Question

Integrals,

Can you please help me solve these integrals http://screencast.com/t/WYlDUCGzJkv

I think they need "intergration by parts" which I am not familiar with

Answer 1

okay we have formula for integration by parts

\[\int\limits_{}^{}u(x).v(x)=u(x).\int\limits_{}^{}v(x) -\int\limits_{}^{} du(x)/dx . \int\limits_{}^{}v(x)\]

Answer 2

Yes, when you have integration by parts you must use the formula,
\[\int u\cdot dv=u\cdot v-\int v \cdot du\]Now you should select which functions are u and dv. Usually you select u as the function that is easier to derive and dv as the funtion that is easier to integrate.

Answer 3

where u(x) and v(x) are functions in terms of x .. and there is dx in every integral sign..

Answer 4

can you please help me on the first ones

Answer 5

we have a priority order for which functions would be u(x) and v(x)

this order is
1.inverse trigonometric functions
2.logarithmic functions
3.arthritic functions (polynomials)
4.trigonometric functions
5.exponential functions

Answer 6

For example, in the first integral, I would do,
\[u=x\Rightarrow du=dx\\
dv=e^{-2x}dx\Rightarrow v=-\frac{1}{2}e^{-2x}\]Then,
\[\int x e^{-2x}dx=-\frac{x}{2}e^{-2x}-\left(\frac{-1}{2}\right)\int e^{-2x}dx=\\
=-\frac{x}{2}e^{-2x}-\frac{1}{4} e^{-2x}\]

Answer 7

or simply remember I LATE

Answer 8

John_ES explained it well but if you want we can slove it did you saw formula wrote?

Answer 9

hold on I am trying the first one to grasp the concept

Answer 10

i meant formula i wrote.. okay that's good..

Answer 11

May the most obscure step is this,
\[u=x\Rightarrow du=dx\]But this is differentiation. For other case,
\[u=x^2\Rightarrow du=2xdx\]The other part,
\[dv=e^{-2x}dx\Rightarrow \int dv=\int e^{-2x}dx=-\frac{e^{-2x}}{2}\]

Answer 12

okie i don't want to confuse you what i told you was the stuff wrote in my maths book...

may be you have different form of integration by parts... this notes would definitely help you...

http://tutorial.math.lamar.edu/Classes/CalcII/IntegrationByParts.aspx

Answer 13

I solved the second by my self :DDD Are the others the same ??

Answer 14

Yes, they are the same.

Answer 15

Thank you guys !!!!!!!!!!!

Answer 16

you are most welcome...!

Answer 17

If you want to check your results you can do it in wolfram alpha.

http://www.wolframalpha.com/

For example, for the first integral, write there,
Integrate[x*Exp[-2x],x]

And you'll find the answer to check it, in cas you don't have the solutions,

http://www.wolframalpha.com/input/?i=Integrate%5Bx+Exp%5B-2x%5D%2Cx%5D

You're welcome.

Answer 18

Don't forget to add a constant in this indefinite integrals.

Answer 19

I have the solutions So I check it easily !! :)

Answer 20

Better ;)

Answer 21

By the way, the notes @prince90p are very good.

Answer 22

thank you paul dawkins is best tutor..! i really lucky to have this notes.!!