Intro to Integrals

**alleysan** · Jan 7th 2010, 10:16 PM

Hi,
I'm stressing out because I've been on this problem for hours and I'm still getting nowhere.

The question asks:
"""
Let R denote the region that lies below the graph of $f(x)=2+4x^2$ on the interval [-1, 2].

Estimate the area of R using six approximating rectangles of the same width and
(a) left endpoints,
(b) right endpoints
"""

my prof doesn't really have notes on what to do and I'm really confused since it wants an estimate, and not an exact answer...

For (a), since the area of one block is:
$(3/n)(2+4(i(3/n))^2)$
but I'm not even sure if that's right...

Please help me! Student in distress!

**Prove It** · Jan 7th 2010, 10:32 PM

Originally Posted by alleysan

Hi,
I'm stressing out because I've been on this problem for hours and I'm still getting nowhere.

The question asks:
"""
Let R denote the region that lies below the graph of $f(x)=2+4x^2$ on the interval [-1, 2].

Estimate the area of R using six approximating rectangles of the same width and
(a) left endpoints,
(b) right endpoints
"""

my prof doesn't really have notes on what to do and I'm really confused since it wants an estimate, and not an exact answer...

For (a), since the area of one block is:
$(3/n)(2+4(i(3/n))^2)$
but I'm not even sure if that's right...

Please help me! Student in distress!

It helps if you draw the graph and the rectangles, to give you a picture of what is happening...

If the region is $[-1, 2]$ , this is a distance of $3$ units.

Since you need $6$ subintervals, that means that each will be $\frac{1}{2}$ unit. This is going to be the length of each rectangle.

How do you know the width? You determine the $x$ co-ordinate and its corresponding $y$ . The $y$ co-ordinate represents the width.

So in your case:

$y = 2 + 4x^2$ .

If you are using the left-hand endpoints, then the first $x$ co-ordinate will be $-1$ . So your $y$ co-ordinate is $2 + 4(-1)^2 = 6$ .

So $A_1 = L\times W$

$= \frac{1}{2}\times 6$

$= 3\,\textrm{units}^2$ .

The second rectangle will be $\frac{1}{2}$ a unit more on the $x$ axis.

So $x = -\frac{1}{2}$ and $y = 2 + 4\left(-\frac{1}{2}\right)^2 = 3$ .

So $A_2 = L \times W$

$= \frac{1}{2}\times 3$

$= \frac{3}{2}\,\textrm{units}^2$ .

The next rectangle will be at $x = 0$ and $y = 2 + 0^2 = 2$ .

So $A_3 = L \times W$

$= \frac{1}{2} \times 2$

$= 1\,\textrm{unit}^2$ .

Once you have all the areas you need, you add them up.

For the right-hand estimate, you start at the right-hand endpoint and go to the left $\frac{1}{2}$ a unit each time.

**alleysan** · Jan 7th 2010, 10:42 PM

Ah crap,

I overcomplicated it.

Thought I was supposed to use something the prof taught us.

Thank you for explaining everything very well!

**Prove It** · Jan 7th 2010, 10:56 PM

Other methods that give approximations to the area under the curve (and in many cases, better approximations) are the midpoint rule and the trapezoidal rule.

Rectangle method - Wikipedia, the free encyclopedia

Trapezoidal rule - Wikipedia, the free encyclopedia

**alleysan** · Jan 7th 2010, 11:07 PM

Oh wow, thanks for the links!

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