Approximating Area

[nycmathguy]

Exercises 1.1
Question 9

Keep in mind that this is a Chapter 1 Question. If you think this question is too advanced for me right now in my studies, please skip.

 

[MathLover1]

too advanced, need integral

but you can do approximate area by finding all points on a curve, calculating distance between them, calculating the area of each right trapezoid (two 90-degree angles), and add them all

 

[nycmathguy]

too advanced, need integral

but you can do approximate area by finding all points on a curve, calculating distance between them, calculating the area of each right trapezoid (two 90-degree angles), and add them all

I guess Ron Larson expects students to calculate "...the area of each right trapezoid (two 90-degree angles), and add them all."

 

[MathLover1]

yes

 

[nycmathguy]

yes

Why else would he introduce the integral so early in the textbook? As you said, we can skip this question. Good to know what's coming up.

 

[HallsofIvy]

Finding the exact area would require an integral but that is not what this exercise is asking. Also this exercise specifically says "use the rectangles", not trapezoids.
The first approximation has four rectangles each with width 1 so areas 5, 5/2, 5/3, and 5/4. The total area of the rectangles is 5+ 5/2+ 5/3+ 5/4= 60/12+ 30/12+ 20/12+ 15/12= 125/12= 10 and 5/12.

The second is the same except that now there are 8 rectangles each width width 1/2.

 

[nycmathguy]

Finding the exact area would require an integral but that is not what this exercise is asking. Also this exercise specifically says "use the rectangles", not trapezoids.
The first approximation has four rectangles each with width 1 so areas 5, 5/2, 5/3, and 5/4. The total area of the rectangles is 5+ 5/2+ 5/3+ 5/4= 60/12+ 30/12+ 20/12+ 15/12= 125/12= 10 and 5/12.

The second is the same except that now there are 8 rectangles each width width 1/2.

Sir, I posted Calculus 1 questions thinking that I could handle two math textbooks at the same time. I was wrong. I will stick to one course at a time.

 

[Country Boy]

This is NOT a Calculus problem! It is a geometry problem, asking you to calculate the areas of rectangles given width and height.

 

[MathLover1]

This is NOT a Calculus problem! It is a geometry problem, asking you to calculate the areas of rectangles given width and height.

The roots of calculus lie in some of the oldest geometry problems on record. By definition, definite integral is the sum of the product of the lengths of intervals and the height of the function that is being integrated with that interval, which includes the formula of the area of the rectangle.

 

[nycmathguy]

The roots of calculus lie in some of the oldest geometry problems on record. By definition, definite integral is the sum of the product of the lengths of intervals and the height of the function that is being integrated with that interval, which includes the formula of the area of the rectangle.

There is a connection between calculus and geometry. We can say that about trigonometry and other courses as well.

 

[nycmathguy]

This is NOT a Calculus problem! It is a geometry problem, asking you to calculate the areas of rectangles given width and height.

There is a connection between calculus and geometry. We can say that about trigonometry and other courses as well.

 

[Country Boy]

The roots of calculus lie in some of the oldest geometry problems on record. By definition, definite integral is the sum of the product of the lengths of intervals and the height of the function that is being integrated with that interval, which includes the formula of the area of the rectangle.

First, I would NOT say "by definition". The integral has to be defined in a way that does not refer to "rectangles" since many applications of integration have nothing to do with rectangles or geometry.

What "rectangles" would you use to find the Lesbesque integral, from 0 to 1, of f(x) defined to be 1 if x is irrational and 0 if x is rational?

Finally, even for a Riemann integral, the integral is NOT "the sum of the product of the lengths of intervals and the height of the function that is being integrated with that interval"! It is the limit of such sums as the width of the rectangles go to 0 and the number of rectangles goes to infinity.

 

[MathLover1]

"by definition" of definite integral: the difference between the values of the integral of a given function f(x) for an upper value b and a lower value a of the independent variable x

If you’re integrating a function f(x), the Lebesgue integral is just a way to evaluate along the range of f(x) instead of the domain (which is the Riemann method).

The “Cauchy integral” is a concept from complex analysis that has a very limited application. It’s really a misnomer to call it an “integral”, because it’s actually a formula in which an integral is evaluated that defines a rule about holomorphic functions (whatever those are). You will probably never learn about any of this unless you become a math major.

Rectangular integration is a numerical integration technique that approximates the integral of a function with a rectangle. It uses rectangles to approximate the area under the curve. Rectangles would used to approximate the integral; each smaller rectangle has the width of the smaller interval.

Usually, integration using rectangles is the first step for learning integration. At its most basic, integration is finding the area between the x axis and the line of a function on a graph - if this area is not "nice" and doesn't look like a basic shape (triangle, rectangle, etc.) that we can easily calculate the area of, a good way to approximate it is by using rectangles.

 

[nycmathguy]

"by definition" of definite integral: the difference between the values of the integral of a given function f(x) for an upper value b and a lower value a of the independent variable x

If you’re integrating a function f(x), the Lebesgue integral is just a way to evaluate along the range of f(x) instead of the domain (which is the Riemann method).

The “Cauchy integral” is a concept from complex analysis that has a very limited application. It’s really a misnomer to call it an “integral”, because it’s actually a formula in which an integral is evaluated that defines a rule about holomorphic functions (whatever those are). You will probably never learn about any of this unless you become a math major.

Rectangular integration is a numerical integration technique that approximates the integral of a function with a rectangle. It uses rectangles to approximate the area under the curve. Rectangles would used to approximate the integral; each smaller rectangle has the width of the smaller interval.

Usually, integration using rectangles is the first step for learning integration. At its most basic, integration is finding the area between the x axis and the line of a function on a graph - if this area is not "nice" and doesn't look like a basic shape (triangle, rectangle, etc.) that we can easily calculate the area of, a good way to approximate it is by using rectangles.

I hope my questions did not start a war between you and Country Boy.

 

[MathLover1]

I hope my questions did not start a war between you and Country Boy.

what war you are talking about, we just exposing some differences between European and US standards in definition of integral

 

[nycmathguy]

what war you are talking about, we just exposing some differences between European and US standards in definition of integral

Maybe I'm reading too much into this....

 

[Country Boy]

I will take exception to
"If you’re integrating a function f(x), the Lebesgue integral is just a way to evaluate along the range of f(x) instead of the domain (which is the Riemann method)."

The difference between the Riemann integral and the Lebesque integral is not between domain and range. The Riemann integral divides the domain into intervals that are taken to be smaller and smaller (going to 0). The Lebesque integral uses a more abstract "measure" that gives a measure to non- interval sets- but still in the domain.

For example, if f(x)= 0 for x rational and f(x)= 1 for x irrational, the Riemann integral of f from 0 to 1 does not exist while the Lebesque integral does and is equal to 1.

The Lebesque integral is more general than the Riemann integral- if the Riemann integral of a function exists so does the Lebesque integral and the two integrals are the same but the Lebesque integral of a function may exist when the Riemann integral does not.

 

[nycmathguy]

I will take exception to
"If you’re integrating a function f(x), the Lebesgue integral is just a way to evaluate along the range of f(x) instead of the domain (which is the Riemann method)."

The difference between the Riemann integral and the Lebesque integral is not between domain and range. The Riemann integral divides the domain into intervals that are taken to be smaller and smaller (going to 0). The Lebesque integral uses a more abstract "measure" that gives a measure to non- interval sets- but still in the domain.

For example, if f(x)= 0 for x rational and f(x)= 1 for x irrational, the Riemann integral of f from 0 to 1 does not exist while the Lebesque integral does and is equal to 1.

The Lebesque integral is more general than the Riemann integral- if the Riemann integral of a function exists so does the Lebesque integral and the two integrals are the same but the Lebesque integral of a function may exist when the Riemann integral does not.

Good information for extra study notes.