Another area under curve problem

[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?

 

[SammyS]

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?

 

[lionely]

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 =/

 

[SammyS]

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?

 

[lionely]

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

 

[SammyS]

-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.

 

[lionely]

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?

 

[SteamKing]

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

 

[HallsofIvy]

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

 

[lionely]

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.

 

[Karnage1993]

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.

 

[iRaid]

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)

 

[SammyS]

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 $\displaystyle \ \ \int_{0}^{3} x^2(3-x)\,dx\ .$

 

[lionely]

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.

 

[Karnage1993]

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!

 

[SammyS]

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) . )​

 

[lionely]

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) . )​

Is the answer 23.3?

 

[SammyS]

Is the answer 23.3?

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

 

[lionely]

For the one you posted?

 

[lionely]

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

 

[lionely]

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

 

[SammyS]

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.

 

[SammyS]

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.