How do you find the area between #y=e^x# and #y=e^-x# and x=1?

Question

Isabella Ackerman · Accepted Answer

Start by finding the intersection point of the two functions.

#e^x = e^-x#

#e^x - e^-x = 0#

#e^x - 1/e^x= 0#

#e^(2x) - 1 = 0#

#e^(2x)= 1#

#2xlne = ln1#

#2x= 0#

#x = 0#

We also know through end behaviour of the function that #y= e^x# will be above #y = e^-x#. So, we determine the area of #y= e^x# in the interval #0 ≤ x ≤1# and then subtract the area of #y= e^-x# in the interval #0 ≤ x ≤1#.

#int_0^1(e^x - e^-x)dx = e^x + e^-x|_0^1 = e^1 +1/e - (1+ 1) = -2 + 1/e + e#

This can be approximated to #1.086" u"^2#.

Hopefully this helps!

Emma Aldridge · Answer

undefined To find the area between the curves $ y = e^x $ and $ y = e^{-x} $ from $ x = 1 $, you need to calculate the definite integral of the difference of the two functions from $ x = 1 $ to the point of intersection.

First, find the point of intersection by setting the two functions equal to each other:

$$ e^x = e^{-x} $$

Solve for $ x $ to find the intersection point.

Once you have the intersection point, integrate the function $ e^x - e^{-x} $ from $ x = 1 $ to the x-coordinate of the intersection point. This integral will give you the area between the curves from $ x = 1 $ onwards. If the intersection point is to the right of $ x = 1 $, integrate from $ x = 1 $ to that point. If it's to the left, you'll integrate from the intersection point to $ x = 1 $ and take the absolute value of the result.