#$&* course MTH 272 10/15 about 3:15pm 008. `query 8
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Given Solution: `a 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):I attempted to take the integral of this but had a difficult time with the sin/cos properties. I have had a class with this before but I am not sure you wanted us to find it in this fashion anyways. ???Am I correct? Do I need to review this????
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Given Solution: `a 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. ** STUDENT QUESTION not seeing were (x^2 + 4) / x = x + 4 / x. What happened to the power of two INSTRUCTOR RESPONSE Division by x is the same as multiplication by 1/x, so the distributive law applies. We obtain • (x^2 + 4) / x = (x^2 / x) + (4 / x), which simplifies to x + (4 / x). To detail the distributive law a little more: (x^2 + 4) / x = (x^2 + 4) * (1/x) = x^2 * (1/x) + 4 * (1/x) = x^2 / x + 1 / x. = x + 1 / x. By the order of operations x^2 / x + 1 / x means (x^2 / x) + (1 / x), which is equal to x + (1 / x). The parentheses aren't necessary but they can be used to clarify the expression. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `q 5.4.5 (previously Extra Problem (formerly 5.4.20) (was 5.4.16) ) Integrate 3x^2+x-2 from x = 0 to x = 3 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Int.(3x^2 + x - 2) = x^3 +(x^2)/2 - 2x + c Substitute the values and subtract: (27+4.5-6) - (0) = 25.5 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `q 5.4.7 ((previously 5.4.28 (was 5.4.24) (was 5.4.20) ) Integrate sqrt(2/x) from 1 to 4 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Int. (2/x)^(1/2) = 2^(1/2) int. x^-(1/2) = ‘sqrt(2) * (x^(1/2)/(1/2) + c = 2*’sqrt(2) * ‘sqrt(x) + c Substitute the values and subtract 2*’sqrt(2)*’sqrt(4) - 2*’sqrt(2)*’sqrt(1) = 4*’sqrt(2) - 2*’sqrt(2) = 2*’sqrt(2) = 2.83 approx. ???Did I do the integration correctly. I did not use substitution??? confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):I think I worked the problem correctly but please see question in solution. ------------------------------------------------ Self-critique Rating:3
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `q 5.4.15 (previously 5.4.66 (was 5.4.56) ) ave val of 1/(x-3)^2 from 0 to 2 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Int. (x-3)^-2 u=(x-3) u’=(1) Int. (u)^-2 u’ = u^(-1)/-1 + c = -1/(x+3) + c Substitute the value and subtract then divide the difference by the length (-1/(2-3)) - (-1/0-3) = 1 - (1/3) = (2/3) (2/3)/2 = 2/6 = 1/3 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qDoes the average value make sense in terms of the graph? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It does make sense. In a very simplistic way, you are only looking at the graph from x=0 to x=2. At x=2, the y value is 1. Without going into as much detail as the given solution, the graph does not increase at a large value until it is well past x=1 but before x=2. This would imply that it would be closer to the y value for x=0(.11) than the y value for x=2(1.0). I understand even if this does not make sense. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We can see this by constructing the graph: The function 1 / (x-3)^2 can be obtained from the graph of 1 / x^2, by shifting the graph 3 units to the right. The vertical asymptote shifts to x = 3, and the function increases at an increasing rate (i.e., it is concave up) as it approaches the asymptote. Its values at x = 0 and x = 2 are .11 and 1.00, and the shape of the graph keeps the value closer to .11 than to 1.0 for most of the interval. We therefore expect that the average value will lie between .11 and 1.00, but be closer to .11. The average value was found to be 1/3. Since 1/3 = .33, which is closer to .11 than to 1.00, our average value makes sense in terms of the graph It's important to know the shapes of basic graphs and to be able to construct graphs using transformations. So the above solution is preferred. However it would also be acceptable to see this by evaluating the function at a few points between x = 0 and x = 2. For example When x = 0, f(x) = 1 / (0 - 3)^2 = .11. When x = 1, f(x) = 1 / (1 - 3)^2 = .25. When x = 2, f(x) = 1 / (2 - 3)^2 = 1.00 The graph is increasing all the way from x = 0 to x = 2. So the average value certainly lies between the x = 0 value .11 and the x = 2 value 1.00. Halfway between x = 0 and x = 2 the value of the function has increased from .11 to only .25, so it is clear that the average value will be closer to .11 than to 1.00. The average value of the function is 1/3, which is approximately equal to .33. This is very consistent with the values given above. We can obtain further confirmation by finding the x value at which the function reaches its average value of 1/3: We solve the equation 1 / (x-3)^2 = 1/3. Inverting both sides we 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 function value 1.268 occurs within our interval; 4.732 does not (this value occurs on the other side of the vertical asymptote, where the function is decreasing). x = 1.268 lies closer to x = 2 than to x = 0, indicating that the function reaches its average value of 1/3 during the second half of the interval. Between x = 1.278 and x = 2, the function's value will then triple to 1.00. A numerical synopsis of function values on the interval from x = 0 to x = 2: At x = 0 the value is .11. At x = 1, halfway through the interval, the function value is .25. At x = 1.278, approx., the function reaches its average value of 1/3. The function then increases fairly rapidly to 1.00. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!