#$&* course mth 272 Question: `q5.2.36 (was 5.2.2( (previously 5.2.36 (was 5.2.34) ) integral of x^2 (1-x^3)^2 by formal substitution. What is the integral of the given function?
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Given Solution: `a If we let u = 1 - x^3 then u ' = - 3 x^2 and the x^2 in our integrand is - u ' / 3. (1-x^3)^2 is u^2, so the integrand is - u ' / 3 * u^2 = -1/3 u^3 u ' . So the integral is you have -1/3 u^2 du. The integral of u^2 u ' is 1/3 u^3. Thus the integral of -1/3 u^2 u ' is -1/3 of 1/3 u^3, or -1/9 u^3. So your integral should be -1/9 u^3 = -1/9 (1-x^3)^3. The general antiderivative is -1/9 ( 1 - x^3)^3 + c. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qWhat is the derivative of your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: -1/9 (1-x^3)^3 3(1-x^3)^2 -1/9(-3x^2)*3(1-x^3)^2=x^2(1-x^3)^2 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q 5.2.54 (was 5.2.4) (previously 5.2.54 (was 5.2.52)) find x | dx/dp = -400/(.02p-1)^3, x=10000 when p=100 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: dx=-400*dp*(.02p-1)^-3 x=10000 u^-2+c x=10000 (.02p-1)^-2+c 10,000 * -2 * .02 (p-1)^-3 = -400 (.02 p - 1)^-3 10000=10000(.02*100-1)^-2+c c=0 x = 10,000 * (.02 p - 1)^-2 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The equation rearranges to dx = -400 * dp * (.02 p - 1)^-3. An antiderivative of the left-hand side could be just x. An antiderivative of dp * (.02 p - 1)^-3 is found using u = .02 p - 1, so du = .02 dp and dp = du / .02 = 50 du. Thus the right-hand side becomes -400 * 50 u^-3 du = -20000 u^-3 du, with antiderivative 20000 / 2 * u^-2 + c = 10,000 u^-2 + c. So we have x = 10,000 * u^-2 + c = 10,000 * (.02 p - 1)^-2 + c. Note that dx / dp is therefore 10,000 * -2 * .02 (p-1)^-3 = -400 (.02 p - 1)^-3, consistent with the original equation. Since x = 10,000 * (.02 p - 1)^-2 + c and x = 10,000 when p = 100 we have 10,000 = 10,000 * (.02 * 100 - 1)^2 + c 10,000 = 10,000 / 1^2 + c 10,000 = 10,000 + c so c = 0. The solution is therefore x = 10,000 * (.02 p - 1)^-2 + 0 or just x = 10,000 * (.02 p - 1)^-2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q 5.3.4 (was 5.3.1) (previously 5.3.04 (was 5.3.04)) integral of e^(-.25 x) by Exponential Rule YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: e^(-.25x) u=-.25x du/dx =-.25 du=-.25 dx=du/(-.25)=-4du dx= -4e^u du -4e^u+c=-4e^(-.25x)+c -4(-.25e^-.25x)=e^-.25x e^u/(du/dx)=e^(-.25x)/(-.25)=-4e^(-.25x) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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). *&*& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q 5.3.10 (was 5.3.2) (previously 5.3.10 (was 5.3.10)) integral of 3(x-4)e^(x^2-8x) by Exponential Rule YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: u=x^2-8x du/dx=2x-8 x-4 = 1/2(2x-8) so 3(x-4) = 3/2 du/dx 3(x-4)e^(x^2-8x) is 3/2 e^u du/dx 3/2 e^u 3/2 e^(x^2-8x) + c confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a if u=x^2 - 8x then du / dx = 2x - 8 x-4 = 1/2(2x-8) so 3(x-4) = 3/2 du/dx. Thus 3(x-4)e^(x^2-8x) is 3/2 e^u du/dx. The general antiderivative of e^u du/dx is e^u + c, so the integral of 3/2 e^u du/dx is 3/2 e^u. Substituting x^2 - 8x for u we have 3/2 e^(x^2-8x) + c. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qproblem 5.3.16 (was 5.3.3) (previously 5.3.16) integral of 1/(6x-5) by Log Rule YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: du/dx = 6 u = 6x - 5 du = 6 dx dx = 1/6 du ln(u) * du/6 = 1/6 ln(u) du 1 / (6x - 5) = 1/6 * 1 / (6x-5) = 1 / [ 6(6x-5) ] confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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) ]. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q 5.3.22 (was 5.3.5) (previously problem 5.3.22 (was 5.3.20)) integral of x/(x^2+4) by Log Rule YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: u = x^2 + 4 du/dx = 2x 1/2 * ( 2x / (x^2+4) ) = 1/2 (1/u du/dx) 1/2 ln(u) + c = 1/2 ln |x^2+4| + c confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a If we let u = x^2 + 4 we get du/dx = 2x so that the x in the numerator is 1/2 du/dx. The integral of x / (x^2 + 4) is the integral of 1/2 * ( 2x / (x^2+4) ) = 1/2 (1/u du/dx). The general antiderivative is therefore 1/2 ln(u) + c = 1/2 ln |x^2+4| + c. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qWhat is the derivative of your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ln(x^2+4) * (1/2) is 1/2 * 2x * 1 / (x^2 + 4) or x / (x^2 + 4) antiderivative ln(x^2+4) * (1/2) + c confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q 5.3.28 (was 5.3.7) (previously 5.3.28 (was 5.3.24) (was 5.3.24) ) integral of e^x/(1+e^x) by Log Rule YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: u = 1 + e^x du/dx = e^x 1 / (1 + e^x) * e^x ln |u| + c = ln | 1 + e^x | + c antiderivative confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a let u = 1 + e^x. Then du/dx = e^x. We are therefore integrating 1 / (1 + e^x) * e^x, which is 1/u du/dx. The antiderivative is ln |u| + c = ln | 1 + e^x | + c. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q 5.3.9 (previously 5.3.46 (was 5.3.34) (was 5.3.34) ) integral of (6x + e^x) `sqrt( 3x^2 + e^x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: u^(1/2) du/dx (6x + e^x) `sqrt( 3x^2 + e^x) = `sqrt(u) * du/dx 2/3 u^(3/2) = 2/3 (3x^2 + e^x)^(3/2) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q 5.3.58 (previously 5.3.11) (previously 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: dp = -125 e^(-t/20) dt p = 2500 e^(-t/20) + c 2500 = 2500 e^(-0/20) + c c=0 p = 2500 e^(-t/20) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!