Mathematics, 18.07.2019 20:30 francisco42002

Approximate area under the curve f(x) =-x^2+2x+4 from x=0 to x=3 by using summation notation with six rectangles and use the the right endpoint value for x to calculate the height

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Answer from: leahphi

Summation notation:

$\frac{1}{2}\sum_{k=1}^6f((.5k))$

or after using your function part:

$\frac{1}{2}\sum_{k=1}^6(-(.5k)^2+2(.5k)+4)$

After evaluating you get 11.125 square units.

Step-by-step explanation:

The width of each rectangle is the same so we want to take the distance from x=0 to x=3 and divide by 6 since we want 6 equal base lengths for our rectangles.

The distance between x=0 and x=3 is (3-0)=3.

We want to divide that length of 3 units by 6 which gives a length of a half per each base length.

We are doing right endpoint value so I'm going to stat at x=3. The first rectangle will be drawn to the height of f(3).

The next right endpoint is x=3-1/2=5/2=2.5, and the second rectangle will have a height of f(2.5).

The next will be at x=2.5-.5=2, and the third rectangle will have a height of f(2).

The fourth rectangle will have a height of f(2-.5)=f(1.5).

The fifth one will have a height of f(1.5-.5)=f(1).

The last one because it is the sixth one will have a height of f(1-.5)=f(.5).

So to find the area of a rectangle you do base*time.

So we just need to evaluate:

$\frac{1}{2}f(3)+\frac{1}{2}f(2.5)+\frac{1}{2}f(2)+\frac{1}{2}f(1.5)+\frac{1}{2}f(1)+\frac{1}{2}f(.5)$

or by factoring out the 1/2 part:

$\frac{1}{2}(f(3)+f(2.5)+f(2)+f(1.5)+f(1)+f(.5))$

To find f(3) replace x in -x^2+2x+4 with 3:

-3^2+2(3)+4

-9+6+4

To find f(2.5) replace x in -x^2+2x+4 with 2.5:

-2.5^2+2(2.5)+4

-6.25+5+4

2.75

To find f(2) replace x in -x^2+2x+4 with 2:

-2^2+2(2)+4

-4+4+4

To find (1.5) replace x in -x^2+2x+4 with 1.5:

-1.5^2+2(1.5)+4

-2.25+3+4

4.75

To find f(1) replace x in -x^2+2x+4 with 1:

-1^2+2(1)+4

-1+2+4

To find f(.5) replace x in -x^2+2x+4 with .5:

-.5^2+2(.5)+4

-.25+1+4

4.75

Now let's add those heights. After we obtain this sum we multiply by 1/2 and we have our approximate area:

$\frac{1}{2}(f(3)+f(2.5)+f(2)+f(1.5)+f(1)+f(.5))$

$\frac{1}{2}(1+2.75+4+4.75+5+4.75)$

$\frac{1}{2}(22.25)$

11.125

Okay now if you wanted the summation notation for:

$\frac{1}{2}(f(3)+f(2.5)+f(2)+f(1.5)+f(1)+f(.5))$

is it

$\frac{1}{2}\sum_{k=1}^{6}(f(.5+.5(k-1)))$

or after simplifying a bit:

$\frac{1}{2}\sum_{k=1}^6 f((.5+.5k-.5))$

$\frac{1}{2}\sum_{k=1}^6f((.5k))$

If you are wondering how I obtain the .5+.5(k-1):

I realize that 3,2.5,2,1.5,1,.5 is an arithmetic sequence with first term .5 if you the sequence from right to left (instead of left to right) and it is going up by .5 (reading from right to left.)

Answer from: Quest

step-by-step explanation:

Answer from: Quest

may i see the picture? it would a lot, xox

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Approximate area under the curve f(x) =-x^2+2x+4 from x=0 to x=3 by using summation notation with si...

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