Start by finding the intersection points.

#{(y = x^2 + 2x + 1), (y= 2x + 5):}#

Substitute the second equation into the first.

#2x+ 5 = x^2 + 2x + 1#

#0 = x^2 + 2x - 2x + 1 - 5#

#0 = x^2 - 4#

#0 = (x + 2)(x- 2)#

#x = -2 and 2#

#:.y = 2x + 5#

#y = 2(-2) + 5 and y = 2(2) + 5#

#y = 1 and y = 9#

The solution set is hence #{-2, 1}# and #{2, 9}#. The next step is doing a basic sketch of the graph. This is not what I'm going to do for this answer, but this is often a necessity when you don't have access to a graphing calculator.

So, our goal is to subtract the area of the lower graph, #y = x^2 + 2x + 1#, from the area of the upper graph, #y = 2x + 5#.

We do this by integrating.

Let's start with finding the area of the upper graph.

#int_(-2)^2(2x + 5)dx#

#=>(x^2 + 5x)|_(-2)^2#

#=> 2^2 + 5(2) - ((-2)^2 + 5(-2))#

#=> 4 + 10 - 4 + 10#

#=> 20#

Now for the lower graph.

#int_(-2)^2(x^2 + 2x + 1)dx#

#=>(1/3x^3 + x^2 + x)|_(-2)^2#

#=> 1/3(2)^3 + 2^2 + 2 - (1/3(-2)^3 + (-2)^2 - 2)#

#=> 8/3 + 4 + 2 + 8/3- 4 + 2#

#=>28/3#

Subtract:

#A = 20 - 28/3 = 32/3#

Hence, the area between #f(x) = x^2 + 2x + 1# and #g(x) = 2x + 5# is #32/3" units"^2#.

Hopefully this helps!