# Another area under curve problem

• lionely
In summary: I could see how the answer given by SammyS could be correct. However, as it stands, the problem does not match the answer key, and so I disagree with SammyS.
lionely

## Homework Statement

Sketch roughly the curve y = x^2(3-x) between x=-1 and x=4. Calculate the area bounded by the curve and the x-axis

. The attempt at a solution

I tried to find the area from x=-1 to x=4 I got 1 1/4
answer in the back of my textbook is 6 3/4

When i find the area from x= 0 to x= 3 I get 6 3/4.

What should I do?

lionely said:

## Homework Statement

Sketch roughly the curve y = x^2(3-x) between x=-1 and x=4. Calculate the area bounded by the curve and the x-axis

. The attempt at a solution

I tried to find the area from x=-1 to x=4 I got 1 1/4
answer in the back of my textbook is 6 3/4

When i find the area from x= 0 to x= 3 I get 6 3/4.

What should I do?
So, what exactly was it that you integrated?

I integrated x^2(3-x) to 3x^2-(x^4/4) for x= -1 to x=4 and I got 1 1/4

then I tried x= 0 to x=3 and i got 6 3/4 the answer in the back of the book but I don't understand why i got it, I got the answer for the wrong reason =/

Is that the exact wording of the problem.

It appears that the intent is for you to find the area of the region enclosed by y = x2(3-x) and the x-axis.

What were the x-intercepts on your graph?

-1 and 3 and yes I typed the exact question word for word1

lionely said:
-1 and 3 and yes I typed the exact question word for word1

Only the region between x=0 and x=3 is enclosed by y = x2(3-x) and the x-axis.

so therefore the answer is 6 3/4

oh wow sorry -1 is not an intercept.. it's 0(twice) and 3
so basically the x=-1 and x=4 are things to trick me?

Check your indefinite integration. It appears the curve is y = x^2*(3-x), which is
y = 3x^2 - x^3 when expanded

The graph is below the x-axis for x from x=3 to x= 4. The area is given by
$$\int_{-1}^3 3x^2- x^3 dx- \int_3^4 3x^2- x^3 dx$$

#### Attachments

• area.jpeg
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Sorry but I'm confused that would make it negative and that would mean the area is under the x-axis. But my book implies it's above.

According to how you put it, there was no mention of the area being above the x-axis. Usually if this is the case, the question would explicitly say to find the region above the x-axis rather than the area bounded by a curve and the x-axis.

Does this help:
$$\int_{-1}^3 3x^{2}-x^{3}dx-\int_3^4 3x^{2}-x^{3}dx$$

In order to get the AREA of the curve, you have to add the 2 parts of it (where it's positive and negative). If you just do the integral from -1 to 5 you end up with 5/4, which isn't the answer you need.

And I don't get the answer your book does, so either you have the problem written wrong or the books answer is wrong (or I'm wrong lol)

Last edited:
Look at the problem statement again.

The only mention of x=-1 and x=4 are in regards to sketching the graph.

The graph of y = x2(3-x) intersects the x-axis at x=0 and x=3.

Therefore, I think you were expected to find the area below y = x2(3-x) and above y = 0, between x=0 and x=3 .

i.e., you were to find $\displaystyle \ \ \int_{0}^{3} x^2(3-x)\,dx\ .$

I agree with SammyS because the other questions I couldn't do, I got the answers when I integrated for the parts of the graph cutting the x.

You should consider learning from another textbook. The area based on the wording should be what HallsofIvey has. The answer that SammyS gave is only involving the 1st quadrant! You can't simply say the area bounded by the curve and the x-axis ignores the 2nd and 4th quadrants because that is very relevant!

Karnage1993 said:
You should consider learning from another textbook. The area based on the wording should be what HallsofIvey has. The answer that SammyS gave is only involving the 1st quadrant! You can't simply say the area bounded by the curve and the x-axis ignores the 2nd and 4th quadrants because that is very relevant!
Of course I disagree with the statement regarding the area. I did initially interpret the question the same way that HallsfoIvy did. The fact that the integral from x=0 to x=3 gave the area matching the answer key made me see if I could interpret the question in a different light.

As for the textbook, it does appear at first glance that the problem could have (and should have) been worded more clearly. However, without examining the textbook in more detail -- particularly looking at the way the material was presented and looking at the examples given -- I can't make a such judgement regarding the textbook -- good or bad.

Had the problem been just a bit different, as in the following, I think there would have been no confusion. Try this out. (I've just shifted everything down by 2 units.)
Roughly sketch the curve y = x^2(3-x)-2 . Be sure the graph extends at least from x=-1 to x=4. Calculate the area bounded by the curve and the line, y=-2 . (It may be helpful to notice that x^2(3-x)-2 = (1-x)(-2-2 x+x^2) . )​

SammyS said:
Of course I disagree with the statement regarding the area. I did initially interpret the question the same way that HallsfoIvy did. The fact that the integral from x=0 to x=3 gave the area matching the answer key made me see if I could interpret the question in a different light.

As for the textbook, it does appear at first glance that the problem could have (and should have) been worded more clearly. However, without examining the textbook in more detail -- particularly looking at the way the material was presented and looking at the examples given -- I can't make a such judgement regarding the textbook -- good or bad.

Had the problem been just a bit different, as in the following, I think there would have been no confusion. Try this out. (I've just shifted everything down by 2 units.)
Roughly sketch the curve y = x^2(3-x)-2 . Be sure the graph extends at least from x=-1 to x=4. Calculate the area bounded by the curve and the line, y=-2 . (It may be helpful to notice that x^2(3-x)-2 = (1-x)(-2-2 x+x^2) . )​

lionely said:
If I didn't mess up the problem, the answer is 6 ¾ .

For the one you posted?

Maybe I messed up the math cause the roots for the one you posted were weird

Lol sorry didn't notice it was like the same question, but I did get 6 3/4 I just found the area from x=0 and x= 3

lionely said:
For the one you posted?

Yes.

As stated, I merely shifted the graph down 2 units then rather than using the x-axis (the line, y = 0) for the part of the boundary, I used
y = -2, which I thought would make the question clearer. Apparently that didn't work for you.

#### Attachments

• Shifted graph.gif
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lionely said:
Maybe I messed up the math cause the roots for the one you posted were weird
Yes, it did make sketching the function a little tougher.

Finding the intercepts of y = x^2(3-x)-2 and y = -2 was similar to finding the x-intercepts for the original problem.

## What is an "area under curve" problem?

An "area under curve" problem is a mathematical concept that involves finding the area between a curve and the x-axis on a graph. This can be done by using integration techniques to calculate the area. It is commonly used in fields such as physics, engineering, and economics.

## Why are "area under curve" problems important?

"Area under curve" problems are important because they allow us to calculate the total amount or quantity of something over a given period of time. This can be useful in predicting future trends, analyzing data, and making informed decisions in various fields of study.

## What are some real-world applications of "area under curve" problems?

Some examples of real-world applications of "area under curve" problems include calculating the total distance traveled by an object over time, finding the total amount of sales over a period of time, and determining the rate of change of a population over time. These are just a few examples, as "area under curve" problems can be applied to many different scenarios.

## What are some common methods for solving "area under curve" problems?

Some common methods for solving "area under curve" problems include using basic integration techniques, such as the trapezoidal rule or Simpson's rule. Other methods include using calculus, numerical methods, or software programs specifically designed for solving these types of problems.

## How can "area under curve" problems be used to make predictions?

"Area under curve" problems can be used to make predictions by analyzing the data and trends shown by the curve. By calculating the total area under the curve, we can determine the total amount or quantity over a given period of time. This information can be used to make predictions about future trends and outcomes.

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