How Do You Solve These Integral Problems Using U-Substitution?

  • Thread starter Thread starter EasyStyle4747
  • Start date Start date
  • Tags Tags
    Integral
AI Thread Summary
The discussion centers on solving integral problems using u-substitution. For the integral of the square root of (e^t - 3), the correct substitution involves letting e^t = u, which simplifies the expression. In the second integral, x^2[e^(x^3)]dx, the appropriate substitution is u = x^3, leading to an easier integration process. Participants clarify the importance of identifying the correct substitutions and adjusting for the derivatives accordingly. Overall, understanding the structure of the integrals is key to applying u-substitution effectively.
EasyStyle4747
Messages
24
Reaction score
0
Ok, ill try to explain what i did:

1.
integral of [square root of( e^t-3)dt]

Sorry i didn't know how to do the square roots symbol and the integral symbol. Anyways, I tried to set e^t-3 as u and then got dt=du/e^t. Then I plugged in dt and couldn't go on after that.

2. integral of x^2[e^(x^3)]dx

so uh... u substitution? but then what?

Ok basically, i think i don't know what your supposed do when u need to do a u-sub with e^something. Your help would be most appreciated.
 
Physics news on Phys.org
1. I don't have any problem with the square root sign but is that e^(t-3) or
(e^t)-3?
If it is e^(t-3) then sqrt(e^(t-3))= (e^(t-3))^(1/2)= e^((t-3)/2). Let u= (t-3)/2 so
du= (1/2)dt or dt= 2du. The integral becomes 2 integral e^u du.

2. Yes, a substitution- seeing that "complicated" x^3 in the exponent and x^2 multiplying, you should immediately think of u= x^3 (NOT u= e^something- the e is not the problem!). Then du= 3 x^2 dx or (1/3)du= x^2 dx so the integral becomes
(1/3) integral e^u du.
 
Well, thinking of u = e^{x^3} wouldn't hurt in this case regardless :smile:
 
thnx, i see how the second one works now.

For the first one, its actually square root of (e^t)-3. Sorry i didnt make that clear.
 
Then make the substitution
e^{t}=u
,then the substitution
u=v^{2}+3

Daniel.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top