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MooseofDoom

Member

Registered: 2008-08-10

Posts: 13

Need Help with Area

If anyone can help it would be great!

AREAS

For the following functions, you will estimate the area underneath the curves between x=1 and x=3. You will be using rectangles to help you estimate this area.

Each rectangle you use in your estimation will have equal widths. The number of rectangles use will be given. EX. USING 8 RECTANGLES IN THE INTERVAL FROM X=1 TO X=5 WILL RESULT IN THE WIDTHS BEING 0.5 IN LENGTH. The base of each rectangle will be the x-axis and the upper left corner of the rectangle will make contact with the curve. The upper right corner will either be above or below the curve. The left had will determine the height.

Procedure for estimating an area. Function: y=2x+1 from x=1 to x=3 using 4 rectangles. Widths of each rectangle will be 0.5. Height of the first rectangle is determined by using x=1. Height of second rectangle is determined by using x=1.5. Height of the third rectangle is determined by using x=2. And the height of the fourth rectangle is determined by using x=2.5. Add all the areas of each rectangle to get estimation. Notice that this estimation is an underestimate based on all rectangles being drawn below curve, leaving some "empty space" not accounted for.

Find the approximate areas for each function from x=1 to x=3 using 8 rectangles of equal width.
1. y=-x²+4x
2. y=x²-4x+5
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.
.
7. Approximate the area of the following semicricle y= √ -x²+6x. Use 12 rectangles of equal width. Then, find actual area of semicircle and compare with the estimation.

So what i figured out is that for #1 each rectangle will have a width of 0.25. Then do I plug in the x into y=-x²+4x? Like y=-(1)²+4(1) which would give me y=3. Then I multiply 0.25(3) and get 0.75 for the area of the first rectangle out of 8.... so then just add all of the rectangles together?

Let me know if I am doing it right... I don't know how to do #7 so if someone could explain that would be great!

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MooseofDoom

Member

Registered: 2008-08-10

Posts: 13

Re: Need Help with Area

Can someone please help me by 7AM Pacific Standard Time tomorrow because I have to turn this in.... Thanks in advance.

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π≈ 3.141 592 653 589 793 238 462 643 383 279 502 884 197 169 399 375 105 820 974 944 592 307 816 406 286 208 998 628 034 825 342 117 067 982 148 086 513 282 306 647 093 844 609 550 582 231 725 359 408 128 481 117 450 284 102 701 938 521 105 559 644 622 948 954 930...

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Need Help with Area

You're doing the first one right. Continue by finding f(1.25), f(1.5) and so on, multiplying it all by 0.25 and adding the lot together.

#7 is done in a similar way, you just need to play with it a bit first to see where the boundaries are.

y = √(-x² +6x)
y² = -x² + 6x
(x²-6x) + y² = 0
(x-3)² + y² = 3²

This is the standard form of a circle, so we see that the curve is a circle of radius 3, centred at (3,0). This means that it touches the x-axis at 0 and 6.
You were asked for 12 rectangles, which means you will have interval lengths of 0.5.

Now you can estimate the area of the semi-circle like you did before, using the y=... equation that you were given.
Calculating the actual area is easy now that you know the radius, and so you can compare them as well.

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Why did the vector cross the road?
It wanted to be normal.

MooseofDoom

Member

Registered: 2008-08-10

Posts: 13

Re: Need Help with Area

Hey thanks a lot! Thant really helped.

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π≈ 3.141 592 653 589 793 238 462 643 383 279 502 884 197 169 399 375 105 820 974 944 592 307 816 406 286 208 998 628 034 825 342 117 067 982 148 086 513 282 306 647 093 844 609 550 582 231 725 359 408 128 481 117 450 284 102 701 938 521 105 559 644 622 948 954 930...