Area of region integrals

Question

Find the area of the region enclosed by the graphs of 𝑦=𝑒^𝑥, 𝑦=𝑒^−𝑥, and 𝑦=7.

(Use symbolic notation and fractions where needed.)

Answer 1

Draw the graph.

e^x = 7 when x = ln(7) This is where the graphs of y = e^x and y = 7 intersect.

The graph of y = e^-x is the reflection of the graph of y = e^x in the y-axis.

The region is symmetric with respect to the y-axis.

Area = 2 ∫_{(from 0 to ln7)} (7 - e^x) dx = 2(7x - e^x) _{(from 0 to ln7)} = 14ln7 - 14 + 2 = 14ln7 - 12

Answer 2

If you plot the 3 functions, you can see that the area is an even function, so we can multiply half the area by 2.

To find the intersection point between 7 and e^x:

e^x=7

x=ln(7)

A = 2∫₀^ln(7)(7-e^x) dx = 2[7x-e^x]₀^ln(7) = 2[7ln(7)-e^ln(7)-0+e⁰] = 14ln(7)-14+2 ≈15.24

Note: e^ln(7) = 7 and e⁰ = 1.