# Homework Help: Next number in a sequence (definite integral problem)

1. Mar 23, 2013

### 5ymmetrica1

(definite integral problem) write as an expression

1. The problem statement, all variables and given/known data

$\int$0k 3x2
suggest an expression for definite integrals from x=0 to x=k

2. Relevant equations
see below

3. The attempt at a solution

when k= 0, $\int$0k 3x2 = 0
when k= 1, $\int$0k 3x2 = 1
when k= 2, $\int$0k 3x2 = 8
when k= 3, $\int$0k 3x2 = 27
when k= 4, $\int$0k 3x2 = 64
when k= 5, $\int$0k 3x2 = 125
when k= 6, $\int$0k 3x2 = 216

but now the question is asking me to find an expression for when k = k in $\int$0k 3x2

Last edited: Mar 23, 2013
2. Mar 23, 2013

### Curious3141

I'm assuming you haven't actually learned how to do integration yet?

In that case, does a pattern jump out at you in these numbers: 0, 1, 8, 27, 64, ...?

Hint: think of a BOX.

3. Mar 23, 2013

### 5ymmetrica1

no Im actually mid-way through my course on integration, and have been going fine until this problem.

Ive tried a bunch of patterns
such as...
0= 0x0, 1=1x1, 8=2x4, 27=3x9, 64=4x16, 125=5x25, 216= 6x36

but all I can conclude from this is that the number is the sum of its number in the sequence and the square of that number
(i.e 27 is the 3rd number in the sequence and 32 is 9, therefore 27 = 3x9)

but I'm not sure how to write this as an expression for the upper limit 'k'

can I say $\int$0k 3x2 dx = k(k2)

Last edited: Mar 23, 2013
4. Mar 23, 2013

### SithsNGiggles

What's the product of a number and its square equivalent to? Like Curious said, "think of a BOX."

5. Mar 23, 2013

### 5ymmetrica1

Im not sure what you mean by that.
If its a box, then 3x9 is the dimensions of the box and 27 is the area.

But how does this help me write an expression for
$\int$0k 3x2 dx

6. Mar 23, 2013

### Curious3141

A more explicit hint, then:

x*x = x SQUARED (x2)

x*x*x = ??? (x?)

7. Mar 24, 2013

### 5ymmetrica1

so you mean $\int$0k 3x2 dx = k3

8. Mar 24, 2013

### 5ymmetrica1

rather than starting another post I also have two more sets of these problems which I have been struggling with finding the pattern.

for 3x2

when k= -1, ∫ -1k 3x2 = 0
when k= 0, ∫ -1k 3x2 = 1
when k= 1, ∫ -1k 3x2 = 2
when k= 2, ∫ -1k 3x2 = 9
when k= 3, ∫ -1k 3x2 = 28
when k= 4, ∫ -1k 3x2 = 65
when k= 5, ∫ -1k 3x2 = 126
when k= k, ∫ -1k 3x2 = ??

so the sequence is [0,1,2,9,28,65,126....]
I believe the answer for this is (k)3 +1

and also for

when k= -2, ∫ -2k 3x2 = 0
when k= -1, ∫ -2k 3x2 = 7
when k= 0, ∫ -2k 3x2 = 8
when k= 1, ∫ -2k 3x2 = 9
when k= 2, ∫ -2k 3x2 = 16
when k= 3, ∫ -2k 3x2 = 35
when k= 4, ∫ -2k 3x2 = 72
when k= k, ∫ -2k 3x2 = ??

so the sequence is [0,7,8,9,16,35,72....]

I need to find the expression for 'k' in each of these but I can't find any pattern in the numbers when applying the same technique as I did in the first part of my question.

Last edited: Mar 24, 2013
9. Mar 24, 2013

### Curious3141

Right.

10. Mar 24, 2013

### Curious3141

Right also.

Another way to write it is $k^3 + 1^3$, correct? The cube of 1 is of course also 1. I'm writing it this way to give you a big hint for the next part.

Where did that "65" (in bold) come from? It shouldn't be there, should it?

From the way I wrote the last expression, can you now figure this one out? It's all got to do with cubes, and it's a simple pattern.

11. Mar 24, 2013

### 5ymmetrica1

oh woops the 65 is from copying the first section so disregard that.

thanks for the help curious, I'm getting the result for the last one of k3 + 23 does that look right to you?

12. Mar 24, 2013

### Curious3141

Yes, correct, altho' you can of course write it more simply as $k^3 + 8$.

13. Mar 25, 2013

### 5ymmetrica1

thanks again curious!

I just have one last question.
Im asked to consider the function y=4x3+3x2 and use my findings from the first part of this post to suggest a function that will generate the definite integral for y=4x3+3x2, from x=a to x=k

then Im asked to write an integral expression for the same function, which I can do
$\int$ak4x3+3x2 = $\int$ak4x3 + $\int$ak3x2

but I'm not sure about the part asking me to use the information Ive found to create a function that will generate the definite integral. Could you possibly give me any clues into how to go about this question?

14. Mar 25, 2013

### Curious3141

OK, you know how to split the integral up.

Let's tackle the second part because the integrand (the expression you're integrating) is the same as what you worked on before. Only the lower bound is different.

When your lower bound was -1, the expression was $k^3 + 1^3$, also expressible as $k^3 - (-1)^3$.

When your lower bound was -2, the expression was $k^3 + 2^3$, also expressible as $k^3 - (-2)^3$.

Can you suggest what the expression might be for a lower bound of a?

For the first integral, you know the form has to be similar to this. But note that the integrand has a one higher exponent (power) for x now. So maybe it follows a similar pattern, just with one power higher? Try checking.

Just curious (that's my name, after all ) about this course you're taking - are they doing this prior to teaching you how to do the actual integration? Because it seems a very weird way of introducing the topic, in my opinion. Do you even have a conceptual understanding of what it means to take a definite integral? An indefinite one? Have you covered differentiation, even?

Anyway, notice that the integrands are always of the form $nx^{n-1}$, which always has the integral $x^n$. Actually, it's $x^n + c$, where c is an arbitrary constant, but this cancels out when you're doing definite integrals. Hopefully, you'll cover all this rigorously soon, because I can't, for the life of me, see why they're teaching your course this way.

15. Mar 25, 2013

### 5ymmetrica1

so would the expression be k3 - a3

and for 4x3

I'm getting these results

$\int$0k4x3dx = k4

$\int$-1k4x3dx = k4 - 1

$\int$-2k4x3dx = k4 - 24

so if my first expression is right would this be k4 - a4 for this one?

and to answer your question about integration, so far Ive been taught about upper and lower sums, definite integrals and yes we've done differentiation, such as the product and chain rule, implicit differention and finding differentials from the first principles. But I agree it is a strange order to teach calculus in.

Last edited: Mar 25, 2013
16. Mar 25, 2013

### Curious3141

No! Careful with the sign. Think about it, when a = -1, does your expression give you the expected result?

This is correct, so your wrong sign in the expression above may have just been a typo. On the other hand, you may be missing the complete picture here. Remember (for example), $k^4 - 2^4$ is equivalent to $k^4 - (-2)^4$. You should be able to see how this ties in to the earlier expression.

A bit of background may help: with definite integration, the last step always involves a subtraction. Evaluate the final expression at the upper bound and the lower bound. Then take the former minus the latter. It's always subtraction, not addition, so this should give you a big clue.

Well, I guess your instructors know best what they're planning.

17. Mar 25, 2013

### 5ymmetrica1

yes it was a typo, thanks again for your excellent help curious!

18. Mar 25, 2013

### Curious3141

19. Mar 25, 2013

### 5ymmetrica1

have gone through your edit and tried to implement it into my expression.
can I say that $\int$ak 4x3+3x2dx = (k4 - a4) - (k3 - a3)

or am I still misunderstanding something?

Last edited: Mar 25, 2013
20. Mar 25, 2013

### Curious3141

Each of the "split" integrals is correct, i.e. $k^4 - a^4$ and $k^3 - a^3$.

My question is: why are you subtracting one of those from the other, when you should be adding?

What I said (subtract lower bound value from upper bound value) applies to a single integral. You'd correctly split the integrals into two, so you should just work out each individual integral (which you did correctly), then add the two expressions together.

Do you get it now?

21. Mar 25, 2013

### 5ymmetrica1

Yes I understand now, I thought that was the case but I was confused by the subtraction statement in your previous post, but it was an error on my part and not in your explanation.

thanks again for all the help curious.

22. Mar 25, 2013

### Curious3141

You're welcome.