lisasayzhi
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Daniel_Feldman
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Re: 2 More Very Tricky Definite Integral Problems
For problem 5, think about how you originally derived the Fundamental theorem of calculus. You have a function, continous and differentiable on an interval [a,b]. You then partition it into n subintervals, each having a width of delta(x)=(b-a)/n. \(\displaystyle \int f(x)dx\) from a to b is the limit, as n approaches infinity, of the sequence from i=1 to n, of f(Xi)(delta(x)), where Xi is the ith subinterval.
In this case, you have a=b, so delta(x)=(a-a)/n=0
Then you get the limit, as n approaches infinity, of the sequence from i=1 to n, of f(Xi)(0)=0, and so \(\displaystyle \int f(x)dx=0\)
6. Are you familiar with the formula for finding the average value of a function? If you are, that's where I would start.
lisasayzhi said:Okay, these are the hardest problems on my homework that are giving me the most trouble. I'm not sure what to do for either of them, so I don't know how to show you much work yet, but I did put a little explanation under each one. If this is not sufficient for a hint at least, I'm very sorry. Please just let me know if you need more work than this, and I'll just have to do my best.
5. Show \(\displaystyle \int f(x)dx=0\) from a to a without using the word area.
For this one, I understand that it is true... but I can't figure out how to "show" it. Wouldn't this basically be saying that the integral at one point, a is 0? How do I show that mathematically? All I can think to write is that according to the FTC, \(\displaystyle \int^b_a f(x)dx = F(b) - F(a)\) so, since a-a=0, any function involving a minus the same function is also 0. I'm not sure if this is very coherent though... Sorry, I'm confused.
6. Is there \(\displaystyle c \in (0,\pi)\) such that \(\displaystyle cos^8 c = \frac{1}{\pi} \int_{2\pi}^{3\pi} cos^8x dx\)?
Our professor didn't really explain these types of problems... so would anyone be able to explain to me what it's asking... or point me to a website that could clear up these types of problems?
For problem 5, think about how you originally derived the Fundamental theorem of calculus. You have a function, continous and differentiable on an interval [a,b]. You then partition it into n subintervals, each having a width of delta(x)=(b-a)/n. \(\displaystyle \int f(x)dx\) from a to b is the limit, as n approaches infinity, of the sequence from i=1 to n, of f(Xi)(delta(x)), where Xi is the ith subinterval.
In this case, you have a=b, so delta(x)=(a-a)/n=0
Then you get the limit, as n approaches infinity, of the sequence from i=1 to n, of f(Xi)(0)=0, and so \(\displaystyle \int f(x)dx=0\)
6. Are you familiar with the formula for finding the average value of a function? If you are, that's where I would start.
For #6, I would use the Mean Value theorem for integrals.
c is some number in the interval from 0 to pi.
So we have:
\(\displaystyle \frac{1}{\pi}\int_{2{\pi}}^{3{\pi}}(cos^{8}(x))dx=\)
\(\displaystyle \frac{1}{{\pi}}cos^{8}(c)(3{\pi}-2{\pi})=cos^{8}(c)\)
Now, evaluate your integral. Set \(\displaystyle cos^{8}(c)\) equal to it and
solve for c. That will give you the number whose existence is guaranteed
by the Mean-Value theorem.
c is some number in the interval from 0 to pi.
So we have:
\(\displaystyle \frac{1}{\pi}\int_{2{\pi}}^{3{\pi}}(cos^{8}(x))dx=\)
\(\displaystyle \frac{1}{{\pi}}cos^{8}(c)(3{\pi}-2{\pi})=cos^{8}(c)\)
Now, evaluate your integral. Set \(\displaystyle cos^{8}(c)\) equal to it and
solve for c. That will give you the number whose existence is guaranteed
by the Mean-Value theorem.
lisasayzhi
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Hello lisasayzhi.
Do you have access to Maple, a nice calculator, something other than
doing this monstrosity by hand. It could be done, but it would be arduous.
One way I would attack it would possibly be by letting:
\(\displaystyle cos^{2}(x)=\frac{1}{2}+\frac{1}{2}cos(2x)\)
This gives:
\(\displaystyle \frac{1}{16}\int_{2{\pi}}^{3{\pi}}(1+cos(2x))^{4}dx\)
=\(\displaystyle \int_{2{\pi}}^{3{\pi}}(\frac{1}{16}+\frac{1}{4}cos(2x)+\frac{3}{8}cos^{2}(2x)+\frac{1}{4}cos^{3}(2x)+\frac{1}{16}cos^{4}(2x))dx\)
WHEW!!!, see what I mean. You can keep breaking it down, but it would
be a long and winding road. If I weren't so lazy right now, I might tackle
it. But anyway, my TI spit out the answer in about 2 seconds. \(\displaystyle \frac
{35}{128}\)
Next, set \(\displaystyle cos^{8}(c)=\frac{35}{128}\)
Now, solve for c. Can you do that?.
Do you have access to Maple, a nice calculator, something other than
doing this monstrosity by hand. It could be done, but it would be arduous.
One way I would attack it would possibly be by letting:
\(\displaystyle cos^{2}(x)=\frac{1}{2}+\frac{1}{2}cos(2x)\)
This gives:
\(\displaystyle \frac{1}{16}\int_{2{\pi}}^{3{\pi}}(1+cos(2x))^{4}dx\)
=\(\displaystyle \int_{2{\pi}}^{3{\pi}}(\frac{1}{16}+\frac{1}{4}cos(2x)+\frac{3}{8}cos^{2}(2x)+\frac{1}{4}cos^{3}(2x)+\frac{1}{16}cos^{4}(2x))dx\)
WHEW!!!, see what I mean. You can keep breaking it down, but it would
be a long and winding road. If I weren't so lazy right now, I might tackle
it. But anyway, my TI spit out the answer in about 2 seconds. \(\displaystyle \frac
{35}{128}\)
Next, set \(\displaystyle cos^{8}(c)=\frac{35}{128}\)
Now, solve for c. Can you do that?.