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course Mth 174
9/24 1 am
Section 7.21) 7.2.6 Find the integral Int(xe^(-2x) dx).
u = x, du = dx, v = e^(-2x), v = -1/2e^(-2x)
= x*-1/2e^(-2x) int(1*-1/2e^(-2x) dx)
= x/2e^(-2x) 1/4e^(-2x) + c
2) 7.2.10 Find the integral Int(q^4 ln(4q) dq).
u =ln(4q), du = 1/q dq, v = q^4, v = 1/5q^5
= ln(4q)*(1/5)q^5 int(1/q*(1/5)q^5 dq)
= (q^5/5)ln(4q) (q^6/30)ln(x)
3) 7.2.12 Find the integral Int(sin^2(x) dx).
u =sinx, du = cosx, v = sinx, v= -cosx
Int(sin^2(x) dx) = [sinx*-cosx] Int(cosx*-cosx)dx
=-sinxcosx - Int(-cos^2x dx)
= -sinxcosx + Int(1-sin^2x dx)
= -sinxcosx + Int(1)- Int(sin^2x dx)
!!!Adding Int(sin^2x dx) to both sides!!!!
2 Int(sin^2(x) dx) = -sinxcosx + x
!!!Divide Eq by 2!!!!!
Int(sin^2(x) dx) = (-sinxcosx)/2 + x/2 + c
4) 7.2.16 Find the integral Int((t+2)sqrt(2+3t) dt).
u = 2 + 3t, du = 3dt, t = (u-2)/3, dt = du/3
1/3Int(((u-2)/3 + 2)`sqrt(u) du) = 1/3Int(((u/3) + 4/3)u^(1/2) du)
= 1/3 Int((1/3)u^(3/2)+(4/3)u^(1/2) du)
= 1/3[(2/15)u^(5/2)+(8/9)u^(3/2)
= (2/45)u^(5/2) + (8/27)u^(3/2) + c
=(2/45)(2+3t)^(5/2) + (8/27)(2+3t)^(3/2) + c
5) 7.2.18 Find the integral Int(z^2/e^z dz).
you can get this with two applications of integration by parts, letting u = z^2 in the first one, which reduces the power of z in the integrand by 1; the second application should then be obvious
6) 7.2.20 Find the integral Int(3y/sqrt(4+y) dy).
u = 4+y, du = dy, y = u-4, dy = du
= Int(3(u-4)/sqrt(u) du)
= Int(3u-12)u^(-1/2) du)
= Int(3u^(1/2)-12u^(-1/2) du)
= 2u^(3/2)-24u(1/2) + c
should be = 2u^(3/2)+24u(1/2) + c
= 2(4+y)^(3/2)-24(4+y)(1/2) + c
= (8+2y)^(3/2) + (-96-24y)^(1/2) + c
7) 7.2.24 Find the integral Int(arctan 4z dz).
u = arctan 4z , du = (4/(1+4z)^2, v = 1, v = z
du = (4/(1+(4z)^2)) dz
= z arctan 4z Int((4/(1+4z)^2*z)
= z arctan 4z Int((4/(1+4z)^2*z)
= Int(4z/(16z^2+4z+4z+1))
?????= z arctan 4z - Int(1/(16z^2+4z+1)) = z arctan 4z - ln(16z^2+4z+1)????
8) 7.2.27 Find the integral Int(x^5 cos(x^3) dx).
???Had some question as to my solution being correct????
u = x^5, du = 5x^4, v = cos(x^3), v =sin(x^3)/3x^2
= (x^5sin(x^3))/(3x^2) - Int((5x^4sin(x^3))/3x^2 dx)
= (1/3)x^3sin(x^3) (5/3) Int(x^2sin(x^3))
!!!!Next sub: u = x^3, du = 3x^2 dx!!!!!
= (1/3)x^3sin(x^3) (5/3) Int(1/3sin(u))
= (1/3)x^3sin(x^3) (5/9) Int(sin(u))
= (1/3)x^3sin(x^3) (5/9)-cos(u)
=(1/3)x^3sin(x^3) + (5/9)cos(x^3) + c
alternatively let u = x^3 so du = 3 x^2 dx. Then x^5 dx = 1/3 u du, and you integrate 1/3 u cos(u) du.
9) 7.2.30 Evaluate the definite integral Int(ln t dt, 2, 4).
Int(ln t dt) = t ln(t) t
For t = 2, = approx. -.6137
For t = 4, = approx. 1.5452
Difference is approx. 2.1589
10) 7.2.36 Evaluate the definite integral Int(arccos z dz-1, 1).
u = arccos, du = -1/`sqrt(1-x^2), v = 1, v = x
z arcos z Int(-x/`sqrt(1-x^2))
= z arcos z Int([-1/`sqrt(1-x^2)]*x)
!!!u = z^2, du = 2z dz!!!!!
= z arcos z Int((1/2)-1/`sqrt(1-u))
int(1/sqrt(1-u) du ) = -sqrt(1-u)
= z arcos z (1/2)Int(-1/`sqrt(1-x^2))
= z arcos z (1/2)arccos z
At z = -1
=approx. -4.7124
At z = 1
= 0
So that 0 (-4.7124) = 4.7124
the final answer is pi; no need for numerical approximation
11) 7.2.40 In number 3) you found Int(sin^2(x) dx) using integration by parts most likely, if not then do so. Redo this integral using the identity sin^2(x) = (1 - cos(2x))/2. Explain any differences in the form of the solution found by the two methods.
????I cannot figure out sub needed to solve this equation??????
an antiderivative of cos(2x) is 1/2 sin(2x), so very little substitution is required
12) 7.2.42 Find the integral Int(e^x(sin x) dx) (Do integration by parts twice and go from there).
???Do not understand what is needed for this problem
let u = e^x, v = sin(x); the integral reduces to e^x sin x - int(e^x cos x).
Do the same with int(e^x) cos(x).
So what does int(e^x sin(x) ) equal? You get an equation with the expression int(e^x sin(x)) in two places. Put all such expressions, everything else on the other side of the equation.
What do you find?
13) 7.2.50 Given f(0) = 6, f(1) = 5, f '(1)=2. Find Int(x * f '(x), x, 0, 1).
u = x, du =dx, v = f(x), v = f(x)
I think this is the right approach but cant seem to get there.
from what you say an antiderivative is u v - int(v du) = x f(x) - int(f(x) dx); doesn't seem to get us anywhere
OK, there's an error in the statement of the question.
Should be int(x * f ''(x), x, 0, 1).
This becomes x f ' (x) - int(f '(x) dx). The integral of f ' is just f (f is clearly an antiderivative of f ').
Thus we have antiderivative x f '(x) - f (x).
Between x = 0 and x = 1 the antiderivative changes by
(1 * f ' (1) - f (1)) - (0 * f ' (0) - f (0) ) = 1 * 2 - 5 - (0 * f ' (0) - 6) - etc.
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Good overall, but check my notes for answers to some of your questions, and a couple of possible errors.
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