#$&* course 272 12/9 3pm If your solution to a stated problem does not match the given solution, you should self-critique per instructions at
.............................................
Given Solution: `a Let u = x^3 so du = 3x^2 dx. The x^2 dx in the integral is just 1/3 du. You therefore have the integral of 1/3 ( ln(u) )^2 du. The table should have something for ( ln(u) ) ^ n. In any case the integral of ln(u)^2 with respect to u is u ln(u)^2 - 2 u ln(u) + 2 u. With the substitution u = x^3 you would be integrating (ln u)^2 * du/3, which would give you u [ 2 - 2 ln u + (ln u)^2 ] / 3, which translates to x^3 ( 2 - 2 ln(x^3) + (ln(x^3) ) ^ 2 ) / 3. DER: int( (ln(u)^2) = u·LN(u)^2 - 2·u·LN(u) + 2·u. Then for increasing powers of n int( ln(u)^n) gives us: u·LN(u)^3 - 3·u·LN(u)^2 + 6·u·LN(u) - 6·u then u·LN(u)^4 - 4·u·LN(u)^3 + 12·u·LN(u)^2 - 24·u·LN(u) + 24·u and u·LN(u)^5 - 5·u·LN(u)^4 + 20·u·LN(u)^3 - 60·u·LN(u)^2 + 120·u·LN(u) - 120· &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qQuery problem 6.3.52 (7th edition 6.4.46) use table to integrate x ^ 4 ln(x) then check by integration by parts YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: integrate x ^ 4 ln(x) u=x du= dx formula 41: u^n ln(u) du u^(n+1)/ (n+1)^2 [-1 + (n+1) ln(u) + c x^(5)/ (5)^2 [-1 + (5) ln(x) + c x^(5)/ 25 [4 ln(x) + c Check with integration by parts: integrate x ^ 4 ln(x) u=ln(x) dv=x^4 dx du=1/x v=x^5/5dx uv- int(vdu) = ln(x)*x^5/5 - int ( (x^5/5) * (dx/x)) so x^(n+1) / (n+1)(ln(x)-1(n+1) which matches equation from int table. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a Integration by parts on x^n ln(x) works with the substitution u = ln(x) and dv = x^n dx, so that du/dx = 1/x and v = x^(n+1) / n, giving us du = dx / x and v = x^(n+1) / (n + 1). Thus our integral is u v - int( v du) = ln(x) * x^(n+1) / (n + 1) - int ( x^(n+1)/(n+1) * dx / x) = ln(x) * x^(n+1)/( n + 1) - int(x^n dx) / (n+1) = ln(x) * x^(n+1) / (n + 1) - x^(n+1) / (n+1)^2 = x^(n+1) / (n+1) ( ln(x) - 1(n+1)). This should be equivalent to the formula given in the text. For n = 4 we get x^(4 + 1) / (4 + 1) ( ln(x) - 1 / (4 + 1)) = x^5 / 5 (ln(x) - 1/5). *&*& (x^5/25)(4 ln x) + C Using integration by parts: (x^5/5) ln x - (x^5/25) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qQuery problem 6.4.63 profit function P = `sqrt( 375.6 t^2 - 715.86) on [8,16]. ( 8th edition problem is 6.2.61 and the function is sqrt(.000645 t^2 + .1673) on (2, 5) ) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: P= = `sqrt( 375.6 t^2 - 715.86) [8,16] Integration table formula: 21 sqrt (u^2 +/- a^2) = ½(u(sqrt (u^2 +/- a^2)) + a^2 ln u + sqrt(u^2 +/- a^2) + C u^2 = 375.6t^2 u=sqrt (375.6)* t = 19.38t approx. a^2 = 715.86 a= sqrt(715.86) * t = 26.755 approx so ½(19.38(sqrt (375.6 +/- 715.86)) + 715.86 ln 19.38 + sqrt(375.6 +/- 715.86) + C = int (sqrt (375.6 t^2 - 715.86) [8,16] = 1850/ 8 = 230 approx. Substituting our values of a and u we find that integral(sqrt(375.6 t^2 - 715.86), t from 8 to 16) will be about 1850. Dividing this by the length 8 of the interval gives us the average value, which is about 1850 / 8 = 230. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a To get the average net profit integrate the profit function over the given interval and divide by the length of the interval. The integrand is sqrt(375.6 t^2 - 715.86), which is of the form `sqrt( u^2 +- a^2). • By the table the integral is 1/2 u sqrt(u^2 +- a^2) -+ a^2 ln(u + sqrt(u^2 +- a^2)) + c. u^2 = 375.67 t^2 so u = `sqrt(375.67) * t = 19.382 t approx. Similarly a = `sqrt(715.86) = 26.755 approx.. Substituting our values of a and u we find that integral(sqrt(375.6 t^2 - 715.86), t from 8 to 16) will be about 1850. Dividing this by the length 8 of the interval gives us the average value, which is about 1850 / 8 = 230. ** THE FUNCTION IS CLOSE TO THE LINEAR FUNCTION 19.4 t. The 715.86 doesn't have much effect when t is 8 or greater so the function is fairly close to P = 19 t. This approximation is linear so its average value will occur at the midpoint t = 12 of the interval. At t = 12 we have P = 19 * 12 = 230, approx. 8TH EDITION INTEGRAL The function given in the 8th edition leads to the integral int( sqrt(.000645 t^2 + .1673) dt, t from 2 to 5), which is of the form `sqrt( u^2 +- a^2). u^2 = .000645 t^2 so u = `sqrt(.000645) * t = .08 t approx. Similarly a = `sqrt(.1673) = .41 approx.. • By the table the integral is 1/2 u sqrt(u^2 +- a^2) -+ a^2 ln(u + sqrt(u^2 +- a^2)) + c. u^2 = 375.67 t^2 so u = `sqrt(375.67) * t = 19.382 t approx. Similarly a = `sqrt(715.86) = 26.755 approx.. Substituting our values of a and u we find that integral(sqrt(.000645 t^2 + .1673), t from 2 to 5) will be about 1850. Dividing this by the length 8 of the interval gives us the average value, which is about 1850 / 8 = 230. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. Feel way more confident using the integral table now after being able to check my answer using parts of integration. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: