Definite Integrals

[tkvictim]

Alright, I read the rules of the forum but I couldn't even ATTEMPT to do this and pretend that I have any clue what I'm doing. [My calc teacher is bogus] We're doing definite integrals right now, and I have a few problems that I desperately need help in!
If you could demonstrate all the steps you've taken to arrive at your answer, that'd be awesome.
I illustrated them so it's easier to refer to.

1. directions read: Use the Midpoint rule with the given value of n to approximate the integral. Round the answer to four decimal places.

2. directions read: Express the limit as a definite integral on the given interval
(that is a 6, not a b)

3. directions read: use the form of the definition of an integral to evaluate.

4. Directions read: Evaluate the integral by interpreting it in terms of area.

5.

 

[mmm4444bot]

Please understand that these boards are not an on-line classroom. Few volunteers here have time to type up weeks worth of lessons. The material needed to answer these exercises already exists in a bazillion places on-line and in your local libraries (lessons, examples, explanations, graphic aids). If you don't understand your instructor, seek out alternate authors. :idea:

The two exercises labeled (4) are very basic.

Do you realize that definite integrals represent area underneath a curve?

If a total area is 12 square units, and we take away 3.6 square units, what's left?

What's the area of half a circle, when the radius is 4 units?

What's the area of a right triangle with legs 2 and sqrt(2) ?

On exercise (2), your limit statement is incomplete; it should be the limit as n approaches infinity, right?

Those "i"s are subscripts, too. Have you seen the notation x[sub:3hnb67wr]i[/sub:3hnb67wr]* ? It's a subscripted variable representing the test values in each subinterval.

Do you have a textbook that explains the definition of a definite integral in terms of this limit?

On exercise (1), there are only four subintervals to work with (n = 4); find the midpoints of these subintervals, multiply each one by the function value there, and sum it all up. Your textbook must have an explanation with examples. Look up "midpoint rule" in your book's index.

If you need help understanding some things in your textbook, then those are the questions that you should ask here first. That makes sense, yes?

I'm not sure why you're stuck on any of these, but I suspect that you're not engaged and looking for an easy way out with this assignment.

(Some victims are bogus, too. You claim that your situation "desperate". Why? Maybe you need to invest more time. Maybe you need intensive face-to-face tutoring combined with a good review of pre-calculus concepts. Maybe you truthfully don't care. At this point, I'm not sure how to help you because I don't know what you actually need.)

 

[galactus]

Here is one I can outline to help you see how to use a Riemann sum.

\(\displaystyle \int_{0}^{2}(2-x^{2})dx\)

The subintervals are of length \(\displaystyle {\Delta}x=\frac{b-a}{n}=\frac{2-0}{n}=\frac{2}{n}\). This is like the dx in your integral.

Using the right point method:

\(\displaystyle x_{k}=a+k{\Delta}x=\frac{2k}{n}\)

Thus, rectangle k has area \(\displaystyle f(x_{k}){\Delta}x=\left(2-(\frac{2k}{n})^{2}\right)\cdot \frac{2}{n}\)

Using some algebra, we find the sum of the areas is \(\displaystyle \sum_{k=1}^{n}f(x_{k}){\Delta}x=\frac{1}{n}\sum_{k=1}^{n} 4-\frac{8}{n^{3}}\sum_{k=1}^{n}k^{2}\)

But, making use of the identity, \(\displaystyle \sum_{k=1}^{n}k^{2}=\frac{n(n+1)(2n+1)}{6}\) we get:

\(\displaystyle 4-\left(\frac{4}{n}+\frac{4}{3n^{2}}+\frac{8}{3}\right)\)

Now, take the limit as \(\displaystyle n\to {\infty}\) and we see all the terms tend to 0 except \(\displaystyle 4-\frac{8}{3}=\frac{4}{3}\)

and that is the area under the curve the same as if you integrated the easy way.

That is what a Riemann sum is. It shows how an integral works. We find the area of the infinite number of rectangles.

As these rectangles, n, become more numerous, we find the area under the curve.

 

[BigGlenntheHeavy]

\(\displaystyle 2) \ Riemann \ Sums\)

\(\displaystyle \lim_{n\to\infty}\sum_{i=1}^{n}f(c_i) \Delta x_i \ = \ \int_{a}^{b}f(x)dx\)

\(\displaystyle Hence, \ \lim_{n\to\infty}\sum_{i=1}^{n}\bigg[4-3[f(c_i)]^{2}+6[f(c_i)]^{5}\bigg]\Delta x_i \ = \ \int_{a}^{b}\bigg[4-3[f(x)]^{2}+6[f(x)]^{5}\bigg]dx\)

\(\displaystyle Now, \ Hard \ way, \ to \ wit:\)

\(\displaystyle \Delta x \ = \ \frac{2-0}{n} \ = \ \frac{2}{n} \ and \ f(c_i) \ = \ 0+\frac{2i}{n} \ = \ \frac{2i}{n}, \ limit \ of \ integration \ = \ (0,2)\)

\(\displaystyle Ergo, \ \lim_{n\to\infty}\sum_{i=1}^{n}\bigg[4-3\bigg(\frac{2i}{n}\bigg)^{2}+6\bigg(\frac{2i}{n}\bigg)^{5}\bigg]\frac{2}{n}\)

\(\displaystyle = \ \lim_{n\to\infty}\sum_{i=1}^{n}\bigg[4-3\bigg(\frac{4i^{2}}{n^{2}}\bigg)+6\bigg(\frac{32i^{5}}{n^{5}}\bigg)\bigg]\frac{2}{n}\)

\(\displaystyle = \ \lim_{n\to\infty}\bigg[\frac{2}{n}\sum_{i=1}^{n}4-\frac{6}{n}\sum_{i=1}^{n}\frac{4i^{2}}{n^{2}}+\frac{12}{n}\sum_{i=1}^{n}\frac{32i^{5}}{n^{5}}\bigg]\)

\(\displaystyle = \ \lim_{n\to\infty}\bigg[\frac{2(4n)}{n}-\frac{24}{n^{3}}\sum_{i-1}^{n}i^{2}+\frac{384}{n^{6}}\sum_{i=1}^{n}i^{5}\bigg]\)

\(\displaystyle = \ \lim_{n\to\infty}\bigg[8-\frac{24}{n^{3}}\bigg(\frac{n(n+1)(2n+1)}{6}\bigg)+\frac{384}{n^{6}}\bigg(\frac{n^{2}(n+1)^{2}(2n^{2}+2n-1)}{12}\bigg)\bigg]\)

\(\displaystyle Note: \ \sum_{i=1}^{n} i^{2} \ = \ \bigg(\frac{n(n+1)(2n+1)}{6}\bigg) \ and \ \sum_{i=1}^{n}i^{5} \ = \ \bigg(\frac{n^{2}(n+1)^{2}(2n^{2}+2n-1)}{12}\bigg)\)

\(\displaystyle = \ \lim_{n\to\infty}\bigg[8-\frac{4}{n^{2}}(2n^{2}+3n+1)+\frac{32}{n^{4}}(n^{2}+2n+1)(2n^{2}+2n-1)\bigg]\)

\(\displaystyle = \ \lim_{n\to\infty}[8-8-\frac{12}{n}-\frac{4}{n^{2}}+\frac{32}{n^{4}}(2n^{4}+6n^{3}+5n^{2}-1)]\)

\(\displaystyle = \ \lim_{n\to\infty}\bigg[0-\frac{12}{n}-\frac{4}{n^{2}}+64+\frac{192}{n}+\frac{160}{n^{2}}-\frac{32}{n^{4}}\bigg] \ = \ 64\)

\(\displaystyle Easy \ way:\)

\(\displaystyle \int_{0}^{2}(4-3x^{2}+6x^{5})dx \ = \ 64\)