integration and negative area

Since the area is under the curve, the area bounded under the x axis and the curve from -1 to +1 is a negative area, +4/3.

I mean -4/3

"If i reverse the sign of teh answer because they are under the x axis i can add them and get 4/3"
not your problem.
Your problem is in the last part of your correct answer.
-(-1) is +1

If the integration is done correctly all the - signs come out correctly and automatically without the need to say anything like the curve is under the x axis.
You are to be commended for graphing the problem, thus giving you a means to check for mistakes.

Hi

I see my error. It is
{1/3(1)^3 - (1)} - {1/3(-1)^3 - (-1)} = -2/3 - (2/3) = -4/3

I believed that because this is under the x axis it is negative area so you reverse the sign. But aburr you said: "If the integration is done correctly all the - signs come out correctly and automatically without the need to say anything like the curve is under the x axis." How should i have done this question to get an answer of 4/3 and not an answer of -4/3 and then reverse the sign?

You now have corrected the arithmetic error, and as a result you got a negative answer for the definite integral; you got the correct answer without forcibly changing the sign. I don't see a reason to try to get an answer of 4/3 and then somehow reason it out to change the sign.

But, if you are asking this just to become more familiar with calculus, and want a related problem, then here it is:

Let y' = -y == 1 - x^2
Now you have a parabola that goes to -infinity instead of +infinity; and now the area from -1 to +1 is positive. But this is a different function. So it is not clear why you want to ask the question differently.

Sorry i dont understand what you mean. Lets assume i never made the arithmetic error in the first place for now and imagine i got an answer of -4/3

My book says that  if you calculate the area under the x axis the value you get will be negative (which my answer was) and so you reverse the sign as area can never be negative. That makes sense to me.

However aburr says you don't need to look if your area is under the x axis and your signs will take care of themselves. But that's not what my book says. And i don't think its what you said either in 26469172/3. Unless you are both saying area can be negative?

Perhaps there is more context in your book so that "so you reverse the sign as area can never be negative" makes some sense. Some physical problems only make sense if the outcome is positive.

Some examples come to mind. If you have a gym court with a parabola on it and line (the x-axis) passing through it, and you have to paint the area bounded by the line and the parabola; and you arbitrarily chose the line to be the x-axis, and the y-axis polarity chosen such that you have y = x^2 - 1.

Of course you cannot purchase -4/3 m^2 worth of paint. So, you realize from the drawing that you have to change your sign from negative to positive and purchase a positive amount of paint.

But, re: positive and negative area. nothing wrong with that concept in calculus. For example consider the curve y = sin(x), where x is in degrees, and the range of x is [0..360]. From [0,180], the area of the curve is positive; from [180, 360], the area is negative. So, from [0..360], the area is 0.

Now if you had to purchase paint to handle the two intervals, you could recognize that the first half area magnitude (i.e., absolute value of area) is the same as in the 2nd half. So, you would integrate from  [0,180] to get your positive area, and multiply by 2 to get the total area to be painted in.

> Unless you are both saying area can be negative?

In http:#26469173 I corrected my mistake in http:#26469170. So, http:#26469170 should have read
     "the curve from -1 to +1 is a negative area, -4/3."

"so you reverse the sign as area can never be negative"   That is not correct
As the book said the area under the curve is negative. In integration the area under the axis is always negative. If you want to define it otherwise, go ahead. You will confuse students, require all sorts of arbitrary rules, and make a general mess, as the many comments to your additional questions show.

>>"so you reverse the sign as area can never be negative"

Yes, we've had this problem before, in integrating between a and b and between b and a. Whatever the curve and where it is, the integration finally results in the summation of very small areas whose value is the product of two infinitesinmal values, which are signed!

> My book says that  if you calculate the area under the x axis the value you get will be negative (which my answer was) and so you reverse the sign as area can never be negative.

If your book says this, then either you are leaving out some context, or the book is wrong. So, if there is anything in the book's question that could shed light on this, then write out the entire question. Maybe that will help explain the apparent contradiction between what we and the book are saying.

Hi
I'm don't think I'm leaving out any context. This is the first chapter in a general maths book on integration. I don't know if its right or wrong as I've never done integration before. I'll write out what it says...

The Meaning of A negative Result

Consider the area bounded by the curve y = 4x^3, the x axis and the lines
a) x = 1 and x = 2
b) x = -2 and x = -1
 (there is a diagram showing these bits shaded in)
The curve is symmetrical about the origin so the 2 shaded areas are equal

Consider a)
(book then integrates it) answer = 15
Consider b)
(book then integrates it)  answer = -15

The integral has a negative value because from -2 to -1 the value of y which gives the length of the strip is negative. Area cannot be negative. The minus sign simply means the area is below the x aixis. The actual area is 15 square units

After this the book gives a set of questions, one of which is the question i posted.
=>find the area below the x-axis and above the curve y = x^2-1

The book didn't actually say to reverse the sign but it took an answer of -15 and made it 15 so I must have made that interpretation myself.