# 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?

##### 1 Answer

I got about

Here are the steps that I would follow:

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

**UPPER/LOWER BOUNDS**

We can tell that

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

*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

#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