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I've made a slight error here.  

I can't edit so I will just explain.

dx=-2y dy

I forgot the negative sign so the answer should all be in brackets with a minus sign out the front.

I might as well simplify the answer while I am at it :)

\(\int\;x^2\;\sqrt{2-x}\;dx\\ =-\left[ \frac{8(2-x)^{1.5}}{3}-\frac{8(2-x)^{2.5}}{5}+\frac{2(2-x)^{3.5}}{7}\right]+c\\ =-\frac{2}{3*5*7}\left[ \frac{5*7*4(2-x)^{1.5}}{1}-\frac{3*7*4(2-x)^{2.5}}{1}+\frac{3*5*(2-x)^{3.5}}{1}\right]+c\\ =-\frac{2}{105}\left[ \frac{140(2-x)^{1.5}}{1}-\frac{84(2-x)^{2.5}}{1}+\frac{15(2-x)^{3.5}}{1}\right]+c\\ =-\frac{2}{140}\left[ 140(2-x)^{1.5}-84(2-x)^{2.5}+15(2-x)^{3.5}\right]+c\\ =-\frac{2(2-x)^{1.5}}{140}\left[ 140-84(2-x)+15(2-x)^{2}\right]+c\\ =-\frac{2(2-x)^{1.5}}{140}\left[ 140-168+84x+15(4-4x+x^2)\right]+c\\ =-\frac{2(2-x)^{1.5}}{140}\left[ 140-168+84x+60-60x+15x^2\right]+c\\ =\frac{-2(2-x)^{1.5}}{140}\left[ 15x^2+24x+32 \right]+c\\\)

3)

\(\int\frac{x}{\sqrt{x+5}}\;dx\\ let\;\;u=x+5\\ so\;\;x=u-5\\ and\;\;dx=du\\~\\ =\int\;\frac{u-5}{\sqrt u}\;du\\ =\int\;u^{0.5}-5u^{-0.5}\;du\\ =\frac{u^{1.5}}{1.5}-\frac{5u^{0.5}}{0.5}+c\\ =\frac{2u^{1.5}}{3}-\frac{10u^{0.5}}{1}+c\\ =\frac{2u^{1.5}}{3}-\frac{30u^{0.5}}{3}+c\\ =\frac{2u^{0.5}}{3}\left[{u}-15\right]+c\\ =\frac{2\sqrt{x+5}}{3}\left[x+5-15\right]+c\\ =\frac{2\sqrt{x+5}}{3}\left[x-10\right]+c\\ =\frac{2\sqrt{x+5}\;(x-10)}{3}+c\\ \)

b=7

2)      \(\int x(sec^2(2x))\;dx\)

I did this with integration by parts.

\(\boxed{\int v \frac{du}{dx}dx=uv-\int u\frac{dv}{dx}dx}\\~\\ v=x \qquad \frac{du}{dx}=sec^2(2x)\\ \frac{dv}{dx}=1 \qquad u=\frac{tan(2x)}{2} \\~\\ \int\;x(sec^2(2x))\;dx\\ =\frac{xtan(2x)}{2}-\int\;1*\frac{tan(2x)}{2}\;dx\\ =\frac{xtan(2x)}{2}-\frac{1}{2}\int tan(2x)\;dx\\ =\frac{xtan(2x)}{2}-\frac{1}{2}\int \frac{sin(2x)}{cos(2x)}\;dx\\ =\frac{xtan(2x)}{2}-\frac{1}{2}\frac{ln(cos(2x)}{-2}+c\\ =\frac{xtan(2x)}{2}+\frac{ln(cos(2x))}{4}+c\\\)

1.     integral of x^2(2-x)^0.5 dx

\(\int\;x^2\sqrt{2-x}\;dx\\~\\ \)

2-x has to be positive so I am going to substitute it for  y^2

\(2-x=y^2\\ 2-y^2=x\\ x=2-y^2\\ \frac{dx}{dy}=-2y\\ dx=2y\;dy \)

\(\int\;x^2\;\sqrt{2-x}\;dx\\ =\int \;(2-y^2)^2*\sqrt{y^2}\;*2y\;dy\\ =\int \;(4-4y^2+y^4)*y\;*2y\;dy\\ =\int \;2y^2(4-4y^2+y^4)\;dy\\ =\int \;(8y^2-8y^4+2y^6)\;dy\\ =\frac{8y^3}{3}-\frac{8y^5}{5}+\frac{2y^7}{7}+c\\~\\ =\frac{8(2-x)^{1.5}}{3}-\frac{8(2-x)^{2.5}}{5}+\frac{2(2-x)^{3.5}}{7}+c\\~\\\)

I haven't had much practice with these but that looks alright :)

I think so anyway.

Don't be Lazy again like this MWizzard.  I wasted a lot of time today puzzling over your incomplete question !!

Why don't you use a calculator?

27:00 - 15:16 = 11.84

but learn to d this yourself kid, if you dont, you will regret it... a lot

(trust me, i'm speaking from experiene)

If the following image represents the situation:

then the solution is as follows (where V = volume and S  surface area):

.

This might help answer your question Chris:

It isn't defined for values of t = 0, -1, -2, ... etc though, where you get division by zero.

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If you are talking about the sum of the first n-1 positive integers then it is given by:

SUM(n-1) = (n-1)*n/2

so SUM(6-1) = (6-1)*6/2 → 5*3 → 15

(i.e.    1 + 2 + 3 + 4 + 5 = 15)

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Hmm... I was too lazy to illustrate. So that is why you have asked. I had that question too.

Is that a part of the question or what? 

 If it is then the question appears to be relatively trivial.

There are 8 faces on the longer face and 4 faces on the shorter face.

Are you sure that this question is presented properly and nothing is missing?

I like the way you did this one, Rom....I have not seen this before.......very crafty !!!!

Suppose that f is a function such that 3f(x)-5xf(1/x)=x-7 for all non-zero real numbers x. Find f(2010)

\(3f(2010) - 5(2010)f\left(\dfrac 1 {2010}\right) = (2010)-7\)

\(3f(2010)-10050f\left(\dfrac 1 {2010}\right) = 2003\)

we can also write

\(3f\left(\dfrac 1 {2010}\right) - \dfrac {5}{2010}f(2010)=\dfrac 1{2010}-7=-\dfrac{14069}{2010}\)

\(let f(2010)=x, f\left(\dfrac 1 {2010}\right)=y\)

\(3x-10050y=2003\\ -\dfrac{5}{2010}x + 3y = -\dfrac{10469}{2010}\)

Solving this we get

\(x=2896\\\\ y = \dfrac{1337}{2010}\)

and thus

\(f(2010) = 2896\)

Part b is straightforward.......we would end up with a profit of 28*$20  = $560  for 28 days, if we bought the same items every day

However.....note that on every 5th day.......we would not actually have to buy any cups, because for the four previous days.....we would have purchased 5 extra cups each day

So....on those days, our daily expense would be $4 less because we would not have to buy any cups on these "5th days".....and this would happen 5 times in 28 days...this would reduce our monthly expenses by 5 * $4   = 20

So....assuming that we charged the same for every cup of coffee.....the totlal profit for 28 days  would be:

23 days * [ 51 - 31] + 5 days * [51 - 27 ] =  $580

There are infinite possibilities.....here's one

A = 2

B = -2

C = 6

So,  A + B + C  = 6

[20000 + 1000(2) + 300 + 10(-2) + 6]  = 22286

And 22286/ 22 =  1013

Well......if you have $50 to spend.. and you want to earn a $20 profit per day....then.........

Revenue - Expenses  = Profit   ......so.....

Revenue - $31   = $20      add 31 to both sides

Revenue  = $51

So....you need to earn $51......so each cup of coffee will be sold for :

51/20  = $2.55 per cup

3308.33

ok then. Let's say I start with $50 instead of $30. What will happen?

Properties:

A number divisible by 2 is even.

A number divisible by 3 has the sum of digits that can be divided exactly by 3.

A number divisible by 5 ends with either 5 or 0.

A number divisible by 6 is divisible by 2 and 3.

A number divisible by 9 has the sum of digits that can be divided exactly by 9.

A number divisible by 10 ends with 0.

It is NOT divisible by 2,3,5,6,9,10.

Well....let's see.....for just one day for 20 people, you need:

20 cups, 40 cups of milk, 40 tsp coffee and 60 tsp sugar

The cups will cost $4......however....you'll be better off by buying the 25 cups for $4 [you get 5 more cups instead of buying two of the 10 cup packages at $2 each]

The 40 cups of milk will cost $2

The 40 tsp of coffee  will cost $14

And the 60 tsp of sugar will cost $3 for 10 tsp  + $8 for 50 tsp   = $11

So your minimum expenses for one day   =  $[ 4  + 2 + 14 + 11]   =  $31

Since you only have $30 to spend.....you'll either have to plan on serving fewer customers or you'll have to cut down on the sugar, milk and/or coffee  in each cup.....!!!!