1 Answer | Add Yours
To answer this, we need to first consider what this is asking for: we need to figure out the area between 1 and `x` under the curve `t^2`. We then take this value to be `A(x)`. In the end, `A(x)` is only a very tedious function to calculate (before you learn more calculus).
Let's start by examining the graph of `y = t^2` as the question asks.
All we are asked to do are to approximate a few values of `A(x)` by calculating the area between the curve and the (here) `t`-axis from 1 to `x`. This request actually gives us a very convenient start point!
If we are asked to calculate an area from 1 to 1 (here, then ` x=1`), we already know that area to be zero. We know this because a line segment has no area! So, on our graph of `A(x)`, we have the following start point:
Now, let's go forward. For simplicity's sake, we'll model everything as a trapezoid and use the following effective formula:
`A(x) = (x^2+1)/2*(x-1)`
Remember that the effective base is the average of the two, which we have calculated by using the two function values at 1 and `x`. Also, the height will necessarily be the distance between 1 and `x` (`x-1`).To visualize, this relation results from the following approximate drawing (here, `x=2`):
Notice that our trapezoid is an alright approximation when we're close to 1, but that it will be worse the further away we are from 1. Also, notice that based on our formula that when we go to the left of 1, then our area becomes negative. This result actually is correct! If a function dips below the line or is positive to the left of your area's start point, the convention is to make it negative.
So, let's continue using our formula to find a few points close to 1 to approximate our function:
A(-1) = -2
A(-0.5) = -0.9375
A(0) = -0.5
A(0.5) = -0.3125
A(1) = 0
A(1.5) = 0.8125
A(2) = 2.5
Now, let's just put all of these points into our approximated graph
Connect the dots and you'll have your approximated function!
We’ve answered 301,735 questions. We can answer yours, too.Ask a question