A short personal integral

(5 pts)

For this problem, you’re going to write out two approximations to an integral using just 4 terms in your approximation. The formulae for the approximating sums are in our textbook and also at the bottom of this webpage.

Note: We are interested in what the approximation looks like written out long hand - not so much what the decimal approximation is. Be sure to clearly indicate what a, b, \Delta x, and x_i are, as well as the sum of four terms that forms your final answer. A picture indicating what the partition looks like would be a good thing, too.

To find the exact integral and types of approximations, choose your name from the following drop down menu:

My integral is \int_1^2\sin(\pi x)dx and I will be estimating using a right and trapezoidal sum.

Trapezoidal Sum

First, I identified f(x)=\sin(\pi x), \Delta x=\frac{1}{4}, and x_i=1.25.
For \Delta x I used the formula \Delta x=\frac{b-a}{n} with b=2, a=1, and n=4.
For x_i is used the formula x_i=a+i \Delta x

Using the trapezoidal sum formula, we get;

Right Sum

First, I identified f(x)=\sin(\pi x), \Delta x=\frac{1}{4}, and x_i=1.25.
For \Delta x I used the formula \Delta x=\frac{b-a}{n} with b=2, a=1, and n=4.
For x_i is used the formula x_i=a+i \Delta x

Using the right sum formula we get;

@rmeares This looks great! I made a couple of small edits to make the typesetting just a little more groovy.:

• I replaced each sin with \sin so that each sin(x) looks like \sin(x) and
• I typeset the larger, full line formulae in display style.

You can accomplish that second thing by entering

$$
your larger formula here
$$

instead of

$your should be larger but still small formula here$

So, for example

\int_{0}^{\infty} e^{-x^2} \, dx = \frac{\sqrt{\pi}}{2}

will look like

I estimated the integral

by using right and trapezoidal sums.

After assigning the values and plugging them in, I got:

As the left reimann sum. I then assigned the values to the trapezoidal sum formula:

My integral is \int_0^{2}\sqrt {1+x^4}dx I used both left and a midpoint sum

For my midpoint I did this:

For my Left I did this:

When we write

then the index variable i takes on integer values from start to finish. For example,

Note that the number of terms is determined by the bounds on i. So…

• @bjohnso9 Your sum has 3 terms and
• @jcastel1 Your sum has 5 terms.

But I’m pretty sure they’re both supposed to have 4 terms.

The integral I received was \int_{-1}^{1} \sqrt{1+x^4} \, dx and I had to use both a right and trapezoidal sum.

To find my four terms using Right sum I used the formula \Delta x = \frac{b-a}{n} with a=-1, b=1, n=4 which gave me \Delta x = \frac{1-(-1)}{4} = \frac{1}{2} and x_i= a + i\,\Delta x which gave me x_i= (-1) + 1\,\frac{1}{2} = \frac{1}{2} or 0.5.

My four terms are:
\sqrt{1+(0.5)^4} *\frac{1}{2}+ \sqrt{1+(1.0)^4} *\frac{1}{2}+ \sqrt{1+(1.5)^4} *\frac{1}{2}+ \sqrt{1+(2.0)^4} *\frac{1}{2}

To find my four terms using Trapezoidal sum I used the formula \Delta x = \frac{b-a}{n} with a=-1, b=1, n=4 which gave me \Delta x = \frac{1-(-1)}{4} = \frac{1}{2} and x_i= a + i\,\Delta x which gave me x_i= (-1) + 1\,\frac{1}{2} = \frac{1}{2} or 0.5.

My four terms are:
\frac{\sqrt{1+(0.5)^4} + \sqrt{1+(0.5)^4}}{2}*\frac{1}{2}+ \frac{\sqrt{1+(1.0)^4} + \sqrt{1+x(1.0)^4}}{2}*\frac{1}{2}+\frac{\sqrt{1+(1.5)^4} + \sqrt{1+(1.5)^4}}{2}*\frac{1}{2}+\frac{\sqrt{1+(2.0)^4} + \sqrt{1+(2.0)^4}}{2}*\frac{1}{2}

My integral was:
\int_0^1\sin(\pi x)dx

and I wanted to find the left sum and mid point

The left sum was
\sum_{i=1}^{4} \sin(\pi{(\frac{i-1}{4}))}*\frac{1}{4} = \frac{1+\sqrt2}{4}

The Midpoint sum was
\sum_{i=1}^{4} \sin(\pi(\frac{{(\frac{i}{4})+(\frac{i-1}{4})}}{2}))*\frac{1}{4} = 0.65328148

I was given

So the right sum is

And the trapezoidal sum is

I was given the integral

And was asked to give the right and trapezoidal sum.

To begin, I used the trig half angle identities to get

Then I found that \Delta x=1, and x_i=i (x_i=a+i\Delta x, x_i=0+i*1)

So for the trapezoidal sum, I found

For the right sum, I found

My integral:

I was asked to estimate the value of the integral using both a left sum with four terms and a midpoint sum with four terms.

The left sum:

The midpoint sum:

I was asked to estimate:

using right and trapezoidal sums.

First, I defined my terms:
\ n=4, ∆x=1/2, a=-1, b=1, and \ x_i=-1+i/2

Then I plugged it in to the right hand sum to get:

this comes out to:
\ (1/2)(cos(-1/16)+cos(0)+cos(1/16)+cos(1)) ≈ 1.768

Then for trapezoidal sum:

This comes out to:
\ \frac{1}{4} (cos(-1)+2cos(-1/16)+2cos(0)+2cos(1/16)+cos(1)) ≈ 1.768

I couldn’t figure out how to make the squares and trapezoids, but here’s a graph of the function:

I’m not sure why you’re using trig identities here.

My integral is

estimated using left and midpoint sums with four terms.

a = 0, b = 2, \Delta x= \frac {1}{2}\ , and x_i= \frac {i}{2}\

For the left sum I got:

For the midpoint sum I got: