# Approximating Area By Three Inscribed Rectangles

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• January 17th 2011, 08:13 PM
PPSSAGW
Approximating Area By Three Inscribed Rectangles
I'm working on a tricky problem. Hopefully I got it right.

If $\int_0^6 (x^2-2x+2) dx$is approximated by three inscribed rectangles of equal width on the x-axis. Determine the approximation of this integral.

After I graphed it, I found that the minimum is actually at 1. So the height of the first and second rectangles is 2. Therefore, their total area: $2*2+2*2=8$.

The third rectangle's height: $4^2-2*4+2=10. 10*2=20$.

Total area: $20+8=28$.

Did I do it correctly? Hopefully I got this one right. Thanks everyone.
• January 17th 2011, 08:14 PM
dwsmith
Quote:

Originally Posted by PPSSAGW
I'm working on a tricky problem. Hopefully I got it right.

If $\int_0^6 (x^2-2x+x) dx$is approximated by three inscribed rectangles of equal width on the x-axis. Determine the approximation of this integral.

After I graphed it, I found that the minimum is actually at 1. So the height of the first and second rectangles are 1. Therefore, their total area: $2*2+2*2=8$.

The third rectangle's height: $4^2-2*4+2=10. 10*2=20$.

Total area: $20+8=28$.

Did I do it correctly? Hopefully I got this one right. Thanks everyone.

I think you should use a Riemann Sum.
• January 17th 2011, 08:15 PM
PPSSAGW
Quote:

Originally Posted by dwsmith
I think you should use a Riemann Sum.

This problem says inscribed rectangles, so I had to use rectangles that are inscribed.
• January 17th 2011, 08:16 PM
dwsmith
Quote:

Originally Posted by PPSSAGW
This problem says inscribed rectangles, so I had to use rectangles that are inscribed.

Is there a picture you can upload?
• January 17th 2011, 08:18 PM
PPSSAGW
Quote:

Originally Posted by dwsmith
Is there a picture you can upload?

http://www.wolframalpha.com/input/?i=plot+x^2-2x%2B2

This is the graph.
• January 17th 2011, 09:03 PM
Prove It
Are you sure the function isn't supposed to be $\displaystyle x^2 - 2x + 1$, not $\displaystyle x^2 - 2x + x$?
• January 17th 2011, 10:18 PM
CaptainBlack
Quote:

Originally Posted by PPSSAGW
I'm working on a tricky problem. Hopefully I got it right.

If $\int_0^6 (x^2-2x+x) dx$is approximated by three inscribed rectangles of equal width on the x-axis. Determine the approximation of this integral.

After I graphed it, I found that the minimum is actually at 1. So the height of the first and second rectangles is 2. Therefore, their total area: $2*2+2*2=8$.

The third rectangle's height: $4^2-2*4+2=10. 10*2=20$.

Total area: $20+8=28$.

Did I do it correctly? Hopefully I got this one right. Thanks everyone.

Lets assume you mean:

$\displaystyle \int_{x=0}^6 x^2-2x+1 \ dx$

The integrand is a parabola opening upwards. It has a minimum of $0$ at $x=1$.

The three rectangle have equal widths so they extend from $x=0$ to $2$, $x=2$ to $4$ and $x=4$ to $6$.

So the first inscribed rectangle has height $0$, the second has height $f(2)$ and the third height $f(4)$ , where $f(x)=x^2-2x+1$.

This is much easier to see if you draw a diagram.

CB
• January 18th 2011, 02:11 PM
PPSSAGW
Quote:

Originally Posted by CaptainBlack
Lets assume you mean:

$\displaystyle \int_{x=0}^6 x^2-2x+1 \ dx$

The integrand is a parabola opening upwards. It has a minimum of $0$ at $x=1$.

The three rectangle have equal widths so they extend from $x=0$ to $2$, $x=2$ to $4$ and $x=4$ to $6$.

So the first inscribed rectangle has height $0$, the second has height $f(2)$ and the third height $f(4)$ , where $f(x)=x^2-2x+1$.

This is much easier to see if you draw a diagram.

CB

I really meant $x^2-2x+x$. Can you please check my answer again? Thanks a lot.
• January 18th 2011, 02:17 PM
CaptainBlack
Quote:

Originally Posted by PPSSAGW
I really meant $x^2-2x+x$. Can you please check my answer again? Thanks a lot.

Why is it not simplified?

If that is what you meant, then sketch the graph of the function and explain what you think "an inscribed rectangle to the curve with a side on the x-axis from 0 to 2" might mean

CB
• January 18th 2011, 05:27 PM
SammyS
Quote:

Originally Posted by PPSSAGW

That's the graph:

$y = x^2-2x+2$
• January 18th 2011, 07:16 PM
PPSSAGW
I'm very sorry. I meant $x^2-2x+2$

I apologize for all the inconvenience that I have created.
• January 18th 2011, 08:35 PM
SammyS
Quote:

Originally Posted by PPSSAGW
I'm working on a tricky problem. Hopefully I got it right.

If $\int_0^6 (x^2-2x+2) dx$is approximated by three inscribed rectangles of equal width on the x-axis. Determine the approximation of this integral.

After I graphed it, I found that the minimum is actually at 1. So the height of the first and second rectangles is 2. Therefore, their total area: $2*2+2*2=8$.

The third rectangle's height: $4^2-2*4+2=10.\quad 10*2=20$. (spacing added by SammyS)

Total area: $20+8=28$.

Did I do it correctly? Hopefully I got this one right. Thanks everyone.

Hi PPSSAGW.

I'm not sure what you mean by "inscribed".

What you have found is the Left Riemann Sum, with n=3.

Yes, you have done that correctly.
• January 19th 2011, 12:06 AM
CaptainBlack
Quote:

Originally Posted by PPSSAGW
I'm very sorry. I meant $x^2-2x+2$

I apologize for all the inconvenience that I have created.

If you had made any effort to make sure you had the question down correctly and some attempt to understand what had been said in earlier posts in this thread you would not need any further help!

See attachment.