course Mth 272 Assignment 7,8,9 While doing assignment another error signed appeared and kept logging me in and out, sorry it the questions are not organized properly. ??????????????assignment #007?{?h?????????Applied Calculus I
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11:29:45 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?
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RESPONSE --> u = 1-x^3 so derivative is -3x^2 -u^/3 * u^2 = -1/3 u^3 u' integral of u^2u' --> 1/3 u^3 so -1/9 u^3 = -1/9(1-x^3)^3
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11:30:03 ** 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. **
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RESPONSE --> the antiderivative is -1/9 ( 1 - x^3)^3 + c
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11:30:48 What is the derivative of your result?
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RESPONSE --> -1/9 (-3x^2) * 3(1-x^3)^2 = x^2 (1-x^3)^2
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11:31:00 ** 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. **
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RESPONSE --> answer was found by using the chain rule.
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11:33:27 5.2.54 (was 5.2.52 find x | dx/dp = -400/(.02p-1)^3, x=10000 when p=100
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RESPONSE --> -400 * 50 u^-3 du = -20000 u^-3 du and antiderivative is 20000 / 2 * u^-2 + c = 10,000 u^-2 + c so equals x = 10,000 * u^-2 + c = 10,000 * (.02 p - 1)^-2 + c plug in the values given and c=0 so x = 10,000 * (.02 p - 1)^-2 + 0
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11:33:35 ** 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. **
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RESPONSE --> ok
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11:36:43 5.3.04 (was 5.3.04 integral of e^(-.25 x) by Exponential Rule
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RESPONSE --> u is -.25 and du/dx is -.25 so the integral is e^(-.25 x) / (-1/4) = -4 e^(-.25 x)
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11:37:18 ** 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). *&*&
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RESPONSE --> substitution can also be used.
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11:39:08 5.3.10 (was 5.3.10 integral of 3(x-4)e^(x^2-8x) by Exponential Rule
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RESPONSE --> u=x^2 - 8x and du / dx = 2x - 8 x-4 = 1/2(2x-8) so 3(x-4) = 3/2 du/dx and then 3/2 e^u du/dx antiderivative of e^u du/dx is e^u + c so the integral is 3/2 e^u
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11:39:25 ** 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. **
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RESPONSE --> you have to substitute back in hence Substituting x^2 - 8x for u we have 3/2 e^(x^2-8x) + c
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11:40:52 problem 5.3.16 integral of 1/(6x-5) by Log Rule
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RESPONSE --> u = 6x - 5 and du = 6 dx so dx = 1/6 du and the integral 1/2 * ln(u) * du/6 = 1/3 ln(u) du and then 1/3 * 1 / u = 1/3 * 1 / (6x-5) = 1 / [ 3(6x-5) ]
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11:40:59 ** 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 1/2 * ln(u) * du/6 = 1/3 ln(u) du The integral of 1/3 ln(u) du is 1/3 * 1 / u = 1/3 * 1 / (6x-5) = 1 / [ 3(6x-5) ]. **
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RESPONSE --> ok
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????w???????l? assignment #007 ?{?h?????????Applied Calculus I 06-17-2006
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11:50:49 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?
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RESPONSE --> .
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11:50:53 ** 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. **
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RESPONSE --> .
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11:50:55 What is the derivative of your result?
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RESPONSE --> .
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11:50:58 ** 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. **
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RESPONSE --> .
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11:51:01 5.2.54 (was 5.2.52 find x | dx/dp = -400/(.02p-1)^3, x=10000 when p=100
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RESPONSE --> .
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11:51:04 ** 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. **
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RESPONSE --> .
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11:51:08 5.3.04 (was 5.3.04 integral of e^(-.25 x) by Exponential Rule
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RESPONSE --> .
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11:51:10 ** 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). *&*&
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RESPONSE --> .
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11:51:14 5.3.10 (was 5.3.10 integral of 3(x-4)e^(x^2-8x) by Exponential Rule
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RESPONSE --> .
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11:51:17 ** 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. **
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RESPONSE --> .
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11:51:22 problem 5.3.16 integral of 1/(6x-5) by Log Rule
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RESPONSE --> .
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11:51:25 ** 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 1/2 * ln(u) * du/6 = 1/3 ln(u) du The integral of 1/3 ln(u) du is 1/3 * 1 / u = 1/3 * 1 / (6x-5) = 1 / [ 3(6x-5) ]. **
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RESPONSE --> .
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12:34:52 problem 5.3.20 integral of x/(x^2+4) by Log Rule
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RESPONSE --> antiderivative is 1/2 ln(u) + c = 1/2 ln |x^2+4| + c b/c u=x^2 and du/dx = 2x and the integral is 1/2 * ( 2x / (x^2+4) ) = 1/2 (1/u du/dx)
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12:34:58 ** 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. **
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RESPONSE --> ok
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12:37:58 What is the derivative of your result?
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RESPONSE --> 1/2 * 2x * 1 / (x^2 + 4) or x / (x^2 + 4)
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12:38:00 ** 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. **
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RESPONSE --> ok
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12:49:16 5.3.24 (was 5.3.24 integral of e^x/(1+e^x) by Log Rule
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RESPONSE --> antiderivative is ln |u| + c = ln | 1 + e^x | + c
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12:49:41 ** 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. **
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RESPONSE --> u = 1 + e^x then du/dx = e^x. and then integrate: integrating 1 / (1 + e^x) * e^x, which is 1/u du/dx.
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13:00:15 5.3.34 (was 5.3.34 integral of (6x + e^x) `sqrt( 3x^2 + e^x)
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RESPONSE --> u = 3x^2 + e^x and du = 6x + e^x and integral of square root.. 2/3 u^(3/2) + c = 2/3 (3x^2 + e^x)^(3/2) + c
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13:00:32 ** 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. **
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RESPONSE --> another way to solve: 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)
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13:09:07 5.3.54 (was 5.3.52 dP/dt = -125 e^(-t/20), t=0, P=2500 and interpretation. Give your complete solution.
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RESPONSE --> k
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13:13:04 ** 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. **
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RESPONSE --> integrate both sides: p = 2500 e^(-t/20) + c the final solution is p = 2500 e^(-t/20)
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|??^????????C?}??assignment #008 ?{?h?????????Applied Calculus I 06-17-2006
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14:37:51 5.4.4 (was 5.4.4 integrate `sqrt(9-x^2) from -3 to 3
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RESPONSE --> integral: 9x - 1/3 x^3
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14:38:51 ** The graph of y = `sqrt( 9-x^2) is a half-circle of radius 3 centered at the origin. We can tell this because any point (x, `sqrt(9-x^2) ) lies at a distance of `sqrt( x^2 + (`sqrt (9-x^2))^2 ) = `sqrt(x^2 + 9-x^2) = `sqrt(9) = 3 from the origin. The area of the entire circle is 9 `pi square units. The region beneath the graph is a half-circle is half this, 9/2 `pi square units, which is about 14.1 square units. This area is the integral of the function from x=-3 to x=3. **SERIOUS STUDENT ERROR: Take the int and get 9x -1/3(x^3) INSTRUCTOR COMMENT: The integral of `sqrt( 9 - x^2) is not 9x - 1/3 x^3. The derivative of 9x - 1/3 x^3 function is 9 - x^2, not `sqrt(9-x^2). **
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RESPONSE --> * the derivative of 9x - 1/3 x^3 function is 9 - x^2, not `sqrt(9-x^2) *
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15:39:02 5.4.13 (was 5.4.10 (x^2+4)/x from 1 to 4
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RESPONSE --> (4^2 / 2 + 4 ln(4) ) - (1^2 / 2 + 4 ln(1) ) = 8 + 4 * 1.2 - (1/2 + 0) = 12
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15:41:28 ** The correct integral is not too difficult to find once you see that (x^2 + 4 ) / x = x + 4/x. There is an addition rule for integration, so you can integrate x and 4/x separately and recombine the results to get x^2/2 + 4 ln(x) + c. The definite integral is found by evaluating this expression at 4 and at 1 and subtracting to get (4^2 / 2 + 4 ln(4) ) - (1^2 / 2 + 4 ln(1) ) = 8 + 4 * 1.2 - (1/2 + 0) = 12 (approx). As usual check my mental calculations. ** STUDENT ERROR: The int is((x^3)/3 + 4x)(ln x) + C INSTRUCTOR CORRECTION: ** That does not work. You can't integrate the factors of the function then recombine them to get a correct integral. The error is made clear by taking the derivative of your expression. The derivative of (x^3/3 + 4x) ln(x) is (x^2 + 4) ln(x) + (x^2/3 + 4). Your approach does not work because it violates the product rule. **
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RESPONSE --> ok
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19:06:38 5.4.20 (was 5.4.16 3x^2+x-2 from 0 to 3
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RESPONSE --> antiderivative is x^3 + x^2/2 - 2x substitute 3 to equal 25.5 and 0 to equal 0 so the change is 25.5
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19:06:42 ** an antiderivative of f(x) = 3 x^2 + x - 2 is F(x) = x^3 + x^2/2 - 2x. Evaluating at 3 we get F(3) = 25.5. At 0 we have F(0) = 0. So the integral is the change in the antiderivative function: F(3) - F(0) = 25.5 - 0 = 25.5. **
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RESPONSE --> ok
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19:07:56 5.4.24 (was 5.4.20 Integrate sqrt(2/x) from 1 to 4
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RESPONSE --> antiderivative is 2 square root of 2x substitute 1 and 4 to get about 2.8
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19:08:15 ** The function can be written as `sqrt(2) / `sqrt(x) = `sqrt(2) * x^-.5. An antiderivative is 2 `sqrt(2) x^.5 = 2 `sqrt(2x). Evaluating at 4 and 1 we get 2 `sqrt(2*4) = 4 `sqrt(2) and 2 `sqrt(2) so the definite integral is 4 `sqrt(2) - 2 `sqrt(2) = 2 `sqrt(2), or approximately 2.8. **
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RESPONSE --> the equations were Evaluating at 4 and 1 we get 2 `sqrt(2*4) = 4 `sqrt(2) and 2 `sqrt(2) so the definite integral is 4 `sqrt(2) - 2 `sqrt(2) = 2 `sqrt(2),
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19:13:49 Extra Problem (was 5.4.52 What is the average value of 5e^(.2(x-10)) from x = 0 to x = 10?
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RESPONSE --> antiderivative is 5 / .2 e^(.2x - 2) substitute to get from 0 to 10
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19:14:15 ** The area under a curve is the product of its average 'height' and its 'width'. The average 'height' is the average value of the function, the area is the definite integral and the 'width' is the length of the interval. It follows that average value = definite integral / interval width. To integrate 5 e^(.2 ( x - 10) ): If you let u = .2x - 2 you get du/dx = .2 so dx = du / .2. You therefore have the integral of 5 e^u du / .2 = (5 / .2) e^u du. The integral of e^u du is e^u. So an antiderivative is 5 / .2 e^u = 5 / .2 e^(.2x - 2). Using the antiderivative 25 e^(.2(x-10)) at 0 and 10 we get about 22 for the definite integral (i.e., the antiderivative function 25 e^(.2(x-10)) changes by 22 between x = 0 and x = 10). The average value (obtained by dividing the integral by the length of the interval) is thus about 22 / 10 = 2.2. ** ERRONEOUS STUDENT SOLUTION: The average value is .4323. INSTRUCTOR COMMENT: This average value doesn't make sense. The function itself has value between 0 and 1 (closer to 1) when x=0 and value 5 when x=10 so its average value is probably greater than .4323. Unless the graph has a serious dip between the point where its value is 1 and the point where its value is 5, its average value would be between 1 and 5 and wouldn't be less than 1. **
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RESPONSE --> Using the antiderivative 25 e^(.2(x-10)) at 0 and 10 we get about 22 for the definite integral (i.e., the antiderivative function 25 e^(.2(x-10)) changes by 22 between x = 0 and x = he average value (obtained by dividing the integral by the length of the interval) is thus about 22 / 10 = 2.2
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19:15:45 Extra Problem (was 5.4.56 ave val of 1/(x-3)^2 from 0 to 2
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RESPONSE --> antiderivative is -1 / (x-3) at values 0 and 2 and to get the average value: 2/3 / (2-0) = 2/3 / 2 = 1/3
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19:15:48 ** An antiderivative of 1 / (x-3)^2 is -1 / (x-3). At 0 and 2 this antiderivative takes values 1/3 and 1 so the integral is 1 - 1/3 = 2/3, the change in the value of the antiderivative. The average value of the function is therefore ave value = integral / interval width = 2/3 / (2-0) = 2/3 / 2 = 1/3. **
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RESPONSE --> ok
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19:15:57 Does the average value make sense in terms of the graph?
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RESPONSE --> ??
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19:16:31 ** When x = 1, f(x) = .25 1 / (x-3)^2 = 1/3. Solve for x. Inverting both sides you get (x-3)^2 = 3 so x-3 = +-`sqrt(3) so x = 3 + `sqrt(3) or x = 3 - `sqrt(3), or approximately x = 4.732 or x = 1.268. The 1.268 makes sense for this interval; 4.732 isn't even in the interval. **
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RESPONSE --> (x-3)^2 = 3 so x-3 = +-`sqrt(3) so x = 3 + `sqrt(3) or x = 3 - `sqrt(3), or approximately x = 4.732 or x = 1.268 1.268 makes sense not the 4.732
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???Q????????o{?s? assignment #009 ?{?h?????????Applied Calculus I 06-17-2006
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20:46:24 5.5.6 (was 5.5.6 area between (x-1)^3 and x-1 from 0 to 2
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RESPONSE --> when graphed it intersects (1,1) and (0,-1) depending on where each equation lies ( x-1 from 0 to 1 and the other 1 to 2) and each of the integrals is equal to .25 meaning .5 is the area
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20:46:30 ** The graphs cross at (0, -1) and (1,1), as we easily find by solving the equation (x-1)^3 = (x - 1). Since (1, 1) lies between the endpoints of our interval we have to be careful about which function lies above which, and we'll have to split the calculation into two separate intervals. The x-1 graph lies above the (x-1)^3 graph from 0 to 1, so the area will be the integral of (x-1) - (x-1)^3 between these limits. The (x-1)^3 graph lies above the (x-1) graph from 1 to 2, so the area will be the integral of (x-1)^3 - (x-1) between these limits. Each integral is equal to .25, so the total area is the sum .25 + .25 = .5 of these areas. ** DER
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RESPONSE --> ok
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20:55:12 5.5.10 (was 5.5.10 describe the region for integral of (1-x^2) - (x^2-1) from -1 to 1
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RESPONSE --> antiderivative: 2x - 2/3 x^3 substitute at -1 and 1 to equal eight thirds which is the area of that region.
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20:55:16 *&*& The graph of 1 - x^2 is an upside down parabola with vertex at (0, 1), intercepting the x axis at x = -1 and x = 1. The graph of x^2 - 1 is a rightside up parabols with vertex at (0, -1), intercepting the x axis at x = -1 and x = 1. The region between the graphs is close to a circle passing thru (-1,0), (0,1), (1,0) and (0, -1), but the region is not exactly circular since it is formed by two parabolas. The graphs aren't vertical at (1,0) and (-1,0), for example, and a circle would be. The parabolas curve in such a way as to stay inside the circular region, so the region between the parabolas will have a bit less area than the circle. The integrand (1-x^2) - (x^2-1) can be simplified to 2 - 2 x^2. An antiderivative would be 2x - 2/3 x^3. Evaluating this at -1 and 1 we obtain integral 8/3. The area of the region is 8/3 = 2.67 approx.. Note that the area of the circle described above would be pi = 3.14, approx., a bit bigger than the area of the region between the parabolas. *&*& STUDENT ERROR: The graph is a circular region centered on the point (0,0) INSTRUCTOR COMMENT: The region is not exactly circular, (for example the graphs aren't vertical at (1,0) and (-1,0), for example), but it's fairly close to the circle. **
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RESPONSE --> ok
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