some people prefer that last form, but I like to integrate v' so the left side is simple:

∫uv dx = u∫v dx − ∫u'(∫v dx) dx Integration by Substitution "Integration by Substitution" (also called "u-substitution") is a method to find an integral, but only when it can be set up in a special way. The first and most vital step is to be able to write our integral in this form:

Note that we have g(x) and its derivative g'(x) Like in this example:

Here f=cos, and we have g=x2 and its derivative of 2x This integral is good to go! When our integral is set up like that, we can do this substitution:

Then we can integrate f(u), and finish by putting g(x) back as u. Like this:

Example:

∫cos(x ) 2x dx 2

We know (from above) that it is in the right form to do the substitution:

Now integrate:

∫cos(u) du = sin(u) + C And finally put u=x2 back again:

sin(x2) + C So

∫cos(x ) 2x dx = sin(x ) + C worked out really nicely! 2

(Well, I knew it would.)

2

This method only works on some integrals of course, and it may need rearranging:

Example:

∫cos(x ) 6x dx 2

Oh no! It is 6x, not 2x. Our perfect setup is gone. Never fear! Just rearrange the integral like this:

∫cos(x ) 6x dx = 3∫cos(x ) 2x dx 2

2

3∫cos(u) du = 3 sin(u) + C Now put u=x2 back again:

3 sin(x2) + C Done! Now we are ready for a slightly harder example:

Example:

∫x/(x +1) dx 2

Let me see ... the derivative of x2+1 is 2x ... so how about we rearrange it like this:

∫x/(x +1) dx = ½∫2x/(x +1) dx 2

Then we have:

2

Then integrate:

½∫1/u du = ½ ln(u) + C Now put u=x2+1 back again:

½ ln(x2+1) + C

Example:

∫(x+1)

3

dx

the derivative of x+1 is ... well it is simply 1. So we can have this:

∫(x+1)

∫(x+1)

3

dx =

3

du = (u4)/4 + C

3

· 1 dx

Then we have:

Then integrate:

∫u

Now put u=x+1 back again:

(x+1)4 /4 + C

In Summary When we can put an integral in this form:

 Then we can make u=g(x) and integrate ∫f(u) du  And finish up by re-inserting g(x) where u is.

Area under a Curve The area between the graph of y = f(x) and the x-axis is given by the definite integral below. This formula gives a positive result for a graph above the x-axis, and a negative result for a graph below the x-axis. Note: If the graph of y = f(x) is partly above and partly below the x-axis, the formula given below generates the net area. That is, the area above the axis minus the area below the axis.

Formul a:

Exampl Find the area between y = 7 – x2 and e 1: the x-axis between the values x = –1 and x = 2.

Exampl Find the net area between y = e 2: sin x and the x-axis between the values x = 0 and x = 2π.

Area between Curves

The area between curves is given by the formulas below.

Formula 1: for a region bounded above and below by y = f(x) and y = g(x), and on the left and right by x = a and x = b.

Formula 2:

for a region bounded left and right by x = f(y) and x = g(y), and above and below by y = c and y = d.

Example 1:1 Find the area between y = x and y = x2 from x = 0 to x = 1.

Example 2:1 Find the area between x = y + 3 and x = y2 from y = –1 to y = 1.

INTEGRATION OF TRIGONOMETRIC INTEGRALS

Recall the definitions of the trigonometric functions.

 The following indefinite integrals involve all of these well-known trigonometric functions. Some of the following trigonometry identities may be needed.  A.)  B.)  C.)

so that

 D.)

so that

 E.)  F.)

so that

 G.)

so that

It is assumed that you are familiar with the following rules of differentiation.       These lead directly to the following indefinite integrals.

o 1.) o 2.) o 3.) o 4.) o 5.)

o 6.)

The next four indefinite integrals result from trig identities and usubstitution.

o 7.) o 8.) o 9.) o 10.) We will assume knowledge of the following well-known, basic indefinite integral formulas :

 

, where

is a constant

 

, where

is a constant

 Most of the following problems are average. A few are challenging. Many use the method of u-substitution. Make careful and precise use of the differential notation and and be careful when arithmetically and algebraically simplifying expressions.

SOLUTION 1 : Integrate

. Use u-substitution. Let

so that , or . Substitute into the original problem, replacing all forms of , getting

(Use antiderivative rule 2 from the beginning of this section.)

Exact Integrals as Limits of Sums

Using the definition of an integral, we can evaluate the limit as goes to infinity. This technique requires a fairly high degree of familiarity with summation identities. This technique is often referred to as evaluation "by definition," and can be used to find definite integrals, as long as the integrands are fairly simple. We start with definition of the integral: Then picking be get,

to we

In some simple cases, this expression can be reduced to a real number, which can be interpreted as the area under the curve if f(x) is positive on [a,b].

Example 1 Find sums.

by writing the integral as a limit of Riemann

In other cases, it is even possible to evaluate indefinite integrals using the formal definition. We can define the indefinite integral as follows:

Example 2 Suppose , then we can evaluate the indefinite integral as follows.

Volumes of Revolution Rotation About the x-axis Integration can be used to find the area of a region bounded by a curve whose equation you know. If we want to find the area under the curve y = x2 between x = 0 and x = 5, for example, we simply integrate x2 with limits 0 and 5. Now imagine that a curve, for example y = x2, is rotated around the x-axis so that a solid is formed. The volume of the shape that

is formed can be found using the formula

Rotation about the y-axis If the body is rotated about the y-axis rather than the x-axis, then we use the formula:

Application of Integration in Real Life 1. Laboratory Systems Integration Application Integration Integration between applications and ERP systems, SAP, MES systems, CAPA and document management systems listed below can be relatively straight forward or extremely complex. Guided by your business needs and pragmatism (the result of working in hundreds of laboratories) we will design and implement the appropriate application integration strategy for your business. Integration Benefits 

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Less remediation and review time

2.

Centroid of an Area by Integration

Typical (straight sided) Problem In tilt-slab construction, we have a concrete wall (with doors and windows cut out) which we need to raise into position. We don't want the wall to crack as we raise it, so we need to know the centre of mass of the wall. How do we find the centre of mass for such an uneven shape?

Tilt-slab construction, also known as tilt-wall or tilt-up Basically the integration concept is used to find the centroid of an area with straight sides, then we'll extend the concept to areas with curved sides. Besides, the concept of moment also involved where the definition of moment of a mass is a measure of its tendency to rotate about a point. Clearly, the greater the mass (and the greater the distance from the point), the greater will be the tendency to rotate. The moment is defined as:

Moment = mass × distance from a point Example 1

In this case, there will be a total moment about O of: (Clockwise is regarded as positive in this work.)

M = 2 × 1 – 10 × 3 = −28 kgm (A)

Centre of Mass

We now aim to find the centre of mass of the system and this will lead to a more general result. Example 2

We have 3 masses of 10 kg, 5 kg and 7 kg at 2 m, 2 m and 1 m distance from O as shown.

We wish to replace these masses with one single mass to give an equivalent moment. Where should we place this single mass? Answer 2 Total moment = 10 × 2 + 5 × 4 + 7 × 5 =75 kg.m If we put the masses together, we have: 10 + 5 + 7 = 22 kg For an equivalent moment, we need: 22 × d = 75 where d is the distance from the centre of mass to the point of rotation. i.e. d=2275≈3.4 m So our equivalent system (with one mass of 22 kg) would have:

(B) Centre of Mass (Centroid) for a Thin Plate 1) Rectangle:

The centroid is (obviously) going to be exactly in the centre of the plate, at (2, 1). 2) More Complex Shapes:

We divide the complex shape into rectangles and find x (the x-coordinate of the centroid) and y (the y-coordinate of the centroid) by taking moments about the y- and x-coordinates respectively. Because they are thin plates with a uniform density, we can just calculate moments using the area. Example 3 Find the centroid of the shape:

We divide the area into 2 rectangles and assume the mass of each rectangle is concentrated at the centre.

Left rectangle: Area = 3 × 2

= 6 sq unit. 1 Center (− 2 ,1) Right rectangle: Area = 2 × 4

= 8 sq unit. Center (2,2) Taking moments with respect to the y-axis, we have:

1

6(− 2 ) + 8(2) = (6+8)x

−3+16 = 14x 13

x = 14

Now, w.r.t the x-axis:

6(1)+8(2)=(6+8) ​y 6+16=14 ​y

y ​​​=

¿1

22 14

4 7

So the centroid is at:

(

13 14

,

4 7

)

We would use this process to solve the tilt slab construction problem mentioned at the beginning of this section.

In general, we can say:

X=

total moments∈x−direction total area

Y=

total moments∈ y−direction total area

This idea is used more extensively in the next section.

(C)

Centroid for Curved Areas

Taking the simple case first, we aim to find the centroid for the area defined by a function f(x), and the vertical lines figure.

x=a∧x=b

as indicated in the following

To find the centroid, we use the same basic idea that we were using for the straight-sided case above. The "typical" rectangle indicated is x units from the yaxis, and it has width Δx (which becomes dx when we integrate) and height y = f(x). Generalizing from the above rectangular areas case, we multiply these 3 values (x, f(x) and Δx, which will give us the area of each thin rectangle times its distance from the x-axis), then add them. If we do this for infinitesimally small strips, we get the x-coordinates of the centroid using the total moments in the xdirection, given by:

x=

total moments total area

=

b

1 a

∫ x f (x ) dx a

And, considering the moments in the y-direction about the x-axis and reexpressing the function in terms of y , we have:

Y=

total moments total area

=

1 a

d

∫ y f ( y) dy c

Notice this time the integration is with respect to y, and the distance of the "typical" rectangle from the x-axis is y units. Also note the lower and upper limits of the integral are c and d, which are on the y-axis. Of course, there may be rectangular portions we need to consider separately. (I've used a different curve for the y case for simplification.) Alternate method: Depending on the function, it may be easier to use the following alternative formula for the y-coordinate, which is derived from considering moments in the x-direction (Note the "dx" in the integral, and the upper and lower limits are along the x-axis for this alternate method).

Y=

total moments total area

= =

1 a 1 a

b

∫ a

b

f (x) x f ( x ) dx 2 2

[f ( x ) ] ∫ 2 dx a

This is true since for our thin strip (width dx), the centroid will be half the distance from the top to the bottom of the strip. Another advantage of this second formula is there is no need to re-express the function in terms of y.

(D)

Centroids for Areas Bounded by 2 Curves

We extend the simple case given above. The "typical" rectangle indicated has width Δx and heighty2 − y1, so the total moments in the x-direction over the total area is given by: b

totalmoments 1 x ​​​= = ∫ x ( y ​2 ​​− y ​1 ​​) dx total area Aa For the y coordinate, we have 2 different ways we can go about it. Method 1: We take moments about the y-axis and so we'll need to re-express the expressions x2 and x1as functions of y. d

total moments 1 y=​ = ∫ x ( x ​2 ​​− x ​1 ​​) dy total area A c

Method 2: We can also keep everything in terms of x by extending the "Alternate Method" given above: b

2 2 total moments 1 [ y 2] −[ y 1] y ​​​= ​ = ∫ dx total area A a 2

Example 4

Find the centroid of the area bounded by y = x3, x = 2 and the x-axis. Answer Here is the area under consideration:

In this case,

3

y=f (x)=x , a=0, b=2.

We find the shaded area first: 2

4 2

[ ]

x 16 A=∫ x dx= = 4 0 4 0 3

=4

Next, using the formula for the x-coordinate of the centroid we have: b

X=

1 ​​∫ x f ( x ) dx A a 2

1 ¿ ∫ x (x 3 )dx 4 0 2

1 ¿ ∫ ( x 4 ) dx 4 0 5 2

[ ]

1 x ¿ 4 5

¿

32 20

¿ 1.6

0

Now, for the y coordinate, we need to find:

this is x ​2 ​​=2(¿¿ this problem) 1 3

x ​1 ​​= y ( this is variable ∈this problem )

c=0, d=8. d

1 y ​​​= ∫ y ( x ​2 ​​− x ​1 ​​) dy A c 8

¿

1 ∫ y (2 ​​− y 1/ 3)dy 4 0

8

1 ¿ ∫ (2 y ​​− y 4 /3 )dy 4 0

[

1 2 3y y− 4 7 ¿

[

7 8 3

]

0

1 3 ×128 64 ​​− 4 7

]

¿ 2.29 So the centroid for the shaded area is at (1.6, 2.29).

3. Moments of Inertia by Integration The moment of inertia is measure of the resistance of a rotating body to a change in motion. The moment of inertia of a particle of mass m rotating about a particular point is given by:

Moment of inertia = md2 Where d is the radius of rotation.

Inertia for a Collection of Particles If a group of particles with masses m1, m2, m3, ... , mn is rotating around a point with distances d1, d2, d3, ...dn, (respectively) from the point, then the moment of inertia I is given by:

I = m1d12 + m2d22 + m3d32 +... + mndn2 If we wish to place all the masses at the one point (R units from the point of rotation) then

d1 = d2 = d3 = ... = dn = R and we can write: I = (m1 + m2 + m3 ... + mn)R2 R is called the radius of gyration. Example 1 Find the moment of inertia and the radius of gyration w.r.t. the origin (0,0) of a system which has masses at the points given: Mass

6

5

9

2

Point

(−3,0)

(−2,0)

(1,0)

(8,0)

Answer 1 The moment of inertia is: I

= 6(-3)2 + 5(-2)2 + 9(1)2 + 2(8)2 = 54 + 20 + 9 + 128 = 211

To find R, we use: I

= (m1 + m2 + m3 ... + mn) R2

211

= (6 + 5 + 9 + 2) R2

So R≈3.097 This means a mass of 22 units placed at (3.1, 0) would have the same rotational inertia about O as the 4 objects.

Moment of Inertia for Area

We want to find the moment of inertia , I y

of the given area, which is rotating

around the y-axis. Each "typical" rectangle indicated has width dx and height is

y 2− y 1 , so its area

( y 2− y 1) dx .

If k is the mass per unit area, then each typical rectangle has mass

k ( y 2− y 1) dx .

The moment of inertia for each typical rectangle is

[k ( y 2− y 1) dx ] x 2 , since

each rectangle is x units from the y-axis. We can add the moments of inertia for all the typical rectangles making up the area using integration: b

I ​y ​​=k ∫ x2 ( y ​2 ​​− y ​1 ) dx a

Using a similar process that we used for the collection of particles above, the radius of gyration Ry is given by:

R ​y ​​= √ where m is the mass of the area. Example 2

I ​y m

For the first quadrant area bounded by the curve

y=1−x ​2 ​​, find: a) The moment of inertia w.r.t the y axis. (Iy) b) The mass of the area c) Hence, find the radius of gyration Answer 2 As usual, first we sketch the part of the curve in the first quadrant. It's a parabola, passing through (1, 1) and (0, 1). A "typical" rectangle is shown.

In this example,

y ​2 ​​=1−x ​2 ​​,∧ y ​1 ​​=0, a=0∧b=1. a) Finding Iy: b

I ​y ​​=k ∫ x2 [ y 2− y 1]dx a

1

¿ k ∫ x 2 [(1−x 2)−(0)] dx 0 1

¿ k ∫ [ x 2−x 4 ]dx 0

[

3

5 1

x x ¿k − 3 5

1 1 ¿k( − ) 3 5

]

0

¿

2k 15

b) The mass of the area, m. Now m = kA, where A is the area. 1

A=∫ (1−x 2 )dx 0

3 1

[ ]

x ¿ x− 3 ¿ 1−

¿

0

1 3

2 3

So m=kA=

2k 3

R ​y ​​= √

I ​y m

¿√

2 k /15 2 k /3

¿√

1 5

≈ 0.447 This means that if a mass of

2k 3

was placed 0.447 units from the y-axis,

this would have the same moment of inertia as the original shape.

Rotation about the x-axis For rotation about the x-axis, the moment of inertia formulae become: d

I ​x ​​= k ∫ y 2 ( x ​2 ​​− x ​1 ) dy c

And

R ​x ​​=

√ ​​I ​x m

4. Work by a Variable Force using Integration

The work (W) done by a constant force (F) acting on a body by moving it through a distance (d) is given by:

W=F×d Example of work done by a constant force

1 N . If you lift the apple 1 m above a table, you have

done approximately 1 Newton meter (Nm) of work.

Work done by a Variable Force If the force varies (e.g. compressing a spring) we need to use calculus to find the work done. If the force is given by F(x) (a function of x) then the work done by the force along the x-axis from a to b is: b

W =∫ F ( x )dx a

Hooke's Law for springs The force (F) that it takes to stretch (or compress) a spring x units from its normal length is proportional ¿ x .

F=kx We can find the spring constant k from observing what force gives what stretch for each spring. This spring constant is also called the stiffness of the spring.

Interactive Appplet Example 1 (a) Find the work done on a spring when you compress it from its natural length of 1 m to a length of0.75 m if the spring constant is k = 16 N/m.

F=16 x We start compressing the spring at its natural length (0 m) and finish at 0.25 m from the natural length, so the lower limit of the integral is 0 and the upper limit is 0.25. So: 0.25

Work= ∫ 16 x dx 0

8 x 2 ¿0.25 0 ¿¿

¿ 0.5 N . m (b) What is the work done in compressing the spring a further 30cm? This time, we start pushing the spring at 0.25 m from the natural length and finish at 0.55 m from the natural length, so the lower limit of the integral is 0.25 and the upper limit is 0.55. 0.55

Work= ∫ 16 x dx 0.25

8 x 2 ¿0.55 0.25 ¿¿ ¿ 1.92 N . m Note: For a spring, b

∫ F ( x) dx a

Requires that

a∧b

are the distance from the natural position of the spring.

5. Electric Charges by Integration The force between charges is proportional to the product of their charges and inversely proportional to the square of the distance between them. So we can write:

f ( x)=

k q ​1 ​​q ​2 2 x

Where q1 and q2 are in coulombs (C), x is in metres, the force is in newtons

k =9 ×109 .

and k is a constant,

It follows that the work done when electric charges move toward each other (or when they are separated) is given by: b

Work=∫ a

k q ​1 ​​q ​2 dx x2

Example An electron has a

1.6 ×10−19 C

negative charge. How much work is done in

separating two electrons from 1.0 pm to 4.0 pm? Answer Recall: "pm" means picometre, or 10 − 12 meters. In this example,

a=1× 10−12 m b=4 × 10−12 m k =9 ×109 q ​1 ​​=q ​2 ​​=1.6 × 10−19 C

So we have b

Work=∫ a

k q ​1 ​​q ​2 dx x2 −12

4 × 10

¿∫

∫ 1 ×10

−12

(9 ×109 )(−1.6 × 10−19 )2 dx x2

¿ ( 2.304 ×10

−28

[ ]

) −1 x

−12

4 ×10

1 ×10

−12

¿ 1.728× 10−16 J

6. Average Value of a Function by Integration The average value of the function y = f(x) from x = a to x = b is given by:

b

y ​ave ​​=∫ a

f ( x )dx b−a

Why? When you see a formula like this for the first time, think about where it comes from and why it should work. Hint: How do we find the average of a set of numbers? What are we really doing each time we find an integral? What does the integral symbol stand for? Example The temperature T (in °C) recorded during a day followed the curve

T =0.001 t 4−0.280 t 2+25 Where t is the number of hours from noon

(−12≤ t ≤ 12)

What was the average temperature during the day? Answer First, we consider the graph of the situation and estimate that the average should be around 14 to 16 degrees.

b

y ​ave ​​=∫ a

f ( x )dx b−a 12

∫ ( 0.001t 4 −0.28 t2 +25)dt ¿ −12

12−(−12)

[

5

3

1 0.001 t 0.28 t ¿ − +25 t 24 5 3

]

12

−12

[

2 0.001 t 5 0.28 t 3 ¿ − +25 t 24 5 3 ¿

]

12

0

1 [ 49.7664−161.28+300] 12

¿ 15.7 ​∘C

7. Force Due to Liquid Pressure by Integration The force F on an area A at a depth y in a liquid of density w is given by

F=wyA The force will increase if the density increases, or if the depth increases or if the area increases. So if we have an unevenly shaped plate submerged vertically in a liquid, the force on it will increase with depth. Also, if the shape of the plate changes as we go deeper, we have to allow for this. So we have:

Now, the total force on the plate is given by b

F=w ∫ xy dy a

Where

x is thelength(¿ m)of theelement of area(expressed ∈terms of y) y is the depth (¿ m) of theelement of area

w is the density of the liquid (¿ N m−3)

(for water , this is w=9800 N m−3) a is the depth at the topof the area∈question(¿ m)

b is thedepth at the bot tom of the area∈question (¿ m)

8. Head Injury Criterion (HIC) part 1: Severity Index

In this section we'll see an example of the average value of a function. Our aim is to find the Head Injury Criterion, a measure of damage to the head. In the 1950s, cars were efficient killing machines. There were no such things as airbags, safety belts, anti-lock braking, crumple zones or plastic knobs. Ralph Nader pressured car manufacturers in the 1960s and 1970s to produce safer cars - and it worked. Normal Braking Normal braking in a street car: 10 ms-2 (or about 1 g). Normal braking in a racing car: 50 ms-2 (or about 5 g). This is due to aerodynamic styling and large tyres with special rubber. When we stop in a car, the deceleration can be either abrupt (as in a crash), as follows:

or more gentle, as in normal braking:

Either way, the area under the curve is the same, since the velocity we must lose is the same. Crash Tests Imagine a car travelling at 48.3 km/h (30 mph). Under normal braking, it will take 1.5 to 2 seconds for the car to come to rest. But in a crash, the car stops in about 150 ms and the life threatening deceleration peak lasts about 10 ms. Crash test experiments include the use of dummies, dead bodies, animals and boxers! Mercedes Benz Crash Test Data - Deceleration of the Head

The Mercedes Benz Company has been a world leader in car safety, and has conducted many crash tests involving dummies, with the aim of reducing injuries for humans. Our head is like a pendulum and so it's the most vulnerable part of our body in a crash. In cars without an airbag, the deceleration is quite violent and lasts a very short time. The Head Injury Criterion (HIC) is very high in such cases, indicating that the occupants' heads will be injured. The A-3 ms Value The A-3 ms value in the following graphs refers to the maximum deceleration that lasts for 3 ms. (Any shorter duration has little effect on the brain.)

If an airbag is present, it will expand and reduce the deceleration forces. Notice that the peak forces (in g) are much lower for the airbag case.

The blue rectangles in these deceleration graphs indicate the most critical part of the deceleration, when the maximum force is exerted for a long duration. With an airbag, you are far more likely to survive the crash. The airbag deploys in 25 ms.

Car design and crash outcomes

This photo was taken just after the car crash in 1997 that killed all occupants, including Princess Diana. Notice how the front crumple zone of the Mercedes did its job while the cabin retained its shape. Unfortunately, it was not enough to save her.

Crumple zones absorb impact forces, so that deceleration is reduced, an in turn, injury is reduced. A model to describe head injuries We aim to describe the risk of head injury in a crash by a number. The two main approaches are the Severity Index and the Head Injury Criterion. The Severity Index The first model developed historically was the Severity Index (SI). It was calculated using the formula: T

SI =∫ {a ( t ) }2.5 dt 0

Where T is the duration of the deceleration during the crash; and a(t) is the deceleration at time t. The index 2.5 was chosen for the head and other indices were used for other parts of the body (usually based on possibly gruesome experiments on human or animal bodies). The Severity Index was found to be inadequate, so researchers developed the Head Injury Criterion.

9. Head Injury Criterion (HIC) pt 2: HIC Index, example Experiments showed researchers that the Severity Index did not accurately describe the likelihood of certain injuries in a crash. They subsequently developed the Head Injury Criterion (HIC), which is based on the average value of the acceleration over the most critical part of the deceleration (shown in the blue rectangles in the Mercedes data before) We met average value of a function earlier in this chapter. The average value `bar(a)` of the acceleration a(t) over the time interval t1 to t2 is given by t2

1 ā= ∫ a(t) d t t 2−t 1 t 1 For the HIC, this was modified (based on experimental data) as follows:

a ( t ) dt t2 1 ¿2.5 t 2−t 1 ∫ t1 HIC= MAX (t 1∨t 2) {(t 2 – t 1)¿

The formula means: The HIC is the maximum value over the critical time period t1 to t2 for the expression in braces, { }. The index 2.5 is chosen for the head, based on experiments.

10. Arc Length of a Curve using Integration Example 1 - Corrugated iron sheeting

Corrugated iron roof. Corrugated iron is used extensively throughout the world as a versatile building material. Bending the material into a regular wave pattern gives it greater strength than if a flat sheet is used. Another example of a light, thin and weak sheet that is made much stronger by having regular folds is corrugated cardboard, used for protecting goods in transit.

The flat sheet is rolled into corrugations, and will be narrower. To make corrugated iron, you need to bend a wide flat sheet into waves. The resulting corrugated sheet is then narrower, of course. The corrugations are commonly in the form of a sine curve. We take a real example of a 4.2" Corrugated Metal Panel, which is a "high profile, wavy style corrugated panel that can be used in almost any roofing, siding, or decorative application." [Source]

Corrugated metal, in the shape of a sine wave This panel has a finished width of 106.7 cm, a period of 10.67 cm (distance from the top of each wave to the top of the next), and has amplitude 1.35 cm (height from the mid-point of the wave to the top of a crest). How wide should the flat sheet be to give us a corrugated sheet of width 106.7 cm? Example 1 Solution We model the corrugations using the curve

y=1.35 sin 0.589 x This has the required amplitude 1.35 and period 10.67. (Within the sine expression, we use 2π/10.67 = 0.589 for the coefficient of x. For background on this, see Period of a sine curve.)

We'll find the width needed for one wave, then multiply by the number of waves.

Approximate answer: Next, let's approximate the length of the curve so we've got a rough idea what our exact length should be. (It's always good practice to estimate your answer first, and in this topic, it helps us understand the concept better). We plot the points O (0, 0), A (2.65, 1.35), B (5.33, 0), C (7.99, -1.35) and D (10.65, 0), which are key points on the curve (at the mid-points, maximum and minimum values), and join the line segments.

We then use Pythagoras' Theorem to find the length OA:

OA=√ ​2.65 ​2 ​​+ 1.35 ​2 ​​​​​=2.97 The distances AB, BC, CD are all equal, so we can say:

OA + AB+ BC +CD =4 × 2.97=11.88 So we expect the curved distance OD to be around 12 cm. Exact value We'll use calculus to find the 'exact' value. But first, some background. We zoom in near the centre of the segment OA and we see the curve is almost straight. For this portion, the curve EF is getting quite close to the straight line segment EF.

For this zoomed-in section, we have: b

Curved length

EF=r ≈∫ √ 12+ 0.572 0.57=1.15 a

Of course, the real curved length is slightly more than 1.15. Let's generalise this. General Form of the Length of a Curve If the horizontal distance is "dx" (or "a small change in x") and the vertical height of the triangle is "dy" (or "a small change in y") then the length of the curved arc "dr" is approximated as:

dr ≈ √ ​dx ​2 ​​+ dy ​2

Now, if we move point E very close to point F, we will have a very good approximation for the length of the curve in that local region. We need to add all those infinitesimally small lengths. We use integration, as it represents the sum of such infinitely small distances. We have for the distance between where x=a to x=b:

length=r =∫ ​a ​b ​​√ ​( dx ) ​2 ​​+ (dy ) ​2 By performing simple surd manipulation, we can express this in more familiar form as follows. The arc length of the curve y = f(x) from x = a to x = b is given by:

length=r =∫ ​a ​b ​​√ ​1+( ​dx ​​dy ​​) ​2 ​​​​​dx Of course, we are assuming the function y=f(x) is continuous in the region x=a to x=b (otherwise, the formula won't work). Example 3 - Golden Gate Bridge cables The central span of the Golden Gate Bridge in San Francisco, USA, is 1280 m long.

Central span of the Golden Gate Bridge. The height of the tower is 152 m from the roadway. What is the length of the main suspension cable between the 2 towers? Answer We first need to model the curve (find an equation that represents the curve accurately). A freely hanging cable takes the form of a catenary. The general form for a catenary is the sum of 2 exponential functions:

ax

y=

a( e +e 2

−ax

)

The Golden Gate Bridge cable is almost a catenary and almost a parabola, but not quite either (because of the weight of the cables, the suspender ropes and the roadway). For the sake of this discussion, we'll assume it is a catenary. For convenience, we'll place the origin at the lowest point of the cable. The required curve (after some guess and check) passing through (-640,152), (0,0) and (640,152) is given by:

y=1280(

e

x 1326

+e 2

−x 1326

−1)

Here is the graph of the above equation. We can see it pases through the required points.

The derivative of our function is

dy 640 e x /1326 + e−x/1326 = ( ) dx 663 2 Using the length of a curve formula, with start point x = -640 and end point x = 640, we have: 640

∫ −640

√ ​1+(

640 e ​x /1326 ​​+ e ​− x /1326 2 ) dx=1326.956 663 2

So the length of the central span of the main cable is 1327.0 m. Of course, the cable continues on both sides of the towers. The total length of each cable is 2,332 m.