Question

he following definite integral can be evaluated by subtracting F(B) - F(A), where F(B) and F(A) are found from substituting the limits of integration.

\[\int_{0}^{4} \frac{4000 x +400 }{(5 x^2 +1 x +4)^5}dx\]

After substitution, the upper limit of integration (B) is :
and the lower limit of integration (A) is :

After integrating,

F(B) =

F(A) =

Answer 1

https://webwork.brocku.ca/webwork2_files/tmp/equations/fc/5ac6a844477ddcb4ba737411ba0e271.png

Answer 2

You have a polynomial raised to a power in the denominator of the integrand so choosing that polynomial as your substitution would be a nice place to start.

Answer 3

I know B is 88 and A is 4. I need help with the last 2.

Answer 4

What is the anti-derivative of the integrand after substitution? Find that and then use the new limits you got (A and B) to find F(A) and F(B).

Answer 5

\[\int_{0}^{4} \frac{4000 x +400 }{(5 x^2 +1 x +4)^5}dx=\int_{0}^{4} \frac{400(10 x +1) }{(5 x^2 +1 x +4)^5}dx=\int_{0}^{4} \frac{400}{(5 x^2 +1 x +4)^5}(10 x +1) dx\]
Let u = \(5x^2+x+4\) , du = (10x+1)dx. Then the integral becomes
\[\int_{a'}^{b'} \frac{400 }{(u)^5}du\]
,where a' and b' are the new limits.

Answer 6

u=5x^2+1x+4.
Du=10x+1
400du=4000x+400

Answer 7

First, integrate, then substitute.

Answer 8

400du/u^5?

Answer 9

Yes, integrate that and use the limits of integration you found earlier.

Answer 10

400du/(1.6u^6)?

Answer 11

http://en.wikipedia.org/wiki/Power_rule

Answer 12

differentiate, no integrate? or am i just to tired

Answer 13

hello

Answer 14

@JohnathanHocker or@Callisto ?

Answer 15

have you learned u-substitution yet?

Answer 16

yeah, thats what i thought it was

Answer 17

what is the goal when doing u-sub?

Answer 18

no idea, sorry.

Answer 19

I just did it.

Answer 20

k it is du = 10x+1 dx

Answer 21

Yes, I know thus far.

Answer 22

and when doing u-substitution you have to check for a and b values
because the range might not be from 0 - 4 anymore

Answer 23

41 and 1?

Answer 24

did you solve or guess it?

Answer 25

subbed into the equations derived

Answer 26

when x = 0
u = ?

when
x = 4
u = ?

that is how you do it then use the new values
also, notice now that you have two 10x+1?
the one callisto did will end up being as the "dx"
when you finally use the u notation

Answer 27

so did my values work?

Answer 28

dunno

Answer 29

oh.

Answer 30

I'm not subbing into the derived equation right?

Answer 31

@nincompoop ?

Answer 32

you are
use the expression you're going to change into u
then compare it to the u

Answer 33

whereas u was 5x^2+x+4?

Answer 34

@nincompoop @Callisto @JonathanHocker any ideas?

Answer 35

@IsTim what you don't get?

Answer 36

To be honest, I feel like I don't understand anything. Lots of people online and offline have explained this to me, but I keep getting lost.

Answer 37

anoyne?

Answer 38

so you don't get what @callisto did?

Answer 39

I managed to get as far as where he stated, but I don't know what to do after (Adn I still don't understand what I did up to that point)

Answer 40

F(x)=\[\int_{0}^{4} \frac{4000 x +400 }{(5 x^2 +1 x +4)^5}dx\]
\[= 400\int\limits_{0}^{4} \frac{ 10x+1 }{ (5x^2 +x+4 )^5}dx\]

Answer 41

So we forget out 400?

Answer 42

^^ we can do that because :-

\(\large \int cf(x) dx = c \int f(x) dx\)

Answer 43

u can pull out 'constant' out of integral

Answer 44

yes only focus on terms inside in integral sign

Answer 45

Also, just for completeness :-

\(\large \int c dx = c \int 1 dx = cx + c_1\)

Answer 46

u need to knw these basic properties before diving into doing complicated indefinite integrals

Answer 47

for make easy this integration we let 5x^2+x+4=u
and differentiate it as du/dx = 10x +1
which can be written as du=(10x+1)dx

Answer 48

I know what you're talking about; I've achieved the same results, I jsut don't know ti use this to figure out constant of integration

Answer 49

actually we're doing a "definite integral",
the constant gets subtracted away when u do : F(B) - F(A)
So, we dont bother about constant of integration in "definite integrals"

Answer 50

Oh no, I forgot to include that in this question. They're looking for that.

Answer 51

So sorry. I reviewed the question and realized I was missing something.

Answer 52

I was wonderin why everyone was telling me what I already know, so I looked through and realized I forgot to ask for constant of integration

Answer 53

OH MY GOD SORRY! LAck of sleep. IT isn't asking for that. I'm doing 2 questions at the same.time. Ignore constant of something.

Answer 54

It's:
After integrating,

F(B) = ?

F(A)=?

Answer 55

That's what I can't figure out.

Answer 56

After integrating, you can put the value of the upper limit to get F(B), and the value of the lower limit to get F(A)

Answer 57

Perhaps you can forget about the definite integral, and do the indefinite integral, shown as below, first:
\[400\int\limits\frac{ 10x+1 }{ (5x^2 +x+4 )^5}dx\]

Answer 58

not gettin the right answers though.

Answer 59

F(B)=-1.66 and F(A)=-0.39

Answer 60

Both in terms of charge and value.

Answer 61

Hello?

Answer 62

I got 0.0226 and 0.465

Answer 63

Would you mind showing us what you have got for the integral BEFORE evaluating it?

Answer 64

That was mechanics...

Answer 65

sorry i don't understand?

Answer 66

wait darnit wrong picture

Answer 67

goddarnit i need sleep

Answer 68

I guess you have not integrated it yet. You just substituted the values in it.

Answer 69

1) You need to integrate\[400\int\limits\frac{ 10x+1 }{ (5x^2 +x+4 )^5}dx\]as usual, the final answer should be in terms of x.
2) Substitute the values (the upper limit and lower limit) back to the result you get. You can drop the constant you got when doing the substitution.

Answer 70

Does that work?

Answer 71

Of course, it does not.

Answer 72

As mentioned above, you need to use u-substitution.

Answer 73

400lnu^5?

Answer 74

c is -2768?

Answer 75

Not ln(u^5).
Use power rule when you have \[\int \frac{1}{x^a}dx\]for a\(\ne\)1.

Answer 76

Sorry, you've confused me a bit more.

Answer 77

argh everyone always leaves

Answer 78

@Callisto @ganeshie8 @gyanu @nincompoop @JonathanHorker any ideas what that means/how to use it?

Answer 79

\[\int x^adx =\frac{1}{a+1}x^{a+1}+C, \text{where } a\ne1 \]

Answer 80

a is dx?

Answer 81

a is a constant, the power of x.