Calculating Area Under the Curve: Precal Exam Prep Tips

[tg22542]

Studying for precal exam.. for some reason I can't figure this out no matter how I try it

Homework Statement

Find the area under the curve for y=x^2 going from x=0 to x=2

Homework Equations

-

The Attempt at a Solution

I first graphed the equation and sketched it on paper. I decided I was going to make each rectangle .5 wide, making 4 rectangles. So deltax= 2-0/4=1/2. Next I plugged all my values into my equation getting:

deltax[(0.25*0.5)+(1*0.5)+(1.5*0.5)]=0.6875 ?
I know this isn't right.. can someone tell me where I'm going wrong.
Thanks!

 

[mfb]

Why do you multiply with deltax twice?

Unrelated to that:
Using the left edge for the rectangles gives a significant error here.

 

[tg22542]

Is my deltax value here right?

I don't know where I'm going wrong and how to fix it

 

[tg22542]

So would my answer just be .5(.25+1+2.25)= 7/4?

 

[mfb]

That looks good.

It is a very rough approximation.

 

[tg22542]

What about y=5x?

I split it into 4 parts, my deltax=1-0/4

1/4(5/4+5/2+15/4) = 15/8

Apparently the answer is 5/2. I have no idea what I'm doing wrong

 

[tg22542]

And apparently the answer for y=x^2 is 8/3 ? ugh

 

[LCKurtz]

And apparently the answer for y=x^2 is 8/3 ? ugh

What were you asked to do in this question? Calculate an approximation using a few rectangles or figure out the exact area? That answer suggests exact area.

 

[tg22542]

That question was kind of confusing, here is another one that is not working for me. The question states exactly:

Find the area under the curve y=5x between x=0 and x=1.

 

[LCKurtz]

Have you learned how to calculate areas with antiderivatives? If so, just do the integrals. If not, please give a word for word statement of the problem.

 

[mfb]

What about y=5x?

I split it into 4 parts, my deltax=1-0/4

1/4(5/4+5/2+15/4) = 15/8

Apparently the answer is 5/2. I have no idea what I'm doing wrong

Can you explain where those numbers come from? There are at least two errors in that calculation.

Have you learned how to calculate areas with antiderivatives?

Based on the first post, I assumed that you (tg22542) had not. Otherwise, you should use those to get the exact area.

 

[LCKurtz]

Have you learned how to calculate areas with antiderivatives? If so, just do the integrals. If not, please give a word for word statement of the problem.

Based on the first post, I assumed that you (tg22542) had not. Otherwise, you should use those to get the exact area.

It's hard to tell from the posts whether he is supposed to calculate the area the hard way by taking the limit of approximating sums or use antiderivatives, since he has done neither.

 

[lurflurf]

The area under a polynomial can be found by taking the limit of the sum if you have not yet reached the fundamental theorem. For example the area under x^3 from x=2 to x=7 can be found by

$$\lim_{n \rightarrow \infty} \sum_{k=1}^n (2+5k/n)^3(5/n)=2385/4$$

the (possibly signed) area under f(x) from x=a to x=b can be found by

$$\lim_{n \rightarrow \infty} \sum_{k=1}^n \mathop{f}(a+(b-a)k/n)((b-a)/n)$$

 

[tg22542]

So for my y=x^2, from x=0 to x=2:

(0+2k/n)^2*(2k/n) =

im actually not sure on how you multiplied that out to be 2385/4

 

[lurflurf]

^Sorry about my typo above that should be

$$Area=\lim_{n \rightarrow \infty} \sum_{k=1}^n (0+2k/n)^2(2/n) \\
=\lim_{n \rightarrow \infty} \sum_{k=1}^n 8k^3/n^3=\lim_{n \rightarrow \infty} (8/3)(1+1/n)(1+1/(2n)) \\
=\frac{8}{3}
$$
I have used
$$\sum_{k=1}^n k^2= \frac{1}{6}n(n+1)(2n+1)$$
Which you may remember, derive, or look up.

Try to do the easier one y=5x
If you ever have one like a x^2+b x+c remember it if linear ie
Area(a x^2+b x+c)=a Area(x^2)+B Area(x)+C Area(1)

Higher powers are a bit tedious if you work them all the way out, but remember that only the constant term matters.

 

[lurflurf]

im actually not sure on how you multiplied that out to be 2385/4

use

$$\sum_{k=1}^n k^3=\left( \sum_{k=1}^n k \right) ^2=\left( \frac{1}{2}n(n+1) \right) ^2$$

 

[Ray Vickson]

Studying for precal exam.. for some reason I can't figure this out no matter how I try it

Homework Statement

Find the area under the curve for y=x^2 going from x=0 to x=2

Homework Equations

-

The Attempt at a Solution

I first graphed the equation and sketched it on paper. I decided I was going to make each rectangle .5 wide, making 4 rectangles. So deltax= 2-0/4=1/2. Next I plugged all my values into my equation getting:

deltax[(0.25*0.5)+(1*0.5)+(1.5*0.5)]=0.6875 ?
I know this isn't right.. can someone tell me where I'm going wrong.
Thanks!

Do you understand that you are computing approximations to the area?

Dividing the interval into 4 parts and summing the areas of 4 rectangles will give you some type of approximation to the area, but not the actual, true value of the area. You can get more accuracy by splitting the interval into 1000 parts and summing the areas of 1000 rectangles, but it will still not give you the true value. Even better, split the interval into 1 million parts and sum the areas of 1 million rectangles; that will get you much closer to the value of area, but still not quite the exact value.

The true value of area is obtained as a limit---roughly speaking, you sum infinitely many infinitely small rectangles.

 

[Ray Vickson]

Studying for precal exam.. for some reason I can't figure this out no matter how I try it

Homework Statement

Find the area under the curve for y=x^2 going from x=0 to x=2

Homework Equations

-

The Attempt at a Solution

I first graphed the equation and sketched it on paper. I decided I was going to make each rectangle .5 wide, making 4 rectangles. So deltax= 2-0/4=1/2. Next I plugged all my values into my equation getting:

deltax[(0.25*0.5)+(1*0.5)+(1.5*0.5)]=0.6875 ?
I know this isn't right.. can someone tell me where I'm going wrong.
Thanks!

Do you understand that you are computing approximations to the area?

Dividing the interval into 4 parts and summing the areas of 4 rectangles will give you some type of approximation to the area, but not the actual, true value of the area. You can get more accuracy by splitting the interval into 1000 parts and summing the areas of 1000 rectangles, but it will still not give you the true value. Even better, split the interval into 1 million parts and sum the areas of 1 million rectangles; that will get you much closer to the value of area, but still not quite the exact value.

The true value of area is obtained as a limit---roughly speaking, you sum infinitely many infinitely small rectangles.

 

[verty]

I think it would be best not to worry about the $y = x^2$ example. Finding the area under $y = 5x$ is much easier if you look at the graph. Riemann sums are best left for when you learn calculus.

If your precalculus book talks about them, it is probably right at the end because it isn't really PRE-calculus.

 

[tg22542]

Yes I know that it's not precal, I've been told that by a few others too hahah.. We learned it like 3 weeks in.. My exam is Wednesday and this is the only real thing that I'm stressing about. I need to figure out how to do it.

 

[NascentOxygen]

here is another one that is not working for me. The question states exactly:

Find the area under the curve y=5x between x=0 and x=1.

Have you plotted the graph for this equation? If not, why not?

 

[tg22542]

I have, I just want to know how the 5 effects my area

 

[tg22542]

I don't know what else to say/ do haha

 

[NascentOxygen]

You are simply finding the area of a right-angled triangle here.