Area Under Curve of y=5x^2+3 from 0 to 2

[tg22542]

Homework Statement

Find the area under the curve of the function y=5x^2+3 from x=0 to x=2

Homework Equations

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The Attempt at a Solution

First I sketched it out on paper, finding that at x=2,y=28. I created a rectangle with l*w of 6 (3*2), would the remaining area not included in the rectangle be 23/3 ? I guess this because it asks from 0 to 2 and the value of 0 is x=3 and value of 2 is x=28.

So my final answer would be 6+23/3 ?

 

[CompuChip]

Why 23/3? The area of a triangle with vertices (0, 3), (2, 3) and (2, 28) would be 1/2 * (28 - 3) * 2 = 25, which is different from 23/3 = 7 + 2/3.

To be honest, I'm a bit lost at the moment as to how you would solve this question without resorting to integration...

 

[tg22542]

Welcome to my precal teacher's logic, she never taught us integration.. So I'm pretty lost with this

 

[CAF123]

Welcome to my precal teacher's logic, she never taught us integration.. So I'm pretty lost with this

Are you being taught approximation methods at the moment? You'll get onto integration later.

 

[tg22542]

Honestly I don't even know what she's doing. Our exam is tomorrow I'm just reviewing now and it didn't make sense how she got her answer of 6+8/3.

She literally gives us constants for values above rectangles such as x^2 being 8/3, then she gets us to create a rectangle from the x to x values given, and it becomes simply l*w then you add the constant.

 

[CompuChip]

Ah, so you have been given that the area below x² between x = 0 and 2 is 8/3?

In that case: look how the graph for the given formula 5x² + 3 can be made from the standard graph of x².
There are two steps, can you describe them geometrically (so in terms of scaling and shifting)?

 

[tg22542]

As in a horizontal translation of 3 units upwards? And the hs of 1/5. But the 5x wouldn't effect the area in this particular case would it?

 

[CompuChip]

Yep, except upwards is called vertical. And note that you have to do them in the correct order: first a horizontal scaling, then the translation; otherwise you would get 5(x + 3) = 5x + 15, instead of 5x + 3.

So the translation is what gives you the rectangle of 2 by 3 units which you had already found. Now let's look at the scaling, which actually does affect the area. Try looking at a straight line y = x between x = 0 and 2. What happens if I apply the scaling there (i.e. switch to y = 5x)?
You could consider such a straight line as a (crude) approximation to the actual curve, which should make it more intuitive to you that it doesn't only hold for straight lines.

 

[tg22542]

your slope will be changed to 1/5 once you apply the 5x

 

[CompuChip]

Yep, but I was not so much interested in the slope, but more in the area when I do that rescaling.
How does the area change if you scale y = x by 1/5?

 

[tg22542]

It would become 5 times greater?

 

[CompuChip]

Exactly.
It's basically a triangle with the same base but 5 times the height.

Now, as I hinted, the same happens to any function. So if you know the area under x² from x = 0 to 2, you can also find the area of 5x².

 

[tg22542]

But I don't know how to find the area of x^2.. :( haha

 

[CAF123]

But I don't know how to find the area of x^2.. :( haha

I think the method provided by CompuChip was suggested on the basis of you already knowing what the area under y=x² from 0 to 2 was. (i.e your teacher had provided that for you). Is this the case?

 

[CompuChip]

She literally gives us constants for values above rectangles such as x^2 being 8/3

At least it is what you said :)

 

[tg22542]

Oh okay I see, but how can I actually figure out that it is 8/3 myself?

 

[CAF123]

Oh okay I see, but how can I actually figure out that it is 8/3 myself?

You would have to compute the integral $$\int_0^2 x^2 dx,$$ but I think you said you have not done this. If you have done approximation methods (I.e something like approximating the area as the sum of finitely many rectangles) then you can certainly obtain an approximation to that 8/3.

 

[tg22542]

Okay, so she gave me the constant as the area under the curve from x=0 to x=2 to be 8/3, this is for y=x^2. So If I had y=5x^2+3 .. Does that mean my area is 5*(8/3) ? Seem right? I know I've been taking long intervals with replying but I have been studying very hard.. so if anyone could answer that would be awesome. And thank you so much for hanging along with me you're really helping a lot.

 

[Mark44]

Okay, so she gave me the constant as the area under the curve from x=0 to x=2 to be 8/3, this is for y=x^2. So If I had y=5x^2+3 .. Does that mean my area is 5*(8/3) ?

No, because you're not taking into account the "+3" part. If y = 5x², then the area under this curve, from x = 0 to x = 2, is 5 * 8/3.

Adding 3 has the effect of shifting the graph of y = 5x² up by 3 units, so what will that do to the area beneath y = 5x² + 3?

Seem right? I know I've been taking long intervals with replying but I have been studying very hard.. so if anyone could answer that would be awesome. And thank you so much for hanging along with me you're really helping a lot.

 

[tg22542]

Well I already solved for the area from x=0 to x=2 by creating a rectangle with a width of 2 and height/vertical length of 3 (due to my shift). So l*w = 6. But I still have my area above this undetermined.. So I know that x^2 has an remainder area above the rectangle of 8/3 as my teacher gave us.. but I don't get how the 5x^2 will effect this

 

[Mark44]

I thought you had this part figured out from what you wrote (5 * 8/3). Each point on the graph of y = 5x² is stretched away from the x-axis by a factor of 5, as compared to the graph of y = x². IOW, each point on the graph of y = 5x² is 5 times as far away from the x-axis as the corresponding point on the graph of y = x².

Now, when you add 3, that will increase the area under the curve by an additional 3*2 = 6 sq. units.

Clear?

 

[tg22542]

So my final area is 30? Or (8/3)*5 + 6

 

[NascentOxygen]

(8/3)*5 + 6

You have made a careless transcription blunder in copying this from your earlier post.

Fix that, and you are right.

 

[tg22542]

Man I seriously don't know.. Throw me a bone here my exam is in 7 hours ;)

 

[CAF123]

so my final area is 30? Or (8/3)*5 + 6

30 ≠ (8/3)*5 + 6

 

[CompuChip]

Maybe you need to take a 5 min break and come back to this.
It seems you have all the pieces but somehow fail to put them together.

Well I already solved for the area from x=0 to x=2 by creating a rectangle with a width of 2 and height/vertical length of 3 (due to my shift). So l*w = 6.

Correct.

But I still have my area above this undetermined..

That area is just the area under the graph of y = 5x² before you shifted it up.

So I know that x^2 has an remainder area above the rectangle of 8/3 as my teacher gave us.. but I don't get how the 5x^2 will effect this

You need to do this the other way around, as I mentioned earlier: first you scale x² to 5x², then you shift up by 3; otherwise you will not get (5x²) + 3 but 5(x² + 3).

For the area, this means that first you need to calculate the area of 5x² from the given value of 8/3. Then you shift that up by 3 units, which will give the additional area of 6 that you have calculated.

So my final area is 30? Or (8/3)*5 + 6

As CAF123 remarked, these are not the same. So with the above hints, which of the two (if any) is it? :)