Approximating areas with Rectangles

Approximating areas with Rectangles - ment the...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
More on Approximating area with rectangles. consider the function: f ( x ) = 16 ° x 2 ; suppose we want to approximate the area under the curve with rectangles on the interval [0 ; 4] with a partition of width 0 : 5 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 0 2 4 6 8 10 12 14 16 x y The area is approximated by using the y ° coordinate of the left endpoint of each subinterval. That is: 1 2 8 X k =1 " 16 ° ° k ° 1 2 ± 2 # = 1 2 h 16 ° (0) 2 i + 1 2 h 16 ° (0 : 5) 2 i + 1 2 h 16 ° (1) 2 i + 1 2 h 16 ° (1 : 5) 2 i + 1 2 h 16 ° (2) 2 i + 1 2 h 16 ° (2 : 5) 2 i + 1 2 h 16 ° (3) 2 i + 1 2 h 16 ° (3 : 5) 2 i = 93 2 = 46 : 5 1
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Now, if we explore the approximation with rectangles whose height is given by the right endpoint of the subinterval: 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 0 2 4 6 8 10 12 14 16 x y We notice that the formulation should be similar to the one used for the left endpoint approach to the approximation. 1 2 7 X k =1 " 16 ° ° k 2 ± 2 # = 1 2 h 16 ° (0 : 5) 2 i + 1 2 h 16 ° (1) 2 i + 1 2 h 16 ° (1 : 5) 2 i + 1 2 h 16 ° (2) 2 i + 1 2 h 16 ° (2 : 5) 2 i + 1 2 h 16 ° (3) 2 i + 1 2 h 16 ° (3 : 5) 2 i = 77 2 = 38 : 5 As before, we can see, that the actual area A is somewhere between 38 : 5 and 46 : 5 : 38 : 5 ± A ± 46 : 5 2
Image of page 2
It is imperative, that we illustrate the fact, that if we choose a better re°ne-
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ment, the approximation whould be much better. Lets choose a partition, such that the base of the rectangle is of length 1 100 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 2 4 6 8 10 12 14 16 x y We can see, that the sum of the area of the approximating rectangles is "close" compared to the previous re&nement. Since the curve is decreasing by choosing the left endpoint of the interval, the sum of the areas of the approximating rectangles, is greater that the actual area. 1 100 400 X k =1 " 16 & & k & 1 100 ± 2 # = 213 733 5000 ± 42 : 746 6 We will explain in the next couple of sections, how to calculate the actual area, which is A = 128 3 ± 42 : 666 666 666 666 666 667 3 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 2 4 6 8 10 12 14 16 x y 4...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern