## Trending News

# [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

MY ANSWERS: PART 1

upper limit: 1479 CORRECT

=-1/8*sqrtudu

lower limit: 3 CORRECT

upper limit: 1479 CORRECT

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

lower limit: 3 CORRECT

### 3 Answers

- MathguyLv 52 months agoFavorite Answer
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.

- Wayne DeguManLv 72 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 sLv 72 months ago
your limits are not valid..as well as your integrand...try again ; remember top goes to top