Finding Left Endpoints without Seeing a Graph

[Jason76]

Dealing with endpoints of vertical rectangles on a graph (primitive integration)

\(\displaystyle f(x) = 4 \cos(x)\)

Using formulas:

\(\displaystyle [a, b]\) - x interval

\(\displaystyle n\) - number of intervals

\(\displaystyle \Delta x = \dfrac{b - a}{n}\) - change in x

\(\displaystyle \sum(\Delta x) [f(a + (i)(\Delta x) -\Delta x)]\) - Formula for right endpoints

Given:

Interval: \(\displaystyle [0,\dfrac{\pi}{2}]\)

\(\displaystyle n = 4\)

\(\displaystyle \Delta x = \dfrac{\dfrac{\pi}{2} - 0}{4} = \dfrac{\pi}{8}\)

Problem:

\(\displaystyle \sum (\dfrac{\pi}{8}) [[f(0 + (1)(\dfrac{\pi}{8}) - \dfrac{\pi}{8})] + [f(0 + (2)(\dfrac{\pi}{8}) -\dfrac{\pi}{8}) ] + [f(0 + (3)(\dfrac{\pi}{8}) - (\dfrac{\pi}{8}) ] + [f(0 + (4)(\dfrac{\pi}{8}) - (\dfrac{\pi}{8})]]\)

\(\displaystyle \sum (\dfrac{\pi}{8}) [[f(0)] + [f(\dfrac{\pi}{8})] + [f(\dfrac{\pi}{4})]] + [f(\dfrac{3\pi}{8})]]\)

\(\displaystyle \sum (\dfrac{\pi}{8}) [[4 \cos(0)] + [4 \cos(\dfrac{\pi}{8})] + [4 \cos(\dfrac{\pi}{4})] + [4 \cos(\dfrac{3\pi}{8})]]\) Any idea on some of these values? What would be the area under the curve?

 

[stapel]

If you're asking about evaluating the cosine terms exactly, this is something you would have learned back in trig or pre-calculus. Try looking up "double-" and "half-angle identities", and using this information to convert the pi-over-8's into pi-over-4's.

 

[DrPhil]

LaTeX hints

Summation Sign \(\displaystyle (\Delta x) [f(a + (i)(\Delta x) -\Delta x)]\)

The summation sign in LaTeX is \sum. To make it large, precede it with \displaystyle. You can take \Delta x outside the sum. When in displaystyle, you can make sure the brackets are of appropriate height by using \left[ and \right]

Lefthand sum = \(\displaystyle \displaystyle \Delta x \sum_{i=1}^{n} \left[f(a + (i-1)\Delta x) \right]\)

 

[Jason76]

The summation sign in LaTeX is \sum. To make it large, precede it with \displaystyle. You can take \Delta x outside the sum. When in displaystyle, you can make sure the brackets are of appropriate height by using \left[ and \right]

Lefthand sum = \(\displaystyle \displaystyle \Delta x \sum_{i=1}^{n} \left[f(a + (i-1)\Delta x) \right]\)

R u sure? My formula also works. Oh wait, it is the same formula (with \(\displaystyle \Delta x\) factored out.

 

[Jason76]

Dealing with endpoints of vertical rectangles on a graph (primitive integration)

\(\displaystyle f(x) = 4 \cos(x)\)

Using formulas:

\(\displaystyle [a, b]\) - x interval

\(\displaystyle n\) - number of intervals

\(\displaystyle \Delta x = \dfrac{b - a}{n}\) - change in x

\(\displaystyle \sum(\Delta x) [f(a + (i)(\Delta x) -\Delta x)]\) - Formula for right endpoints

Given:

Interval: \(\displaystyle [0,\dfrac{\pi}{2}]\)

\(\displaystyle n = 4\)

\(\displaystyle \Delta x = \dfrac{\dfrac{\pi}{2} - 0}{4} = \dfrac{\pi}{8}\)

Problem:

\(\displaystyle \sum (\dfrac{\pi}{8}) [[f(0 + (1)(\dfrac{\pi}{8}) - \dfrac{\pi}{8})] + [f(0 + (2)(\dfrac{\pi}{8}) -\dfrac{\pi}{8}) ] + [f(0 + (3)(\dfrac{\pi}{8}) - (\dfrac{\pi}{8}) ] + [f(0 + (4)(\dfrac{\pi}{8}) - (\dfrac{\pi}{8})]]\)

\(\displaystyle \sum (\dfrac{\pi}{8}) [[f(0)] + [f(\dfrac{\pi}{8})] + [f(\dfrac{\pi}{4})]] + [f(\dfrac{3\pi}{8})]]\)

\(\displaystyle \sum (\dfrac{\pi}{8}) [[4 \cos(0)] + [4 \cos(\dfrac{\pi}{8})] + [4 \cos(\dfrac{\pi}{4})] + [4 \cos(\dfrac{3\pi}{8})]]\)

Here is the answer but computer still not getting it right. Asking for L4 - which i think is the endpoint of the 4th triangle.

\(\displaystyle \sum (\dfrac{\pi}{8}) [[4]+[4]+[4]+[4]]\)

\(\displaystyle \sum [1.6]+[1.6]+[1.6]+[1.6]\)

OR

\(\displaystyle \sum [\dfrac{\pi}{2}]+[\dfrac{\pi}{2}]+[\dfrac{\pi}{2}]+[\dfrac{\pi}{2}]\)

\(\displaystyle L4 = 1.6 = \dfrac{\pi}{2}\) This is the question

Total Area under curve = \(\displaystyle 6.4 = 2\pi\)

 

[DrPhil]

Dealing with endpoints of vertical rectangles on a graph (primitive integration)

\(\displaystyle f(x) = 4 \cos(x)\)

Using formulas:

\(\displaystyle [a, b]\) - x interval

\(\displaystyle n\) - number of intervals

\(\displaystyle \Delta x = \dfrac{b - a}{n}\) - change in x

\(\displaystyle \sum(\Delta x) [f(a + (i)(\Delta x) -\Delta x)]\) - Formula for left endpoints

Given:

Interval: \(\displaystyle [0,\dfrac{\pi}{2}]\)

\(\displaystyle n = 4\)

\(\displaystyle \Delta x = \dfrac{\dfrac{\pi}{2} - 0}{4} = \dfrac{\pi}{8}\)

Problem:

Since you have written every term on the sum explicitly, there is no "sum" operation still to be done ...
The Sum symbols are not appropriate in the next three lines.
\(\displaystyle \sum (\dfrac{\pi}{8}) [[f(0 + (1)(\dfrac{\pi}{8}) - \dfrac{\pi}{8})] + [f(0 + (2)(\dfrac{\pi}{8}) -\dfrac{\pi}{8}) ] + [f(0 + (3)(\dfrac{\pi}{8}) - (\dfrac{\pi}{8}) ] + [f(0 + (4)(\dfrac{\pi}{8}) - (\dfrac{\pi}{8})]]\)

\(\displaystyle \sum (\dfrac{\pi}{8}) [[f(0)] + [f(\dfrac{\pi}{8})] + [f(\dfrac{\pi}{4})]] + [f(\dfrac{3\pi}{8})]]\)

\(\displaystyle \sum (\dfrac{\pi}{8}) [[4 \cos(0)] + [4 \cos(\dfrac{\pi}{8})] + [4 \cos(\dfrac{\pi}{4})] + [4 \cos(\dfrac{3\pi}{8})]]\) Any idea on some of these values? What would be the area under the curve?

Wouldn't "L4" refer to the 4th term in the left-hand Rieman sum?

\(\displaystyle \displaystyle f(a + 3\Delta x) = f\left(\dfrac{3\pi}{8}\right) = \ \cdot\ \cdot\ \cdot\)