Question

Let R be a region in he first quadrant bounded above by `y=x+2` and below by `y=e^x`, how do you find the area of R?

Answer 1

I got about `0.8031` `"u"^2`.

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Here are the steps that I would follow:

1. Recognize which function is above or below. Highly recommend sketching the functions.
2. Define your lefthand and righthand integration bounds.
3. Choose your method of slicing the region up (horizontal or vertical).
4. Integrate.

UPPER/LOWER BOUNDS

We can tell that `y = x+2` is your upper bound and `y = e^x` is your lower bound, since `e^x` grows slowly for small `x`, whereas `x+2` already has a 2-unit vertical boost.

Here's a visual for that:

graph{(x+2-y)(e^x-y) = 0 [-0.5, 1.5, -5, 5]}

When you define your upper/lower bounds, it's so you can take the area under the higher function and subtract the area under the lower function to get the area in between.

FINDING LEFT/RIGHT-HAND BOUNDS

Left-hand bound:

`a = 0` (we are bounded by the `y` axis in the first quadrant!)

Then, find the righthand point of intersection:

`e^x = x + 2`

As it turns out, this is difficult to evaluate except via calculator. We would numerically determine the right-hand intersection to get:

`b = 1.1461932` `larr` right-hand bound

CHOOSING WHICH METHOD

Generally for this kind of problem, you probably would want to enumerate the `dx` slices that make up the region `R` from left to right.

Remember that the integral is really adding up a bunch of super thin rectangles over a given interval, each of which has a height varying with the function itself.

INTEGRATING

Now that we got all that, define `g(x) = x+2` and `f(x) = e^x` and we have:

`int_(a)^(b) g(x) - f(x)dx = int_(0)^(1.1461932) x+2 - e^x dx`

`= |[x^2/2 + 2x - e^x]|_(0)^(1.1461932)`

where we've used the reverse power rule, `F(x) = int x^ndx = 1/(n+1)x^(n+1)` on `x + 2` and the fact that the antiderivative or derivative of `e^x` is itself.

`=> |[x^2/2 + 2x - e^x]|_(0)^(1.1461932)`

`= ((1.1461932)^2/2 + 2(1.1461932) - e^(1.1461932)) - (cancel((0)^2/2)^(0) + cancel(2(0))^(0) - cancel(e^(0))^(1))`

`= (1.1461932)^2/2 + 2(1.1461932) - e^(1.1461932) + 1`

Once you evaluate that you should get `color(blue)(0.8031)` `color(blue)("u"^2)`.