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Mathematics 16 Online
OpenStudy (anonymous):

integral of (e^x)[(f(x))+(f'(x)]

OpenStudy (anonymous):

e^xf(x)

OpenStudy (anonymous):

can you explain the working ?

OpenStudy (anonymous):

yea sure :)

OpenStudy (anonymous):

http://www.twiddla.com/575173 u can use this ,its easier here :)

OpenStudy (anonymous):

e^x f'(x) + e^xf(x) first we will solve the first one e^x f'(x) Now do you know the by parts

OpenStudy (anonymous):

I'm on twidlla

OpenStudy (anonymous):

the terms f(x) and its prime,i dont know how to deal with them,and yes i know the 'by parts'

OpenStudy (anonymous):

e^x (f(x)+f'(x)) if you factorize it is e^xf(x) +e^x(f'(x) now, you can see clearly that this is a derivative of e^x(f(x), because forming the derivative you have to use the product rule. Can you see it now?

OpenStudy (anonymous):

Just look at the product rule, it is: (u,v)' = u'*v + u*v' , u and v being functions of x

OpenStudy (anonymous):

Let look at \[{(e^x*f(x))'=(e^x)'f(x)+e^xf'(x) =e^xf()+e^xf'(x)=e^x(f(x)+f'(x))}\] So we can say that \[e^x(f(x)+f'(x))=(e^xf(x))'\] \[\int\limits(f'(x)dx)=f(x)+C\] Then \[\int\limits(e^x(f(x)+f'(x)))dx=\int\limits((e^xf(x))')dx=e^xf(x)+C\]

OpenStudy (anonymous):

@Ishaan:thanks for all the help,and im sorry about leaving without informing,my power supply got heavily disrupted. @ajahangir and tatiana: thankyou :)

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