Answer:
3.58
Step-by-step explanation:
Given :
y = 2x + 4 -------- eq1
y = 2x - 4 -------- eq2
sanity check : both equations have same slope, so we can conclude that they are both parallel to one another.
Step 1: consider equation 1, pick any random x-value and find they corresponding y-value. we pick x = -2
This gives us y = 2(-2) + 4 = 0
Hence we get a point (x,y) = (-2,0)
Step 2: express equation 2 in general form (i.e Ax + By + C = 0)
y = 2x-4 -------rearrange---> 2x - y -4 = 0
Comparing with the general form, we get A = 2, B = -1, C = -4
Recall that the distance between 2 parallel lines is given by the attached formula (see attached picture).
substituting the values for A, B, C and (x, y) from the previous step:
d = | (2)(-2) + (-1)(0) + (-4) | / √(2² + (-1)²)
d = | -4 + 0 - 4 | / √(4 + 1)
d = | -8 | / √5
d = 8 / √5
d = 3.5777
d = 3.58 (rounded 2 dec. pl)
I'm going to assume that your function is f(x) = 1 + x^2 (NOT x2).
I suspect you're trying to estimate the "area under the curve of f(x) = 1 + x^2. You need to use this or a similar description to explain what you're doing.
Also, you need to specify whether you want "left end points" or "right end points" or "midpoints." Again I must assume you want one or the other (and will assume that you meant "left end points").
First, let's address the case n=3. You must graph f(x) = 1 + x^2 between -1 and +1. We will find the "lower sum," using "left end points." The 3 x-values are {-1, -1/3, 1/3}. Evaluate the function f(x) = 1 + x^2 at these 3 x-values. Keep in mind that the interval width is 2/3.
The function (y) values are {0, 2/3, 4/3}.
Sorry, Michael, but I must stop here and await clarification from you regarding what you've been told to do in this problem. Otherwise too much guessing (regarding what you meant) is necessary. Please review the original problem and ensure that you have copied it exactly as presented, and also please verify whether this problem does indeed involve estimating areas under curves between starting and ending x-values.
