3. Area Between 2 Curves using Integration
by M. Bourne
Area bounded by the curves `y_1` and `y_2`, & the lines `x=a` and `x=b`, including a typical rectangle.
We are trying to find the area between 2 curves, `y_1 = f_1(x)` and `y_2 = f_2(x)`, and the lines `x = a` and `x = b`.
We see that if we subtract the area under lower curve
`y_1 = f_1(x)`
from the area under the upper curve
`y_2 = f_2(x)`,
then we will find the required area. This can be achieved in one step:
Alternative Way to Find The Formula (from first principles)
Another way of deriving this formula is as follows (the thinking here is important for understanding how we develop the later formulas in this section).
Each "typical" rectangle indicated has width `Δx` and height `y_2 − y_1`, so its area is `(y_2 − y_1)Δx`.
If we add all these typical rectangles, starting from `a` and finishing at `b`, the area is approximately:
Now if we let Δx → 0, we can find the exact area by integration:
Summing vertically to find area between 2 curves
Likewise, we can sum vertically by re-expressing both functions so that they are functions of y and we find:
Notice the `c` and `d` as the limits on the integral (to remind us we are summing vertically) and the `dy`. It reminds us to express our function in terms of `y`.
Find the area between the curves `y = x^2 + 5x` and `y = 3 − x^2` between `x = -2` and `x = 0`.
Graphs of `y=x^2 + 5x` and `y=3-x^2`, showing the portion between `-2 < x < 0`.
From the graph, we see that `y=3-x^2` is above `y=x^2 + 5x` in the region of interest, so we'll use:
`y_1=x^2 + 5x`
So we need to find:
`=int_-2^0 [(3-x^2)-(x^2+5x)] dx`
`=int_-2^0 [(-2x^2-5x+3)] dx`
`=10 2/3\ text(sq units)`
Some of the shaded area is above the `x`-axis and some of it is below. Don't worry about taking absolute value - the formula takes care of that automatically.
1. Find the area bounded by `y = x^3`, `x = 0` and `y = 3`.
Graph of `y=x^3`, showing the portion bounded by `x = 0` and `y=3`.
We will use:
and use horizontal elements. (In this example we could have added horizontally as well, but will do it vertically to illustrate the method.)
In this case, `c = 0` and `d = 3`.
We need to express `x` in terms of `y`:
`y = x^3` so `x = y^(1//3)`
`=int_0^3 (y^(1//3)) dy`
`=3.245\ text[sq units]`
2. Find the area bounded by the curves
`y = x^2 + 5x` and `y = 3 − x^2`.
(This is an extension of the Example above.)
Area bounded by the curves `y=x^2 + 5x` and `y=3-x^2`, including a typical rectangle.
We need to use: `A=int_a^b(y_2-y_1) dx`
We note that `y = 3 − x^2` is above `y = x^2+ 5x` so we take
`y_2= 3 − x^2` and `y_1= x^2+ 5x`
Points of intersection occur where:
`x^2 + 5x = 3 − x^2`
`2x^2 + 5x − 3 = 0`
`(x + 3)(2x − 1) = 0`
So `x = -3` or `x = 0.5`
We take vertical elements (indicated by the vertical rectangle in the graph above).
So the area is given by:
`=int_-3^0.5 ([3-x^2]-[x^2+5x]) dx`
`=14.29\ text[sq units]`
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3. Find the area bounded by the curves
`y = x^2`, `y = 2 − x` and `y = 1`.
Area bounded by `y = x^2`, `y = 2 − x` and `y = 1`, including a typical rectangle.
We take horizontal elements in this case.
So we need to solve `y = x^2` for `x`:
`x = ±sqrt(y)`
We need the left hand portion, so `x = − sqrt(y)`.
Notice that `x = 2 − y` is to the right of `x = -sqrt(y)` so we choose
`x_2= 2 − y` and `x_1= -sqrt(y.)`
The intersection of the graphs occurs at `(-2,4)` and `(1,1)`.
So we have: `c = 1` and `d = 4`.
`=19/6\ text[sq units]`
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