Calculus help-definite integrals
Favorites|Homepage
Subscriptions | sitemap
HOME > > Calculus help-definite integrals

Calculus help-definite integrals

[From: ] [author: ] [Date: 12-04-23] [Hit: ]
-To find the point of intersection of any functions,You cant solve for this manually, so youll have to either use a calculator or Newtons Method.Since we know they must intersect between 0 and 1, [e^0 = 1, e^1 = 2.......
Please help with my Calculus extra credit problem. I drew the graph, but I don't understand how to find points of intersection.

Find the area of the region enclosed under the graphs of the functions f(x)= e^x and g(x)= 1/x to the x-axis between x=0 and x=4.

Thanks so much for any help!

-
To find the point of intersection of any functions, you just set them equal to each other:
e^x = 1/x
x∙e^x = 1
x∙e^x - 1 = 0
f(x) = x∙e^x - 1
f '(x) = (x + 1)∙e^x
You can't solve for this manually, so you'll have to either use a calculator or Newton's Method.
Since we know they must intersect between 0 and 1, [e^0 = 1, e^1 = 2.78, 1/1 = 1] we can start with a guess of x_0 = 1/2.
x_1 = 0.57102
x_2 = 0.567156
x_3 = 0.567143
x_4 = 0.567143
...
Therefore, the point of intersection is approximately x = 0.567143.
Of course, i could have simply typed in e^x = 1/x into wolfram and it would've done the same calculations, but this demonstrates Newton's Method.

Just integrate under e^x for 0 to 0.567143, then under 1/x from 0.567143 to 4.

-
the intersections happen when the x and y values of both functions are equal.
that is f(x) = g(x)
and the x's re the same

so e^x = 1/x
1
keywords: integrals,definite,help,Calculus,Calculus help-definite integrals
New
Hot
© 2008-2010 http://www.science-mathematics.com . Program by zplan cms. Theme by wukong .