What is #f(x) = int 5e^(2x)-2e^x+3x dx# if #f(4 ) = 2 #?

1 Answer
Dec 19, 2016

This can be broken down into separate integrals.

#f(x) = int(5e^(2x) - 2e^x + 3x)dx#

#f(x) = int(5e^(2x))dx + int(-2e^x)dx + int(3x)dx#

We know that #int(ae^(nx))dx = a/n e^(nx) + C# and #int(x^n)dx = (x^(n + 1))/(n + 1) + C#. Therefore:

#f(x) = 5/2e^(2x) - 2e^x + 3/2x^2 + C#

We can now solve for #C#.

#2 = 5/2e^(8) - 2e^(4) + 3/2 4^2 + C#

#2 = 5/2e^8 - 2e^4 + 24 + C#

#2 - 5/2e^8 - 2e^4 - 24 = C#

#C = -22 - 5/2e^8 - 2e^4#

This is as simplified a form as we can get without using a calculator to approximate.

Use any calculator to get #C ~= -131.197#.

Hence, #f(x) = 5/2e^(2x) - 2e^x + 3/2x^2 - 131.197#.

Hopefully this helps!