Good Work
only thing:
1) ln(ln(e^4)) = ln(4)
Back again! This time I have five problems that I have completed that, if anyone would be so kind, I would like to have looked over and corrected if they are wrong. Thank you very very much in advance!
Problem 1. dx
My Solution
u = ln(x), du = dx, dx = x du
du = du = ln(u) + C = = ln[ln( )] - ln[ln(e)] = ln[ln( )] - 0 = ln[ln( )]
Problem 2. xsin4x dx
My Solution
u = x, du = dx, dv = sin4x, v = cos4x
dx = dx = = =
Problem 3. dx
My Solution
u = 4 + 9 , du = 18x dx, dx = \frac{1}{18}x du
du = du = du = =
Problem 4. dx
My Solution
dx = dx = dx = =
Problem 5. Calculate the length along the curve x = , y = ,
My Solution (Not so sure about this one.)
= , =
L = dt = dt = dt = dt = cos(t)sin(t) dt
u = cos(t), du = -sin(t) dt, dt = du
du = u du = = = -[0 - ] =
Once again, thank you very much in advance.
In the first problem notice that the limits of integration can also go through the u-substitution process, and you don't need to return to terms of x. Since e and e^4 are in terms of x, and u = ln(x), then:
It isn't always easy or better to convert your limits of integration to the terms of your substitution, but sometimes it does simplify things.
Brian
(Who is only about a semester past this point, so I hope what I've said is all correct!)
It is far simpler just to use substitution. Let u= ln(x) so that du= dx/x. When x= e, u= ln(e)= 1. When , [tex]u= ln(e^4)= 4. The integral becomes
Problem 2. xsin4x dx
My Solution
u = x, du = dx, dv = sin4x, v = cos4x
dx = dx = = =
Problem 3. dx
My Solution
u = 4 + 9 , du = 18x dx, dx = \frac{1}{18}x du
du = du = du = =
Problem 4. dx
My Solution
dx = dx = dx = =
Problem 5. Calculate the length along the curve x = , y = ,
My Solution (Not so sure about this one.)
= , =
L = dt = dt = dt = dt = cos(t)sin(t) dt
u = cos(t), du = -sin(t) dt, dt = du
du = u du = = = -[0 - ] =
Once again, thank you very much in advance.