Integration by substitution

  • #1

Homework Statement



evaluate:

higher limit of 36
lower limit of 0 (36+3x)^1/2 dx

Homework Equations



i thought of using subsititution?

The Attempt at a Solution



g(x)=36+3x
g'(x)=3

when x=0, u=36+3(0)=36
when x=36, u=36+3(36)=144

from lower limit of 36 to higher limit of 144

3(u)^1/2 du= 3(2/3)u^3/2 + C

substitute 36+3x back into u, i get: 2(36+3x)^3/2 + C

=[2(36+3(144))^3/2) + C ] - [ 2(36+3(36))^3/2] + C

= 2(468)^3/2 - 3456 + C

is this correct??
 

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Answers and Replies

  • #2
Office_Shredder
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Your final answer should just be a number, there should be no constant of integration. When you have -[ 2(36+3(36))^3/2] +C it should really be -[ 2(36+3(36))^3/2 + C] and the constants should cancel out
 
  • #3
so it would just be: 2(468)^3/2 - 3456 ?
 
  • #4
Office_Shredder
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Checking it, you made a couple mistakes. First with your substitution. The derivative of [tex] 2 (36+3x)^{3/2}[/tex] is [tex] 9(36+3x)^{1/2}[/tex], which isn't right. Check your integration by substitution again and be careful with plugging in du for dx.

Also, you turned your answer back into x's, but then plugged in the upper and lower bounds for u.
 
  • #5
um...i don't get it..i didn't find i wrote 9(36+3x)^1/2..so what should it be?
and for the x=144,x=36, i do substitute those in right?...
 
  • #6
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When you make a substitution to a new variable of integration, you can either a) fix your limits of integration by putting them through the function or b) leave them as "x=whatever" and then when you are done integrating, put back your original variable and evaluate. But not both, which is what you seem to have done here.

Let me see if I can make that clearer. Consider [itex] \intop_1^2 (x^2-2x+1) dx [/itex].
Now you can just evaluate this:
[tex] \frac{x^3}{3}-x^2+x \vert^1_2 = \frac{1}{3} [/tex].

But let's suppose you decide you want to do a substitution:
[itex] u=x-1, du=dx [/itex]
[tex] \intop_1^2 (x^2-2x+1) dx = \intop^?_? u^2 du [/tex].

Now you have two choices, you can move the limits from x to u by plugging into your substitution:
[tex]\intop^1_0 u^2 du = \frac{u^3}{3} \vert^1_0 = \frac{1}{3} [/tex]
or you can write:
[tex]\intop^{x=2}_{x=1} u^2 du = \frac{u^3}{3} \vert^{x=2}_{x=1} = \left[\frac{x^3}{3}-x^2+x-1 \vert^2_1 \right]= \frac{1}{3}[/tex]

Sometimes it's more convenient to move your limits of integration to the new variable and sometimes it's not. But what you did was convert your limits of integration to the new variable u. Then when you were done integrating, you converted the expression to be evaluated back to x, but plugged in the "u" bounds.
 

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