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Trapezoidal approximation of area under curve Video Lecture - Chemical Engineering
FAQs on Trapezoidal approximation of area under curve Video Lecture - Chemical Engineering
| 1. What is the trapezoidal approximation method for calculating the area under a curve in chemical engineering? |  |
Ans. The trapezoidal approximation method is a numerical integration technique used in chemical engineering to estimate the area under a curve. It involves dividing the curve into trapezoids and calculating the area of each trapezoid using the formula (h/2)*(b1+b2), where h is the height and b1 and b2 are the lengths of the parallel sides of the trapezoid. The sum of the areas of all the trapezoids gives an approximation of the total area under the curve.
| 2. How accurate is the trapezoidal approximation method in chemical engineering? | |
Ans. The accuracy of the trapezoidal approximation method depends on the number of trapezoids used to divide the curve. Generally, the more trapezoids used, the more accurate the approximation will be. However, even with a large number of trapezoids, the trapezoidal approximation method may still introduce some error compared to the actual area under the curve. Other numerical integration techniques, such as Simpson's rule, may provide higher accuracy in certain cases.
| 3. What are the advantages of using the trapezoidal approximation method in chemical engineering? | |
Ans. The trapezoidal approximation method is relatively simple to implement and does not require advanced mathematical techniques. It can provide a reasonably accurate estimation of the area under a curve, especially when the curve is not well-behaved or when analytical solutions are not readily available. The method is widely used in chemical engineering for various applications, including the calculation of heat transfer rates, reaction kinetics, and process optimization.
| 4. Can the trapezoidal approximation method be used for any type of curve in chemical engineering? | |
Ans. The trapezoidal approximation method can be applied to a wide range of curves encountered in chemical engineering. It is particularly useful for curves that do not have a simple mathematical expression or when only discrete data points are available. However, the accuracy of the method may vary depending on the shape of the curve and the number of trapezoids used. In some cases, other numerical integration techniques may be more suitable for highly complex or irregular curves.
| 5. Are there any limitations or considerations when using the trapezoidal approximation method in chemical engineering? | |
Ans. While the trapezoidal approximation method is a useful tool in chemical engineering, it has certain limitations. One limitation is that it assumes a linear interpolation between adjacent data points, which may not accurately represent the true behavior of the curve. Additionally, the method may not be suitable for curves with rapidly changing slopes or when high accuracy is required. It is important to carefully choose the number of trapezoids and consider alternative numerical integration techniques if needed to ensure accurate results.
Text Transcript from Video
For fun, let's try to
approximate the area
under the curve y is
equal to the square root,
the principal root, of x minus
1, between x is equal to 1
and x is equal to 6.
So I want to find
this entire area.
Or I want to at least
approximate this entire area.
And the way I'll do
it is by setting up
five trapezoids of equal width.
So this will be
the left boundary
of the first trapezoid.
This will be its right
boundary, which will also
be the left boundary of
the second trapezoid.
This will be the right boundary
of the second trapezoid.
This is the right boundary
of the third trapezoid.
This'll be the right boundary
of the fourth trapezoid.
And then finally, this
will be the right boundary
of the fifth trapezoid.
And since we're going from 1 to
6, so we're traveling 6 minus 1
in the x direction, and I want
to split it into five sections,
the width of each trapezoid is
just going to be equal to 1.
And so if we say that the width
of a trapezoid is delta x,
we just can now say the
delta x is equal to 1.
So let's set up our trapezoids.
So the first trapezoid is
going to look like that.
It's going to go like that.
Actually, it's going to
be a triangle, not really
a trapezoid.
Then the second trapezoid
is going to look like this.
I guess you could
say a trapezoid where
one of the sides has length
0 turns into a triangle.
And then the third trapezoid
is going to look like this.
And then the fourth trapezoid
is going to look like that.
And then finally, you
have the fifth trapezoid.
So let's calculate the
area of each of these,
and then we will have our
approximation for the area
under the curve.
So let's do trapezoid--
or I really should say
triangle-- this first shape,
whatever you want to call it.
What is the area of
that going to be?
Well, the area of a
trapezoid-- and you'll
see this will just turn into
the area of a triangle-- it's
the average of the
heights of the two
sides of the trapezoid, the
way we've looked at it--
or you could say the average
of the heights of the two
parallel sides, I guess
is the best way to say it.
So f of 1-- that's the
height here-- plus f of 2,
all of that over
2, and then we're
going to multiply it
times our delta x.
Actually, let me do that
in that same red color
to show you that this is the
area of that first trapezoid,
so times delta x.
And as you see right over here,
if you look at it, the f of 1
is just going to be 0.
So you're going to
have f of 2 times--
so it's going to be this height
times this base times 1/2,
which is just the
area of a triangle.
Let's look at the
second trapezoid,
trapezoid two right over here.
What is its area going to be?
Well, it's going to
be f of 2 plus f of 3.
f of 2 is this height.
f of 3 is this height.
So we're taking the average
of those two heights-- divided
by 2; that's the
average of those two
heights-- times the
base, times delta x.
And then let's do
trapezoid three.
I think you're
getting the idea here.
Trapezoid three is
going to be f of 3
plus f of 4 divided
by 2 times delta x.
And then-- let's see.
I'm running out of colors.
This is trapezoid
four right over here.
So plus f of 4 plus f of 5, all
of that over 2, times delta x.
And then we have
our last trapezoid,
which I will do in yellow.
So this is trapezoid
number five.
I'll scroll down a little bit,
get some more real estate.
So it's going to be plus-- I'll
just write the plus over here--
plus f of 5 plus f of 6
over 2 times our delta x.
So let's see how we can
simplify this a little bit.
All of these terms,
we have a 1/2 delta x,
so let's actually
factor that out.
So remember, this is our
approximation of our area.
So area is
approximately-- remember,
this is just a
rough approximation.
It's very clear--
actually, you might say,
this is pretty good
using the trapezoids,
but it is pretty
clear that we are
letting go of some of the area.
We're letting go of that area.
We're letting go of some
of this right over here.
You can barely see it.
Some of this right over
here, you can barely see it.
But we are-- this is going
to be, it looks like,
an underestimate, but it
is a decent approximation.
But let's see if we can
simplify this expression.
So it's approximately
going to be equal to,
I'm going to factor
out a delta x over 2.
And then what I'm
left with-- and I
will switch to a neutral
color-- if I factor out
a delta x over 2, then
I have just an f of 1.
And then I have two f of
2's, so plus 2 times f of 2.
And I'm doing this because
you might see a formula that
looks something like this
in your calculus book,
and it's not some
mysterious thing.
They just really summed up
the areas of the trapezoids.
And then we're going to have
two f of 3's-- plus 2 times f
of 3's.
Plus we're going to have two f
of 4's-- plus 2 times f of 4.
And then we're going to have two
f of 5's-- plus 2 times f of 5.
And then finally, we're going to
have one f of 6-- plus f of 6.
If you were to
generalize it, you
have one of the first
endpoint, the function
evaluated at the first
endpoint, one of it at the very
last endpoint, and then two
of all of the rest of them.
But this is just the
area of trapezoids.
I'm not actually a big fan
of when textbooks write this.
Because when you see this,
it's hard to visualize
the trapezoids.
When you see this,
it's much clearer
how you might visualize that.
But with that out of the way,
let's actually evaluate this.
Lucky for us, the
math is simple.
Our delta x is just 1.
And then we just have to
evaluate all of this business.
f of 1, let's just
remind ourselves
what our original function was.
Our original function was
the square root of x minus 1.
So f of 1 is the square
root of 1 minus 1,
so that is just going to be 0.
This expression
right over here is
going to be 2 times the
square root of 2 minus 1.
The square root of
2 minus 1 is just 1,
so this is just going to be 2.
Actually, let me do it
in that same-- well,
I'm now using the purple
for a different purpose
than just the first trapezoid.
Hopefully, you realized that.
I was just sticking
with that pen color.
Then f of 3.
3 minus 1 is 2--
square root of 2.
So the function evaluated at
3 is the square root of 2.
So this is going to be 2
times the square root of 2.
Then the function
evaluated at 4.
When you evaluate it at 4,
you get the square root of 3.
So this is going to be 2
times the square root of 3.
And then you get 2 times the
square root of 4-- 5 minus 1
is 4.
2 times the square
root of 4 is just four.
And then finally, you get f of
6 is square root of 6 minus 1,
is the square root of 5.
And I think we're now
ready to evaluate.
So let me get my handy TI-85
out and calculate this.
So it's going to
be-- well I'm just
going to calculate--
well, I'll just multiply.
So 0.5 times open
parentheses-- well, it's a 0.
I'll just write it, just
so you know what I'm doing.
0 plus 2 plus-- whoops.
Lost my calculator.
Plus 2 times the square root of
2 plus 2 times the square root
of 3 plus 4-- I'm almost done--
plus the square root of 5--
so let me write
that-- gives me--
now we are ready for our
drum roll-- it gives me--
and I'll just round it-- 7.26.
So the area is
approximately equal to 7.26
under the curve y is equal
to the square root of x
minus 1 between x
equals 1 and x equals 6.
And we did this
using trapezoids.
approximate the area
under the curve y is
equal to the square root,
the principal root, of x minus
1, between x is equal to 1
and x is equal to 6.
So I want to find
this entire area.
Or I want to at least
approximate this entire area.
And the way I'll do
it is by setting up
five trapezoids of equal width.
So this will be
the left boundary
of the first trapezoid.
This will be its right
boundary, which will also
be the left boundary of
the second trapezoid.
This will be the right boundary
of the second trapezoid.
This is the right boundary
of the third trapezoid.
This'll be the right boundary
of the fourth trapezoid.
And then finally, this
will be the right boundary
of the fifth trapezoid.
And since we're going from 1 to
6, so we're traveling 6 minus 1
in the x direction, and I want
to split it into five sections,
the width of each trapezoid is
just going to be equal to 1.
And so if we say that the width
of a trapezoid is delta x,
we just can now say the
delta x is equal to 1.
So let's set up our trapezoids.
So the first trapezoid is
going to look like that.
It's going to go like that.
Actually, it's going to
be a triangle, not really
a trapezoid.
Then the second trapezoid
is going to look like this.
I guess you could
say a trapezoid where
one of the sides has length
0 turns into a triangle.
And then the third trapezoid
is going to look like this.
And then the fourth trapezoid
is going to look like that.
And then finally, you
have the fifth trapezoid.
So let's calculate the
area of each of these,
and then we will have our
approximation for the area
under the curve.
So let's do trapezoid--
or I really should say
triangle-- this first shape,
whatever you want to call it.
What is the area of
that going to be?
Well, the area of a
trapezoid-- and you'll
see this will just turn into
the area of a triangle-- it's
the average of the
heights of the two
sides of the trapezoid, the
way we've looked at it--
or you could say the average
of the heights of the two
parallel sides, I guess
is the best way to say it.
So f of 1-- that's the
height here-- plus f of 2,
all of that over
2, and then we're
going to multiply it
times our delta x.
Actually, let me do that
in that same red color
to show you that this is the
area of that first trapezoid,
so times delta x.
And as you see right over here,
if you look at it, the f of 1
is just going to be 0.
So you're going to
have f of 2 times--
so it's going to be this height
times this base times 1/2,
which is just the
area of a triangle.
Let's look at the
second trapezoid,
trapezoid two right over here.
What is its area going to be?
Well, it's going to
be f of 2 plus f of 3.
f of 2 is this height.
f of 3 is this height.
So we're taking the average
of those two heights-- divided
by 2; that's the
average of those two
heights-- times the
base, times delta x.
And then let's do
trapezoid three.
I think you're
getting the idea here.
Trapezoid three is
going to be f of 3
plus f of 4 divided
by 2 times delta x.
And then-- let's see.
I'm running out of colors.
This is trapezoid
four right over here.
So plus f of 4 plus f of 5, all
of that over 2, times delta x.
And then we have
our last trapezoid,
which I will do in yellow.
So this is trapezoid
number five.
I'll scroll down a little bit,
get some more real estate.
So it's going to be plus-- I'll
just write the plus over here--
plus f of 5 plus f of 6
over 2 times our delta x.
So let's see how we can
simplify this a little bit.
All of these terms,
we have a 1/2 delta x,
so let's actually
factor that out.
So remember, this is our
approximation of our area.
So area is
approximately-- remember,
this is just a
rough approximation.
It's very clear--
actually, you might say,
this is pretty good
using the trapezoids,
but it is pretty
clear that we are
letting go of some of the area.
We're letting go of that area.
We're letting go of some
of this right over here.
You can barely see it.
Some of this right over
here, you can barely see it.
But we are-- this is going
to be, it looks like,
an underestimate, but it
is a decent approximation.
But let's see if we can
simplify this expression.
So it's approximately
going to be equal to,
I'm going to factor
out a delta x over 2.
And then what I'm
left with-- and I
will switch to a neutral
color-- if I factor out
a delta x over 2, then
I have just an f of 1.
And then I have two f of
2's, so plus 2 times f of 2.
And I'm doing this because
you might see a formula that
looks something like this
in your calculus book,
and it's not some
mysterious thing.
They just really summed up
the areas of the trapezoids.
And then we're going to have
two f of 3's-- plus 2 times f
of 3's.
Plus we're going to have two f
of 4's-- plus 2 times f of 4.
And then we're going to have two
f of 5's-- plus 2 times f of 5.
And then finally, we're going to
have one f of 6-- plus f of 6.
If you were to
generalize it, you
have one of the first
endpoint, the function
evaluated at the first
endpoint, one of it at the very
last endpoint, and then two
of all of the rest of them.
But this is just the
area of trapezoids.
I'm not actually a big fan
of when textbooks write this.
Because when you see this,
it's hard to visualize
the trapezoids.
When you see this,
it's much clearer
how you might visualize that.
But with that out of the way,
let's actually evaluate this.
Lucky for us, the
math is simple.
Our delta x is just 1.
And then we just have to
evaluate all of this business.
f of 1, let's just
remind ourselves
what our original function was.
Our original function was
the square root of x minus 1.
So f of 1 is the square
root of 1 minus 1,
so that is just going to be 0.
This expression
right over here is
going to be 2 times the
square root of 2 minus 1.
The square root of
2 minus 1 is just 1,
so this is just going to be 2.
Actually, let me do it
in that same-- well,
I'm now using the purple
for a different purpose
than just the first trapezoid.
Hopefully, you realized that.
I was just sticking
with that pen color.
Then f of 3.
3 minus 1 is 2--
square root of 2.
So the function evaluated at
3 is the square root of 2.
So this is going to be 2
times the square root of 2.
Then the function
evaluated at 4.
When you evaluate it at 4,
you get the square root of 3.
So this is going to be 2
times the square root of 3.
And then you get 2 times the
square root of 4-- 5 minus 1
is 4.
2 times the square
root of 4 is just four.
And then finally, you get f of
6 is square root of 6 minus 1,
is the square root of 5.
And I think we're now
ready to evaluate.
So let me get my handy TI-85
out and calculate this.
So it's going to
be-- well I'm just
going to calculate--
well, I'll just multiply.
So 0.5 times open
parentheses-- well, it's a 0.
I'll just write it, just
so you know what I'm doing.
0 plus 2 plus-- whoops.
Lost my calculator.
Plus 2 times the square root of
2 plus 2 times the square root
of 3 plus 4-- I'm almost done--
plus the square root of 5--
so let me write
that-- gives me--
now we are ready for our
drum roll-- it gives me--
and I'll just round it-- 7.26.
So the area is
approximately equal to 7.26
under the curve y is equal
to the square root of x
minus 1 between x
equals 1 and x equals 6.
And we did this
using trapezoids.
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