what is the total area bounded by y=e^x, y=1, y=2, x=3?

Question

it comes from AP Cal. BC about Math 124,125

Answer 1

y = e^x
when x = 0, y = 1
when x = 3, y = e^3 = 20.085 > 2
when y = 2 , x = ln 2

Total area bounded by y=e^x, y=1, y=2, x=3
= total area bounded by y=e^x, y=1, y=2, x=0, x=3
= total area bounded by y=e^x, y=1, x=0, x=3
- total area bounded by y=e^x, y=2, x=ln2, x=3
= ∫(e^x - 1) dx - ∫(e^x - 2)dx
= [e^x - x] {from x=0 to 3} - [e^x -2x] {from x=ln2 to 3}
= [e^3 - 3 - (e^0 - 0)] - [e^3 - 2(3) - (e^ln2 - 2ln2)]
= [e^3 - 3 - 1] - [e^3 - 6 - 2 + 2ln2]
= e^3 - 4 - e^3 + 8 - 2ln2
= 4 - 2ln2
= 2.6137

Answer 2

We know that e^0 = 1 and e^(ln2) = 2, so from x = 1 to x = ln2 we need to find the area between the graphs of y=e^x and y=1. From x = ln2 to x = 3, we only need to find the area between the graphs of y=2 and y=1.

Let's set up some integrals to show these areas:

∫{0:ln2} e^x - 1 dx + ∫{ln2:3} 2 - 1 dx

Integrating we obtain:

= e^x - x]{0:ln2} + x]{ln2:3}

Plugging in our values we now get:

= e^(ln2) - ln2 - [e^0 - 0] + 3 - ln2

Let's simplify:

= 2 - ln2 -1 + 0 + 3 -ln2
= 4 -2ln2

Using a calculator, you get a value about equal to 2.614, or if you remember that ln2 is about equal to 0.693, you can do this with paper and pencil.

--charlie