course MTH 272 04/08/10, 2200 ᪻ڑeassignment #007
......!!!!!!!!...................................
21:05:08 5.2.36 (was 5.2.34 integral of x^2 (1-x^3)^2 by formal substitution. What is the integral of the given function?
......!!!!!!!!...................................
RESPONSE --> 3x^2 (1-x^3)^2 u=1-x^3 du/dx=-3x^2 du=(-3x^2)dx -1(-3x^2)(1-x^3)^2 -1(u^2)du -1(u^3 / 3) + C -(u^3 / 3) + C -(1-x^3)^3 / 3 + C confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.................................................
......!!!!!!!!...................................
21:05:48 If we let u = 1 - x^3 then u ' = - 3 x^2 and the x^2 in our integrand is - u ' / 3. (1-x^3)^2 is u^2, so the integrand is - u ' / 3 * u^2 = -1/3 u^3 u ' . So the integral is you have -1/3 u^2 du. The integral of u^2 u ' is 1/3 u^3. Thus the integral of -1/3 u^2 u ' is -1/3 of 1/3 u^3, or -1/9 u^3. So your integral should be -1/9 u^3 = -1/9 (1-x^3)^3. The general antiderivative is -1/9 ( 1 - x^3)^3 + c. **
......!!!!!!!!...................................
RESPONSE --> ------------------------------------------------ Self-critique rating #$&*ent: 2
.................................................
......!!!!!!!!...................................
21:06:53 What is the derivative of your result?
......!!!!!!!!...................................
RESPONSE --> -(1/3) 3(1-x^3)^2 (-3x^2) (1-x^3)^2 (3x^2) confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.................................................
......!!!!!!!!...................................
21:07:03 The derivative of -1/9 (1-x^3)^3, using the Chain Rule, is the product of -1/9, 3(1-x^3)^2, and the derivative - 3x^2 of the 'inner function' (1-x^3). Multiplying these factors we get -1/9 (-3x^2) * 3(1-x^3)^2 = x^2 (1-x^3)^2. **
......!!!!!!!!...................................
RESPONSE --> ------------------------------------------------ Self-critique rating #$&*ent: 3
.................................................
......!!!!!!!!...................................
21:14:40 5.2.54 (was 5.2.52 find x | dx/dp = -400/(.02p-1)^3, x=10000 when p=100
......!!!!!!!!...................................
RESPONSE --> (-400)(.02p-1)^-3 dp (-20,000) (-400)(-1/20,000)(.02p-1)^-3 dp (-20,000) ((.02p-1)^-2) / -2 + C (10,000/ (.02p -1)^2 ) + C=x (10,000 / (.02(100) -1)^2) + C= 10,000 10,000 + C =10,000 C=0 x=f(p)= 10,000 (1/(.02p-1)^2) confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.................................................
......!!!!!!!!...................................
21:14:56 The equation rearranges to dx = -400 * dp * (.02 p - 1)^-3. An antiderivative of the left-hand side could be just x. An antiderivative of dp * (.02 p - 1)^-3 is found using u = .02 p - 1, so du = .02 dp and dp = du / .02 = 50 du. Thus the right-hand side becomes -400 * 50 u^-3 du = -20000 u^-3 du, with antiderivative 20000 / 2 * u^-2 + c = 10,000 u^-2 + c. So we have x = 10,000 * u^-2 + c = 10,000 * (.02 p - 1)^-2 + c. Note that dx / dp is therefore 10,000 * -2 * .02 (p-1)^-3 = -400 (.02 p - 1)^-3, consistent with the original equation. Since x = 10,000 * (.02 p - 1)^-2 + c and x = 10,000 when p = 100 we have 10,000 = 10,000 * (.02 * 100 - 1)^2 + c 10,000 = 10,000 / 1^2 + c 10,000 = 10,000 + c so c = 0. The solution is therefore x = 10,000 * (.02 p - 1)^-2 + 0 or just x = 10,000 * (.02 p - 1)^-2. **
......!!!!!!!!...................................
RESPONSE --> ------------------------------------------------ Self-critique rating #$&*ent: 3
.................................................
......!!!!!!!!...................................
21:16:48 5.3.04 (was 5.3.04 integral of e^(-.25 x) by Exponential Rule
......!!!!!!!!...................................
RESPONSE --> integral of e^(-.25 x) u=-.25x du/dx=-.25=-25/100 -100/25 integral e^u(du/dx) dx -4e^-.25x + C confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.................................................
......!!!!!!!!...................................
21:17:08 Simple substitution u = -.25 x gives us du/dx = -.25 so that du = -.25 dx and dx = du / (-.25) = -4 du. Our original integrand e^(-.25 x) dx therefore becomes e^u * (-4 du) = -4 e^u du. Our general antiderivative will be -4 e^u + c, meaning -4 e^(-.25 x) + c. The derivative of -4 e^(-.25 x) + c is -4 ( -.25 e^-.25 x) = e^-.25 x, verifying our result. The General Exponential Rule is equivalent to this: u = -.25 x so du/dx = -.25. Thus the integral is of e^u / (du/dx) = e^(-.25 x) / (-1/4) = -4 e^(-.25 x). *&*&
......!!!!!!!!...................................
RESPONSE --> ------------------------------------------------ Self-critique rating #$&*ent: 3
.................................................
......!!!!!!!!...................................
5.3.10 (was 5.3.10 integral of 3(x-4)e^(x^2-8x) by Exponential Rule
......!!!!!!!!...................................
RESPONSE --> integral of 3(x-4)e^(x^2-8x) u=x^2 - 8x du/dx= 2x - 8 (3/2) integral (2/3)(3)(x-4) e^u (3/2) e^(x^2 - 8x) + C confidence rating #$&*2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.................................................
......!!!!!!!!...................................
21:17:25 if u=x^2 - 8x then du / dx = 2x - 8 x-4 = 1/2(2x-8) so 3(x-4) = 3/2 du/dx. Thus 3(x-4)e^(x^2-8x) is 3/2 e^u du/dx. The general antiderivative of e^u du/dx is e^u + c, so the integral of 3/2 e^u du/dx is 3/2 e^u. Substituting x^2 - 8x for u we have 3/2 e^(x^2-8x) + c. **
......!!!!!!!!...................................
RESPONSE --> ------------------------------------------------ Self-critique rating #$&*ent: 3
.................................................
......!!!!!!!!...................................
21:19:08 problem 5.3.16 integral of 1/(6x-5) by Log Rule
......!!!!!!!!...................................
RESPONSE --> integral of 1/(6x-5) u=6x-5 du/dx=6 (1/6) integral (6/(6x-5)) dx (1/6) ln |6x-5| + C confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.................................................
......!!!!!!!!...................................
21:21:00 du/dx is the derivative of 6x-5, so du/dx = 6 If we let u = 6x - 5 then du = 6 dx so dx = 1/6 du and the integral becomes that of ln(u) * du/6 = 1/6 ln(u) du The integral of 1/6 ln(u) du is 1/6 * 1 / u + c Substituting u = 6x - 5 we get the final result int(1 / (6x - 5) = 1/6 * 1 / (6x-5) = 1 / [ 6(6x-5) ]. **
......!!!!!!!!...................................
RESPONSE --> I cannot understand where the value of 1/2 comes from that is used in the explanation. If u=6x - 5, then du/dx = 6. After multiplying the numerator of 1 by 6, does the entire integral not get divided by 6, or multiplied by 1/6?
.................................................
......!!!!!!!!...................................
21:21:51 problem 5.3.22 (was 5.3.20) integral of x/(x^2+4) by Log Rule
......!!!!!!!!...................................
RESPONSE --> integral of x/(x^2 + 4) u=x^2 + 4 du/dx=2x (1/2)int. (2)(X) / (x^2 + 4) 1/2 ln |x^2 + 4| +C confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.................................................
......!!!!!!!!...................................
21:21:57 If we let u = x^2 + 4 we get du/dx = 2x so that the x in the numerator is 1/2 du/dx. The integral of x / (x^2 + 4) is the integral of 1/2 * ( 2x / (x^2+4) ) = 1/2 (1/u du/dx). The general antiderivative is therefore 1/2 ln(u) + c = 1/2 ln |x^2+4| + c. **
......!!!!!!!!...................................
RESPONSE --> ------------------------------------------------ Self-critique rating #$&*ent: 3
.................................................
......!!!!!!!!...................................
21:22:04 What is the derivative of your result?
......!!!!!!!!...................................
RESPONSE --> 1/2 ( 1/ (X^2 + 4))(2x) (2x/ 2(x^2 + 4)) x/(x^2 + 4) confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.................................................
......!!!!!!!!...................................
21:22:11 The derivative of ln(x^2+4) * (1/2) is 1/2 * 2x * 1 / (x^2 + 4) or x / (x^2 + 4). This confirms that ln(x^2+4) * (1/2) is a solution to the equation. The general antiderivative is of course ln(x^2+4) * (1/2) + c. **
......!!!!!!!!...................................
RESPONSE --> ------------------------------------------------ Self-critique rating #$&*ent: 3
.................................................
......!!!!!!!!...................................
21:22:17 5.3.28 (was 5.3.24) (was 5.3.24 integral of e^x/(1+e^x) by Log Rule
......!!!!!!!!...................................
RESPONSE --> integral of e^x / (1 + e^x) u= 1 + e^x du/dx=e^x ln |1 + e^x| + C confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.................................................
......!!!!!!!!...................................
21:22:22 let u = 1 + e^x. Then du/dx = e^x. We are therefore integrating 1 / (1 + e^x) * e^x, which is 1/u du/dx. The antiderivative is ln |u| + c = ln | 1 + e^x | + c. **
......!!!!!!!!...................................
RESPONSE --> ------------------------------------------------ Self-critique rating #$&*ent: 3
.................................................
......!!!!!!!!...................................
21:22:32 5.3.46 (was 5.3.34) (was 5.3.34 integral of (6x + e^x) `sqrt( 3x^2 + e^x)
......!!!!!!!!...................................
RESPONSE --> integral of (6x + e^x) ( 3x^2 + e^x)^(1/2) u=3x^2 + e^x du/dx=6x + e^x u^(1/2) du/dx u^(3/2) / (3/2) (2/3)u^(3/2) (2/3) (3x^2 + e^x)^(3/2) confidence rating #$&* 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.................................................
......!!!!!!!!...................................
21:22:37 Here are two detailed solutions: (6x + e^x) `sqrt( 3x^2 + e^x) = `sqrt(u) * du/dx = u^(1/2) du/dx. The antiderivative is thus 2/3 u^(3/2) = 2/3 (3x^2 + e^x)^(3/2). Alternatively If u = 3x^2 + e^x then du = 6x + e^x and we have the integral of `sqrt(u) du, which is just 2/3 u^(3/2) + c = 2/3 (3x^2 + e^x)^(3/2) + c. **
......!!!!!!!!...................................
RESPONSE --> At first, I instinctively wanted to use the log rule, but since there was no division, realized that the power rule would be much simpler. ------------------------------------------------ Self-critique rating #$&*ent: 3
.................................................
......!!!!!!!!...................................
21:22:42 5.3.58 (was 5.3.54) (was 5.3.52 dP/dt = -125 e^(-t/20), t=0, P=2500 and interpretation. Give your complete solution.
......!!!!!!!!...................................
RESPONSE --> dP/dt= -125 e^(-t/20), t=0 P=2500 u=-t/20 du=-1/20 2500 int. (1/2500) (-125) e^(-t/20) 2500 e^(-t/20) + C=P 2500 e^(-0/20) + C = 2500 e^0 + C =1 1 + C= 1 C=0 P(t)=2500e^(-t/20) Pop. after 15 days: P(15)=2500e^(-15/20) =1180 Zero pop.: 0=2500e^(-t/20) 0=e^(-t/20) ln 0= ln e (-t/20) undefined = -t/20 The answer of t being undefined does not make sense as to when the population would be 0. confidence rating #$&* 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.................................................
......!!!!!!!!...................................
21:22:48 If dP/dt = -125 e^(-t/20) then dp = -125 e^(-t/20) dt. Integrating both sides we get p = 2500 e^(-t/20) + c ( to integrate the right-hand side start with u = -t / 20, etc. If p = 2500 when t = 0 we have 2500 = 2500 e^(-0/20) + c so 2500 = 2500 + c and c = 0. The final solution is thus p = 2500 e^(-t/20) After 15 days the population is p(15) = 2500 e^(-15/20) = 1000, give or take a couple hundred (you can evaluate the expression). All the trout are considered dead when the population is below 1/2. So you need to solve 1/2 = 2500 e^(-t/20) for t. Dividing both sides of this equation by 2500 then taking the natural log of both sides you get -t/20 = ln( 1/2500 ) so t = -20 * ln (1/2500) = -11 or -12 or so. Thus t is about 200 days, give or take a little. Alternative reasoning of the particular solution: If u = -t/20 then e^u du/dt = e^(-t/20) * -1/20. -125 e^(-t/20) is 2500 * ( -1/20 e^(-t/20) ) = 2500 e^u du/dx. The integral is 2500 e^u + c = 2500 e^(-t/20) + c. If t = 0, P=2500 then 2500 = 2500 e^0 + c = 2500 + c, so c = 0. Thus the particular solution is P = 2500 e^(-t/20). Alternative solution for the time when all trout are dead: 2500 e^(-t/20) < .5 means e^(-t/20) < .0002 so -t/20 < ln(.0002) so -t < ln(.0002) * 20 so -t < -170.34 and t > 170.34. The probability is that all trout are dead by day 171. STUDENT QUESTION: I couldn't figure out the time for all the trout to die because the ln 0 is undefined ** When the population falls below 1/2 of a fish it rounds off to 0 and you assume that all the trout are dead. You can think of this in terms of probability. The function doesn't really tell us the precise number but the probable number. When the probability is againt that last fish being alive we figure that it's most likely dead. **
......!!!!!!!!...................................
RESPONSE --> Similarly to the question of the student listed in the explanation, I was unaware that when the population falls below half, you round it to zero. ------------------------------------------------ Self-critique rating #$&*ent: 2
.................................................