# Using left endpts help with squr(3i/16) how to get the i out

• camboguy
In summary, the conversation discusses the process of approximating the area under a curve using left endpoints. The equations and method used involve finding the change in x, determining the x values for each rectangle, and multiplying them by the corresponding value of f(x) before summing them up. It is noted that this process can be tedious but is necessary for accurate approximation.
camboguy

## Homework Statement

i am have to approximate the area under the curve using left endpoints.

## Homework Equations

y= squr(x+2) on [1,4], n =16

## The Attempt at a Solution

i have the change in x to be 3/16 using (b-a)/n i also found my xsubi to be (1+3i/16) using xsubi = xsubo + i(chang in x). then i got (3/16) E (squr(1+(3i/16)+ 2)) i am stuck on how to get the i out to then apply the formula n(n+1)/2 or n(n+1)(2n+1)/6.

another question similar problem but i get stuck after i don't know how to take the i out of cos(pi(i)/100) i tried multiplying by sec but i still don't get the right answer. do i multiply by sec? to get the i out of cos(pi(i)/100)?

Last edited:
You can't apply those summations to this one, the first one applies to linear functions and the second to quadratics, but for this one all they want is to use left rectangles.

You know the x values you have to set up the rectangles at, and the width (3/16). Now all you have to do is multiply each by f(x) at the point you set up the rectangle at, them sum them all up. Simple, but tedious process.

## 1. How do I use left endpoints to help with squr(3i/16) to get the i out?

Using left endpoints can help with simplifying the expression squr(3i/16) and getting the i out. This is because the left endpoint of a square root represents the smallest possible value, and since i is a complex number, it has no real value. Therefore, using left endpoints allows us to find the smallest possible value for i, making it easier to simplify the expression.

## 2. What is the purpose of using left endpoints in this situation?

The purpose of using left endpoints in this situation is to help us simplify the expression squr(3i/16) and get the i out. By finding the smallest possible value for i, we can make the expression easier to work with and understand.

## 3. Can I use any other method to simplify squr(3i/16) instead of using left endpoints?

Yes, there are other methods that can be used to simplify squr(3i/16), such as using the properties of square roots or converting the complex number into polar form. However, using left endpoints is a specific method that can be helpful in certain situations, such as when trying to get the i out of a complex number.

## 4. Are there any limitations to using left endpoints to simplify squr(3i/16)?

One limitation of using left endpoints is that it may not always give the exact value of i. This is because the left endpoint represents the smallest possible value, but i can have infinitely many values. Therefore, using left endpoints may provide an approximate value for i, but it may not be the exact value.

## 5. In what other situations can using left endpoints be helpful?

Using left endpoints can be helpful in simplifying other expressions that involve complex numbers, such as squr(5i/24) or squr(-2i/3). It can also be useful in finding the smallest possible value for a variable in an equation, such as finding the smallest possible value for x in an equation like x^2 = 25.

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