Area under the curve need help

1. Jun 15, 2013

tg22542

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

1. The problem statement, all variables and given/known data

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

2. Relevant equations

-

3. 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!

2. Jun 15, 2013

Staff: Mentor

Why do you multiply with deltax twice?

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

3. Jun 15, 2013

tg22542

Is my deltax value here right?

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

4. Jun 15, 2013

tg22542

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

5. Jun 15, 2013

Staff: Mentor

That looks good.

It is a very rough approximation.

6. Jun 15, 2013

tg22542

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

7. Jun 15, 2013

tg22542

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

8. Jun 15, 2013

LCKurtz

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.

Last edited: Jun 15, 2013
9. Jun 15, 2013

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.

10. Jun 15, 2013

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.

11. Jun 15, 2013

Staff: Mentor

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

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

12. Jun 15, 2013

LCKurtz

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.

13. Jun 15, 2013

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

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

Last edited: Jun 15, 2013
14. Jun 15, 2013

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

15. Jun 15, 2013

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.

Last edited: Jun 15, 2013
16. Jun 15, 2013

lurflurf

use

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

17. Jun 16, 2013

Ray Vickson

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.

18. Jun 16, 2013

Ray Vickson

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.

19. Jun 16, 2013

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.

20. Jun 16, 2013

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.