#$&* course Mth 272 006. `query 6*********************************************
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Given Solution: `a*& 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): For this one, as I see it worked out I understand it, but while doing it by myself I wasn’t sure if I was supposed to incorporate the chain rule or how I was even supposed to multiple each integral. ------------------------------------------------ Self-critique Rating: 3
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Wow, I feel very stupid. When doing this problem since it gave a solution for f(x) I solved it like a standard algebra problem. ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: `qIs the derivative of your particular solution equal to 1/5 * x - 2? Why should it be? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: No confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. *&*& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didn’t solve the previous question right therefore I was thrown off on the right of the questions that were based off of the previous one. ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: `q5.1.7 (previously 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2x^2+C confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a Since you have the formula for f ''(x), which is the second derivative of f(x), you need to take two successive antiderivatives to get the formula for f(x). The general antiderivative of f''(x) = x^2 is f'(x) = x^3/3 + C. If f'(0) = 6 then 0^3/3 + C = 6 so C = 6. This gives you the particular solution f'(x) = x^3 / 3 + 6. The general antiderivative of f'(x) = x^3 / 3 + 6 is f(x) = (x^4 / 4) / 3 + 6x + C = x^4 / 12 + 6 x + C. If f(0) = 3 then 0^4/12 + 6*0 + C = 3 and therefore C = 3. Thus f(x) = x^4 / 12 + 6x + 3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I’m very confused by this one. I am not sure which figures I’m supposed to be plugging in the function to solve for f(x). ------------------------------------------------ Self-critique Rating: 3
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Given Solution: `a*& The particular solution is f(x) = x^4 / 12 + 6 x + 3. The derivative of this expression is f ' (x) = (4 x^3) / 12 + 6 = x^3 / 3 + 6. The derivative of this expression is f ''(x) = (3 x^2) / 3 = x^2. Thus f '' ( x ) matches the original condition of the problem, as it must. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): My previous answer was wrong therefore this answer was incorrect. ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: `q5.1.10 (previously 5.1.76 (was 5.1.70) ) dP/dt = 500 t^1.06, current P=50K, P in 10 yrs YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): When I tried to solve a problem like this above I get it incorrect because I substituted the variable for the values given and don’t understand why I’m not supposed to perform this action. ------------------------------------------------ Self-critique Rating: 3
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Given Solution: `a 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. ** If you take the derivative of this expression -2/3 ( 3 - x^3)^(3/2) + c, you should get the original expression `sqrt(3-x^3) * 3 x^2. You should do this, and if you don't get the original expression you should self-critique this step by showing the instructor your steps, with your best explanation. 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. ** STUDENT QUESTION what happened to the dervative of 3x^2 INSTRUCTOR RESPONSE You're doing an antiderivative, not a derivative (naturally you know this). The derivative of 3 x^2 isn't relevant to finding the antiderivative. If the 3 x^2 wasn't there, you wouldn't be able to integrate the function: • The derivative of 2/3 * (3 - x^2)^(3/2) isn't sqrt( 3 - x^3). • The derivative of 2/3 * (3 - x^3)^(3/2) is (3 - x^3) * (-3 x^2); the -3 x^2 comes from the chain rule. • So without that 3 x^2 the antiderivative wouldn't be anything like 2/3 ( 3 - x^3)^(3/2). This is the tricky thing about integration at this point in the course. It can be very tricky to 'reverse' the chain rule, but it's gotta be done in one way or the other. These tricks are basically what Chapter 5 is about. Also note the following: • Once you think you have the antiderivative, it's very important to take the derivative and be sure it works out. If the derivative of the antiderivative isn't equal to your original expression, then you haven't integrated correctly. • You learn a whole lot by taking the derivatives of your answers. This 'checking' process gives you immediate feedback and correction. • Very few students learn to integrate without doing this. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I’m having a really hard time finding the antiderivative of the square root of this functions. The square root is really throwing me off and the solutions don’t quite help me understand what’s going on. ------------------------------------------------ Self-critique Rating: 3
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Given Solution: `a 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 FREQUENT STUDENT ERROR The antiderivative is (1/3)x^3 (2(1-x^3)^(1/2) INSTRUCTOR COMMENTS: 1/3 x^3 is an antiderivative of x^2. 2 ( 1 - x^3)^(1/2) is not an antiderivative of 1 / 3 (x^3 - 1), as you will see if you take the derivative of that expression, but even if it was this strategy would not work. We simply cannot integrate the product of two nonconstant functions by integrating them separately in this manner. • The derivative of (1/3)x^3 (2)(1-x^3)^(1/2) is 2 * x^2 * sqrt(1 - x^3) - x^5 / sqrt(1 - x^3). • This is not the original function (1/3)x^3 (2(1-x^3)^(1/2), so (1/3)x^3 (2)(1-x^3)^(1/2) is not an antiderivative of (1/3)x^3 (2(1-x^3)^(1/2). Splitting an expression into a product of two or more expressions and integrating them separately never, never leads to a correct result. If you take the derivative of a result obtained in this manner, it will never equal the original expression. We quickly learn this if we get into the habit of checking our antiderivatives by taking the derivative and comparing with the original function. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think I simplified my answer wrong. ------------------------------------------------ Self-critique Rating: 2
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Given Solution: `a*& If we let u = 1 - x^3 we get du/dx = -3 x^2 so that x^2 = -1/3 du/dx. So the expression 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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think a lot of steps took place in this equation that I have not gone over yet. ------------------------------------------------ Self-critique Rating: 2 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!