course Mth 272 I understand the process of how distance students are supposed to take a test, in a testing center etc. I've never taken an online class and I was wondering when is the first test and how do we go about receiving it? Thanks for your help. _IßSwy
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15:13:14 5.1.40 (was 5.1.30)(was 5.1.34 int of 1/(4x^2)
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RESPONSE --> to use the power rule, the 1/4 must be factored out of the equation first. 1/4 (x^-2) is the result cuz we take the x^2 from the denomenator then the power rule can be used (1/4)/(-1) * (x^-2+1)= -1/4x+c confidence assessment: 3
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15:13:35 *& An antiderivative of 1 / (4 x^2) is found by first factoring out the 1/4 to get 1/4 ( x^-2). An antiderivative of x^-2 is -1 x^-1. So an antiderivative of 1/4 (x^-2) is 1/4 (-x^-1) = 1 / 4 * (-1/x) = -1/(4x). The general antiderivative is -1 / (4x) + c. STUDENT QUESTION: I know I haven't got the right answer, but here are my steps int 1/4 x^-2 dx 1/4 (x^-1 / -1) + C -1/ 4x + C INSTRUCTOR ANSWER: This appears correct to me, except that you didn't group your denominator (e.g. 1 / (4x) instead of 1 / 4x, which really means 1 / 4 * x = x / 4), but it's pretty clear what you meant. The correct expression should be written -1/ (4x) + C. To verify you should always take the derivative of your result. The derivative of -1/(4x) is -1/4 * derivative of 1/x. The derivative of 1/x = x^-1 is -1 x^-2, so the derivative of your expression is -1/4 * -1 x^-2, which is 1/4 x^-2 = 1 / (4x^2). STUDENT ERROR: The derivative By rewriting the equation to (4x^2)^-1 I could then take the integral using the chain rule. ** it's not clear how you used the Chain Rule here. You can get this result by writing the function as 1/4 x^-2 and use the Power Function Rule (antiderivative of x^n is 1/(n+1) x^(n+1)), but this doesn't involve the Chain Rule, which says that the derivative of f(g(x)) is g'(x) * f'(g(x)). The Chain Rule could be used in reverse (which is the process of substitution, which is coming up very shortly) but would be fairly complicated for this problem and so wouldn't be used. **
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RESPONSE --> got it self critique assessment: 3
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15:53:53 5.1.50 (was 5.1.46)(was 5.1.44 particular soln of f ' (x) = 1/5 * x - 2, f(10)=-10. What is your particular solution?
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RESPONSE --> first integrate the function using power rule so (1/5)/(2)* (x^1+1) - 2x= 1/10x^2 - 2x + c so now plug in the numbers given to meet the requirements -10= 1/10* 10^2 - 2*10 + c -10= -10 + c then add c= 0 so the equation is f(X)= 1/10*x^2 - 2x confidence assessment: 3
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15:54:00 An antiderivative of x is 1/2 x^2 and an antiderivative of -2 is -2x, so the general antiderivative of 1/5 x - 2 is 1/5 * (1/2 x^2) - 2 x + c = x^2 / 10 - 2x + c. The particular solution will be f(x) = x^2 / 10 - 2x + c, for that value of c such that f(10) = -10. So we have -10 = 10^2 / 10 - 2 * 10 + c, or -10 = -10 + c, so c = 0. The particular solution is therefore f(x) = x^2 / 10 - 2 x + 0 or just x^2 / 10 - 2 x. **
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RESPONSE --> got it self critique assessment: 3
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16:21:26 Is the derivative of your particular solution equal to 1/5 * x - 2? Why should it be?
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RESPONSE --> It is equal because now we just work backwards to check if the answer was correct essentially. so the deriv or 1/10x^2 - 2x use power rule to get 2(1/10x) = 1/5x and -2x becomes -2 so = 1/5x - 2 confidence assessment: 3
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16:21:32 The derivative of the particular solution f(x) = x^2 / 10 - 2 x is f ' (x) = (x^2) ' / 10 - 2 ( x ) '. Since (x^2) ' = 2 x and (x) ' = 1 we get f ' (x) = 2 x / 10 - 2 * 1 = x / 5 - 2, which is 1/5 * x - 2. The derivative needs to be equal to this expression because the original problem was to find f(x) such that f ' (x) = 1/5 * x - 2. *&*&
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RESPONSE --> got it self critique assessment: 6
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17:01:15 5.1.60 (was 5.1.56)(was 5.1.54 f''(x)=x^2, f(0)=3, f'(0)=6. What is your particular solution?
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RESPONSE --> two antiderivatives must be taken first x^2 = x^3/3 +c just power rule f(0)= 0^3/3 +c = 6 so c= 6 which makes f '(x)= x^3/3 + 6 then take the anti deriv of that power rule again (x^4)/(4)/3 + 6x + c x^4/ 12 +6x + c f(0)= 0^4/12 +6(0) +c= 3 this time so c= 3 and f(x)= x^4/12 +6x + 3 confidence assessment: 3
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17:01:21 Since you have the formula for f ''(x), which is the second derivative of f(x), you need to take two successive antiderivatives to get the formula for f(x). The general antiderivative of f''(x) = x^2 is f'(x) = x^3/3 + C. If f'(0) = 6 then 0^3/3 + C = 6 so C = 6. This gives you the particular solution f'(x) = x^3 / 3 + 6. The general antiderivative of f'(x) = x^3 / 3 + 6 is f(x) = (x^4 / 4) / 3 + 6x + C = x^4 / 12 + 6 x + C. If f(0) = 3 then 0^4/12 + 6*0 + C = 3 and therefore C = 3. Thus f(x) = x^4 / 12 + 6x + 3. **
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RESPONSE --> got it self critique assessment: 3
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17:12:10 Is the second derivative of your particular solution equal to x^2? Why should it be?
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RESPONSE --> working backwards again will find the correct answer x^4/12+6x+3 f '(x)= x^3/3 + 6 f ``(x)= (3x^2)/3 so it= x^2 confidence assessment: 3
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17:12:18 *& The particular solution is f(x) = x^4 / 12 + 6 x + 3. The derivative of this expression is f ' (x) = (4 x^3) / 12 + 6 = x^3 / 3 + 6. The derivative of this expression is f ''(x) = (3 x^2) / 3 = x^2. Thus f '' ( x ) matches the original condition of the problem, as it must.
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RESPONSE --> got it self critique assessment: 3
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18:29:02 5.1.76 (was 5.1.70 dP/dt = 500 t^1.06, current P=50K, P in 10 yrs
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RESPONSE --> first find antideriv of dP/dt power rule = 500t^2.06/ 2.06 + c so when t=0 P= 50,000 50,000= 500*0^2.06/2.06+c c= 50,000 P= 500(10)^2.06/2.06 + 50000 plug in 10 for t P= 27868+ 50000= 77,868 confidence assessment: 3
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18:29:32 You are given dP/dt. P is an antiderivative of dP/dt. To find P you have to integrate dP/dt. dP/dt = 500t^1.06 means the P is an antiderivative of 500 t^1.06. The general antiderivative is P = 500t^2.06/2.06 + c Knowing that P = 50,000 when t = 0 we write 50,000 = 500 * 0^2.06 / 2.06 + c so that c = 50,000. Now our population function is P = 500 t^2.06 / 2.06 + 50,000. So if t = 10 we get P = 500 * 10^2.06 / 2.06 + 50,000 = 27,900 + 50,000= 77,900. ** DER
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RESPONSE --> got it self critique assessment: 3
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20:11:15 5.2.12 (was 5.2.10 integral of `sqrt(3-x^3) * 3x^2
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RESPONSE --> first rewrite the eqn as (3-x^3)^1/2 (3x^2)dx u= 3-x^3 du=-3x^2 since there is no -3x^2 a -1 will haveto be multiplied { u^1/2 (-1du) power rule is applied -1{ (u^3/2)/ (3/2) + c -2/3u^3/2 +c then substitute -2/3(3-x^3)^3/2 + c confidence assessment: 3
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20:12:06 You want to integrate `sqrt(3-x^3) * 3 x^2 with respect to x. If u = 3-x^3 then u' = -3x^2. So `sqrt(3-x^3) * 3x^2 can be written as -`sqrt(u) du/dx, or -u^(1/2) du/dx. The General Power Rule tells you that this integral of -u^(1/2) du/dx with respect to x is the same as the integral of -u^(1/2) with respect to u. The integral of u^n with respect to u is 1/(n+1) u^(n+1). We translate this back to the x variable and note that n = 1/2, getting -1 / (1/2+1) * (3 - x^3)^(1/2 + 1) = -2/3 (3 - x^3)^(3/2). The general antiderivative is -2/3 (3 - x^3)^(3/2) + c. ** DER COMMON ERROR: The solution is 2/3 (3 - x^3)^(3/2) + c. The Chain Rule tell syou that the derivative of 2/3 (3 - x^3)^(3/2), which is a composite of g(x) = 3 - x^3 with f(z) = 2/3 z^(3/2), is g'(x) * f'(g(x)) = -3x^3 * `sqrt (3 - x^3). You missed the - sign. **
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RESPONSE --> got it self critique assessment: 3
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21:15:17 5.2. 18 (was 5.2.16 integral of x^2/(x^3-1)^2
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RESPONSE --> u= x^3-1 du=3x^2 since there is no du 1/3 is multiplied { (1/3du)/ (u^-2) power rule 1/3(-u^-1) + c 1/3 (x^3-1) + c confidence assessment: 3
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21:15:25 Let u = x^3 - 1, so that du/dx = 3 x^2 and x^2 = 1/3 du/dx. In terms of u we therefore have the integral of 1/3 u^-2 du/dx. By the General Power Rule our antiderivative is 1/3 (-u^-1) + c, or -1/3 (x^3 - 1)^-1 + c = -1 / (3 ( x^3 - 1) ) + c. This can also be written as 1 / (3 ( 1 - x^3) ) + c. ** DER
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RESPONSE --> got it. self critique assessment: 3
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00:55:41 5.2.26 (was 5.2.24 integral of x^2/`sqrt(1-x^3)
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RESPONSE --> for this one u= 1-x^3 and du=-3x^2 so -1/3 must be multiplied. { -1/3du / u^1/2 then the power rule -2/3 u^1/2 +c so it = -2/3 sqrt(1-x^3) +c confidence assessment: 3
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00:55:48 *& If we let u = 1 - x^3 we get du/dx = -3 x^2 so that x^2 = -1/3 du/dx. So the exression x^2 / sqrt(1-x^3) is -1/3 / sqrt(u) * du/dx By the general power rule an antiderivative of 1/sqrt(u) du/dx = u^(-1/2) du/dx will be (-1 / (-1/2) ) * u^(1/2) = 2 sqrt(u). So the general antiderivative of x^2 / (sqrt(1-x^3)) is -1/3 ( 2 sqrt(u) ) + c = -2/3 sqrt(1-x^3) + c. *&*& DER
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RESPONSE --> understood self critique assessment: 3
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