Area between two curves problem

  • Thread starter Thread starter zeezey
  • Start date Start date
  • Tags Tags
    Area Curves
Click For Summary
To find the area between the curves y = e^x, y = 2, and the y-axis, the integral of e^x - 2 is used, with the area defined by the bounds of the intersection points. The y-axis serves as a lower bound for the integral, specifically at x = 0. To express x in terms of y, the equation y = e^x can be rearranged, allowing for the integration to be set up correctly. After determining the upper bound from the intersection of the curves, the definite integral can be evaluated. The final area calculated is 4ln(4) - 4.
zeezey
Messages
11
Reaction score
0

Homework Statement


Find the area of the indicated region.
Enclosed by y = e^x, y = 2, and the y-axis

The Attempt at a Solution



So I the area should be integral of e^x - 2 which = e^x - 2x . I'm not sure what to do with the y-axis it mentions?
 
Physics news on Phys.org
zeezey said:

The Attempt at a Solution



So I the area should be integral of e^x - 2 which = e^x - 2x . I'm not sure what to do with the y-axis it mentions?


Firstly where does y=ex cut the y-axis?

You should then know the area between a curve and the y-axis is given by ∫ x dy. So if y=ex, how do you put x in terms of y?
 
umm I'm not sure you mean like dx = e^x dy ?
 
So you have e^x-2 as the integrand, as you mentioned. The mention of the y-axis is to put a lower bound on x (in this case, evaluating your integral from a-to-b by the FTC, the lower bound, a, is x=0)...then the upper bound would be the intersection of the functions. It's always helpful to draw out what you're asked for.

I'll leave it to you to find the upper bound and then evaluate your now definite integral.
 
So I got 4ln(4) - 4 . Is this correct?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 14 ·
Replies
14
Views
2K
Replies
5
Views
2K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
5K
Replies
4
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
13
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K