Question

Integration by parts?

Answer 1

\[\int\limits_{}^{} x ^{2}e ^{10x}dx=\frac{ 1 }{3 }x^2e ^{10x}-\int\limits_{}^{} \frac{ 1 }{ 3 }x ^{3}(10e ^{10x})dx\]

Answer 2

not sure if that is correct

Answer 3

and not sure what to do after expanding the right hand side integral

Answer 4

no take
x^2 as the first function
so that it gets differentiated.. and the power of x will get reduced ..

Answer 5

so take u = x^2?

Answer 6

to me this problem screams integration by parts
so just do what i taught you before

Answer 7

but should i take u as e^10x or x^2?

Answer 8

when you are deciding which to let u =
then remember this acronym
I = Inverse Trig
L = Logarithmic
A = Algebraic
T = Trig
E = exponetial

Answer 9

you Top is your highest priority and bottom is lowest
so in this case let u = x^2

Answer 10

so if u = x^2 then dv = e^(10x)
that means

du = 2x dx
and v = (1/10)e^(10x)

Answer 11

so using the formula again you get:

\[\frac{ 1 }{ 10 } x^2 e^{10x} - \int\limits_{}^{} \frac{ 1 }{ 10 }e^{10x} (2xdx)\]

Answer 12

and that integral on the right simplifies to
\[\int\limits_{}^{} \frac{ 1 }{ 5 }xe^{10x} dx\]

do you understand so far?

Answer 13

yes, I've been doing it along while you were typing.

Answer 14

how to integrate the 1/5xe^(10x)dx though?

Answer 15

well if doing integration by parts once doesnt finsish it then just do it again :)

Answer 16

you from that integral we just do the same method
let u = (1/5)x and dv = e^(10x)dx
so
du = 1/5 and v = 1/10(e^10x)

Answer 17

and use the formula again

Answer 18

so now i will write out the whole thing for you

\[\frac{ 1 }{ 10 }x^2e^{10x} - \left[ \frac{ 1 }{ 50 }x e^{10x} - \int\limits_{}^{} \frac{ 1 }{ 50 }e^{10x}dx \right]\]

Answer 19

so now you can do that integral right there and its just

(1/500)e^(10x)

Answer 20

put it all together and add a +C and your done

Answer 21

wait, shouldn't the integral of (1/5)xe^(10x) dx = (1/2)xe^(10x) - (1/5)e^(10x)?

Answer 22

im not quite sure where you got your answers
but it is another integration by parts where

u = 1/5x dv = e^(10x) dx
du = 1/5dx v = 1/10(e^(10x)

so (u)(v) = (1/50)(x)(e^(10x)
then you subtract the integral of (1/10)(e^(10x))(1/5)(dx)

Answer 23

yeh sorry, i made a mistake doing the integral of dv.

Answer 24

so from the final part of what you typed above, you find the integral of 1/50e^(10x) obviously and then expand the whole brackets?

Answer 25

yea and that should be your answer with a +C at the end

Answer 26

\[\frac{ 1 }{ 10 }e ^{10x}x ^{2} - \frac{ 1 }{ 5 }xe ^{10x}-\frac{ 1 }{ 500 }e ^{10x}+C\]

Answer 27

so is that the answer?

Answer 28

the middle part is a - 1/50 not a 1/5 and the right part is a + not a -

Answer 29

1/50** for the second

Answer 30

yeh you are right

Answer 31

yea so thats the whole problem hope you got it

Answer 32

\[\int\limits_{}^{}x ^{2}e ^{^{10x}}dx = \frac{ 1 }{ 10 }e ^{10x}x ^{2}-\frac{ 1 }{ 5 }e ^{10x}x+\frac{ 1 }{ 500 }e ^{10x} +C\]

Answer 33

-1/50** for the second part

Answer 34

so that should be the final answer right?

Answer 35

yup thats right

Answer 36

yep it was correct! oh man it had so many steps and it took so long...

Answer 37

haha yea it takes a while but you'll get used to it

Answer 38

are you currently busy jay?

Answer 39

not really busy whats up

Answer 40

uh I have more questions I need to ask, and they are all relating to differentiation which I have no idea on how to do since it's more in depth and my lecturers don't explain it well and the textbook is rubbish

Answer 41

what is the question i might know how to do it

Answer 42

are you familar with trig substitution?

Answer 43

no sorry :(

Answer 44

um.. well i show you how to do it

Answer 45

yep

Answer 46

lets go back to the problem
you see a sqrt(x^2 + 13)

so if you look at the triagle: a = 13 and u = x

Answer 47

shouldn't it be 13^2 not just 13 then?

Answer 48

sorry your a = sqrt(13)

Answer 49

so now you can do
\[\sin \theta = \frac{ x }{ \sqrt{x^2 + 13} }\]
or
\[\tan \theta = \frac{ x }{ a }\]
or
\[\cos \theta = \frac{\sqrt{13} }{ \sqrt{x^2 + 13} }\]

Answer 50

so lets use them
\[x = \sqrt{13} \tan \theta\]
\[dx = \sqrt{13} \sec^2 \theta\]

Answer 51

ok but how is a sqrt(13) in the first place?

Answer 52

try leting a = 13 and a = x
then solve for the hypotenuse by using the pythagorean theorem