Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 months ago

# [REPOST] I got the upper and lower limits correct but I am getting the indefinite integral wrong and writing it in terms of you? (Part 1)?

Suppose we want to evaluate the definite integral integral 25(top),16(bottom) rsqrt(2503-4r^2)dr using the substitution, u=2503-r^2

Part 1

Re-write the definite integral in terms of the variable u and remeber to use the limits of integration for the function u=f(r). Then, input the anti derivative of the integrand and the limits of integration you found

upper limit: 1479 CORRECT

=-1/8*sqrtudu

lower limit: 3       CORRECT

upper limit: 1479 CORRECT

=-(u^(3/2))/12

lower limit: 3       CORRECT

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• 2 months ago

I see  ....The part about  must have lower limit smaller is extremely hard to see in first post...... basically drop the - sign by using  integral property...

don't forget to choose a Best Answer....

If  U = 2503 - 4r^2      dU = -8r dr    so   r dr = - dU / 8

Let's fix  your  limits of integration first ........

From your  ORIGINAL problem it  becomes   ∫  - (1/8) U ^1/2  dU    ........the  lower limit   x = 16  is now  U = 1479   [  2053 - 4* (16)^2  = 1479 ]   not  3 !    .......  Upper limit  x = 25 is  now  U = 3,  NOT 1479  !   [ 2053 - 4* (25)^2  =  3  ]

Now  it will work from here.... it can be integrated  as  ∫ [ 1479  to 3 ] ( -1/8) U^1/2 dU     where  1479 is  LOWER limit.......

BUT

if you Must have lower limit smaller  then use  integral property  to do it correctly.

∫ [ a to b ] f(x) dx  = -  ∫ [ b to a ] f(x) dx

{  where  a = lower limit , b = upper  on left side... visa versa on rt. side  }

we have  ∫  [ 1479 to 3 ]   - (1/8) U^1/2  dU  =  - ∫ [ 1479 to 3 ]   (1/8) U^1/2  dU

and that =  by integral property...    + ∫  [ 3 to 1479 ]   (1/8) U^1/2  dU    with  3 the lower limit .....  :)       [[[ multiply integ. by -1 ... flip limits of integ.  ]]]

So  get rid of the - sign on U^1/2  and  U^3/2   in your solution.

In Both Cases  they give  THE SAME ANSWER !    But now  a little nicer , I guess,  since your function is  +  and  smaller limit is at the bottom.

• 2 months ago

u = 2503 - 4r²

so, du = -8r.dr

Hence, dr = -du/8r

Then, ∫ r√(2503 - 4r²) dr => ∫ r√(u)(-du/8r)

Now, the r cancels nicely to leave:

(-1/8) ∫ (√u).du

i.e. (-1/8)[u³/²]/(3/2).du

=> (-1/12)[u³/²]

Now, we need to change the r limits to u limits

so, when r = 25 we have, u = 2503 - 4(25)² => 3

With r = 16 we have, u = 2503 - 4(16)² => 1479

Then, we have (-1/12)[u³/²] with top limit of u = 3 and bottom limit of 1479

Switching the sign, switches the limits thus giving:

(1/12)[u³/²]...with top limit now u = 1479 and bottom limit u = 3

Then, (1/12)[1479³/² - 3³/²] => 4739

:)>

• ted s
Lv 7
2 months ago

your limits are not valid..as well as your integrand...try again ; remember top goes to top