Estimate the area under the graph of f(x)=x^(2)+3x from x=1 to x=5 using 4 approximating rectangles and left endpoints.

 Rn ≈ 0.6345

 Ln ≈ 0.7595

Step-by-step explanation:

The interval from -1 to 2 has a width of (2 -(-1)) = 3. Dividing that into 4 equal intervals means each of those smaller intervals has width 3/4.

It can be useful to use a spreadsheet or graphing calculator to evaluate the function at all of the points that define these intervals:

 x = -1, -.25, 0.50, 1.25, 2

Of course, the spreadsheet can easily compute the sum of products for you.

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The approximation using right end-points will be the sum of products of the interval width (3/4) and the function value at the right end-points:

 Rn = (3/4)f(-0.25) +(3/4)f(0.50) +(3/4)f(1.25) +(3/4)f(2)

 Rn ≈ 0.6345

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The approximation using left end-points will be the sum of products of the interval width (3/4) and the function value at the left end-points:

 Ln = (3/4)f(-1) +(3/4)f(-0.25) +(3/4)f(0.50) +(3/4)f(1.25)

 Ln ≈ 0.7595

It is usually convenient to factor out the interval width, so only one multiplication needs to be done: (interval width)(sum of function values).

Step-by-step explanation:

 Rn ≈ 0.6345

 Ln ≈ 0.7595

Step-by-step explanation:

The interval from -1 to 2 has a width of (2 -(-1)) = 3. Dividing that into 4 equal intervals means each of those smaller intervals has width 3/4.

It can be useful to use a spreadsheet or graphing calculator to evaluate the function at all of the points that define these intervals:

 x = -1, -.25, 0.50, 1.25, 2

Of course, the spreadsheet can easily compute the sum of products for you.

__

The approximation using right end-points will be the sum of products of the interval width (3/4) and the function value at the right end-points:

 Rn = (3/4)f(-0.25) +(3/4)f(0.50) +(3/4)f(1.25) +(3/4)f(2)

 Rn ≈ 0.6345

__

The approximation using left end-points will be the sum of products of the interval width (3/4) and the function value at the left end-points:

 Ln = (3/4)f(-1) +(3/4)f(-0.25) +(3/4)f(0.50) +(3/4)f(1.25)

 Ln ≈ 0.7595

It is usually convenient to factor out the interval width, so only one multiplication needs to be done: (interval width)(sum of function values).

really really sorry if i'm

-12 -5(x + 4) = 3x

-12 - 5x - 20 = 3x

-12 - 20 - 5x = 3x

-32 - 5x = 3x

    + 5x   + 5x

-32 = 8x

8         8

- 4= x