answer：Okay, so I have this problem about two CFL quarterbacks, A and B, and I need to analyze their performance over their careers. The problem is divided into two parts: first, finding the year when each quarterback reaches their peak performance, and second, calculating their total passing yards over the first 10 years and comparing them.Starting with the first part, I need to find the peak performance year for each quarterback. The performance is given by a quadratic function for each, which is a parabola. Since the coefficient of the ( t^2 ) term is negative in both cases (-10 for A and -15 for B), the parabolas open downward, meaning the vertex will be the maximum point, which is the peak performance.For a quadratic function in the form ( P(t) = at^2 + bt + c ), the vertex occurs at ( t = -frac{b}{2a} ). So, I can use this formula to find the year of peak performance for each quarterback.Let me write down the functions again:- Quarterback A: ( P_A(t) = 1500 + 200t - 10t^2 )- Quarterback B: ( P_B(t) = 1200 + 250t - 15t^2 )For Quarterback A, the coefficients are:- ( a = -10 )- ( b = 200 )So, the peak year ( t_A ) is:( t_A = -frac{200}{2 times -10} = -frac{200}{-20} = 10 )Wait, that can't be right. If t is 10, that's the 10th year. But let me check my calculation again.Wait, no, actually, the formula is correct. So for Quarterback A, the peak is at t = 10. Hmm, but let me think about that. If the peak is at t = 10, then that's the 10th year of their career. But the second part of the problem asks for the total over the first 10 years, so maybe that's why it's set up that way.Now for Quarterback B, the coefficients are:- ( a = -15 )- ( b = 250 )So, the peak year ( t_B ) is:( t_B = -frac{250}{2 times -15} = -frac{250}{-30} = frac{250}{30} approx 8.333 )So, approximately 8.333 years. Since we're dealing with years, which are discrete, we can say that the peak occurs between the 8th and 9th year. But since the question asks for the year t, we might need to consider whether to round up or down. But in the context of a mathematical model, we can just state it as approximately 8.33 years, which would be around the start of the 9th year.But let me verify my calculations again to be sure.For Quarterback A:( t_A = -b/(2a) = -200/(2*(-10)) = -200/(-20) = 10 ). Yes, that's correct.For Quarterback B:( t_B = -250/(2*(-15)) = -250/(-30) = 8.333... ). Yes, that's correct.So, Quarterback A peaks at t = 10, and Quarterback B peaks around t = 8.33.Moving on to the second part, calculating the total passing yards over the first 10 years. Since t starts at 0, the first 10 years would be from t = 0 to t = 9, inclusive, because t = 10 would be the 11th year. Wait, actually, the problem says "the first 10 years," so t = 0 to t = 9, which is 10 years. Alternatively, sometimes people count t = 1 to t = 10 as the first 10 years, but the problem says "careers started at t = 0," so t = 0 is the first year. Therefore, the first 10 years are t = 0 to t = 9.Wait, but let me think again. If t = 0 is the first year, then t = 1 is the second year, and so on. So, t = 0 to t = 9 would be 10 years. Alternatively, if t = 1 is the first year, then t = 1 to t = 10 would be 10 years. But the problem says "careers started at t = 0," so I think t = 0 is the first year. Therefore, the first 10 years would be t = 0 to t = 9.But wait, let me check the problem statement again: "Note: You may assume that careers started at t = 0." So, t = 0 is the first year, t = 1 is the second, etc. So, the first 10 years would be t = 0 to t = 9, inclusive. Therefore, we need to calculate the sum of P_A(t) and P_B(t) from t = 0 to t = 9.Alternatively, sometimes in sports, the first year is considered year 1, but since the problem specifies t = 0 as the start, we should go with t = 0 to t = 9.So, to calculate the total passing yards for each quarterback over the first 10 years, we need to compute the sum of P_A(t) from t = 0 to t = 9 and similarly for P_B(t).But since these are quadratic functions, we can find a closed-form expression for the sum instead of calculating each year individually. That would be more efficient.The sum of a quadratic function from t = 0 to t = n can be found using the formula for the sum of squares and linear terms.Given ( P(t) = at^2 + bt + c ), the sum from t = 0 to t = n is:( sum_{t=0}^{n} P(t) = a sum_{t=0}^{n} t^2 + b sum_{t=0}^{n} t + c sum_{t=0}^{n} 1 )We know that:- ( sum_{t=0}^{n} t = frac{n(n+1)}{2} )- ( sum_{t=0}^{n} t^2 = frac{n(n+1)(2n+1)}{6} )- ( sum_{t=0}^{n} 1 = n + 1 )So, for each quarterback, we can plug in a, b, c, and n = 9 into these formulas.First, let's compute for Quarterback A:( P_A(t) = -10t^2 + 200t + 1500 )So, a = -10, b = 200, c = 1500.Sum from t = 0 to t = 9:( S_A = -10 times sum t^2 + 200 times sum t + 1500 times sum 1 )Compute each sum:- ( sum t^2 = frac{9 times 10 times 19}{6} = frac{1710}{6} = 285 )- ( sum t = frac{9 times 10}{2} = 45 )- ( sum 1 = 10 )So,( S_A = -10 times 285 + 200 times 45 + 1500 times 10 )Calculate each term:- ( -10 times 285 = -2850 )- ( 200 times 45 = 9000 )- ( 1500 times 10 = 15,000 )Add them up:-2850 + 9000 = 61506150 + 15,000 = 21,150So, Quarterback A's total passing yards over the first 10 years is 21,150 yards.Now for Quarterback B:( P_B(t) = -15t^2 + 250t + 1200 )So, a = -15, b = 250, c = 1200.Sum from t = 0 to t = 9:( S_B = -15 times sum t^2 + 250 times sum t + 1200 times sum 1 )Using the same sums as before:- ( sum t^2 = 285 )- ( sum t = 45 )- ( sum 1 = 10 )So,( S_B = -15 times 285 + 250 times 45 + 1200 times 10 )Calculate each term:- ( -15 times 285 = -4275 )- ( 250 times 45 = 11,250 )- ( 1200 times 10 = 12,000 )Add them up:-4275 + 11,250 = 6,9756,975 + 12,000 = 18,975So, Quarterback B's total passing yards over the first 10 years is 18,975 yards.Comparing the two totals:- Quarterback A: 21,150 yards- Quarterback B: 18,975 yardsTherefore, Quarterback A had a better first 10 years in terms of total passing yards.Wait, but let me double-check my calculations to make sure I didn't make any arithmetic errors.For Quarterback A:- ( -10 times 285 = -2850 )- ( 200 times 45 = 9000 )- ( 1500 times 10 = 15,000 )- Total: -2850 + 9000 = 6150; 6150 + 15,000 = 21,150. Correct.For Quarterback B:- ( -15 times 285 = -4275 )- ( 250 times 45 = 11,250 )- ( 1200 times 10 = 12,000 )- Total: -4275 + 11,250 = 6,975; 6,975 + 12,000 = 18,975. Correct.Yes, the calculations seem correct.So, summarizing:1. Peak performance years: - Quarterback A: t = 10 - Quarterback B: t ≈ 8.33 (or between 8th and 9th year)2. Total passing yards over first 10 years: - Quarterback A: 21,150 yards - Quarterback B: 18,975 yardsTherefore, Quarterback A had a better first 10 years in terms of total passing yards.

answer：Okay, so I have this problem about South Shields FC and their community program. They have a budget of 50,000 and they want to organize football matches and educational workshops. I need to figure out how to model this with inequalities and then find the maximum number of each they can do without exceeding the budget. Then, in the second part, I have to use an attendance function to find the maximum increase in attendance based on those constraints.Alright, let's start with the first part. They have a budget of 50,000. Each football match costs 1,200, and each workshop costs 800. They want to organize at least 20 football matches and at least 30 workshops. So, I need to create inequalities for this.First, let me define the variables. Let x be the number of football matches and y be the number of educational workshops. The cost for football matches would be 1200x and for workshops, it would be 800y. The total cost should be less than or equal to 50,000. So, the first inequality is:1200x + 800y ≤ 50,000Next, they want to organize at least 20 football matches, so x has to be greater than or equal to 20. Similarly, y has to be greater than or equal to 30. So, the other inequalities are:x ≥ 20y ≥ 30Also, since you can't have a negative number of matches or workshops, x and y should be greater than or equal to zero. But since they already have to be at least 20 and 30, those are the more restrictive constraints.So, the system of inequalities is:1200x + 800y ≤ 50,000x ≥ 20y ≥ 30x ≥ 0y ≥ 0But since x is already ≥20 and y is ≥30, the last two are redundant.Now, the next part is to determine the maximum number of football matches and educational workshops they can organize while staying within the budget.So, we need to maximize x and y, but we have a budget constraint. However, since we have two variables, we can't maximize both at the same time. So, perhaps we need to find the feasible region defined by the inequalities and then find the maximum possible x and y within that region.Alternatively, maybe the question is asking for the maximum number of each, given the constraints. Hmm.Wait, perhaps it's asking for the maximum number of each individually, given the constraints. So, if we want to maximize x, we set y to its minimum, and vice versa.Let me think. If we want to maximize x, we should minimize y. Since y has to be at least 30, let's set y=30. Then, plug that into the budget inequality:1200x + 800*30 ≤ 50,000Calculate 800*30: that's 24,000.So, 1200x + 24,000 ≤ 50,000Subtract 24,000 from both sides:1200x ≤ 26,000Divide both sides by 1200:x ≤ 26,000 / 1200Let me compute that. 26,000 divided by 1200.Well, 1200*21 = 25,20026,000 - 25,200 = 800So, 800 / 1200 = 2/3 ≈ 0.6667So, x ≤ 21.6667But since x has to be an integer (you can't have a fraction of a match), so the maximum x is 21. But wait, we already have a constraint that x must be at least 20. So, x can be up to 21.But wait, 21.6667 is approximately 21 and two-thirds. So, 21 is the maximum integer less than that. So, x can be 21.But hold on, let me check:If x=21, y=30:Total cost is 1200*21 + 800*301200*21: 1200*20=24,000; 1200*1=1,200; so total 25,200.800*30=24,000. Total cost: 25,200 +24,000=49,200, which is under 50,000.If x=22, y=30:1200*22=26,400; 800*30=24,000. Total=50,400, which exceeds the budget.So, x can be at most 21.Similarly, if we want to maximize y, set x to its minimum, which is 20.So, x=20:1200*20=24,000So, 800y ≤ 50,000 -24,000=26,000Thus, y ≤26,000 /800=32.5Since y must be an integer, y=32.Check: x=20, y=32Total cost: 24,000 +800*32=24,000 +25,600=49,600, which is under 50,000.If y=33, then 800*33=26,400; total cost=24,000+26,400=50,400, which is over.So, y can be at most 32.So, the maximum number of football matches is 21, and the maximum number of workshops is 32.But wait, the question says "the maximum number of football matches and educational workshops they can organize while staying within the budget." Hmm, does that mean the maximum total number, or the maximum of each individually? Because if it's the total, we might have a different approach.But the way it's phrased, "the maximum number of football matches and educational workshops," it might mean the maximum of each, given the constraints. So, 21 matches and 32 workshops.Alternatively, if it's asking for the maximum total number, then we need to maximize x + y, subject to the constraints.Wait, maybe I should check that as well.So, if we want to maximize x + y, given 1200x +800y ≤50,000, x≥20, y≥30.We can set up the problem as a linear programming problem.The feasible region is defined by:1200x +800y ≤50,000x ≥20y ≥30x,y ≥0We can graph this or find the corner points.But since it's a two-variable problem, let's find the intersection points.First, let's write the budget equation:1200x +800y =50,000We can simplify this by dividing all terms by 400:3x +2y =125So, 3x +2y=125We can find where this line intersects the constraints x=20 and y=30.First, when x=20:3*20 +2y=125 =>60 +2y=125 =>2y=65 => y=32.5But y must be integer, so y=32 or 33. But since 32.5 is the exact point, but we can't have half workshops. So, the intersection is at (20,32.5)Similarly, when y=30:3x +2*30=125 =>3x +60=125 =>3x=65 =>x≈21.6667So, the intersection is at (21.6667,30)So, the feasible region is a polygon with vertices at:(20,30), (20,32.5), (21.6667,30), and maybe another point where x or y is higher.Wait, but actually, the budget line intersects x=20 at y=32.5 and y=30 at x≈21.6667.But since x and y must be integers, the feasible region's corner points would be at integer coordinates near these points.But perhaps for the purpose of maximizing x + y, we can consider the real intersection points and then check the integer points around them.But since we are dealing with integers, maybe the maximum occurs at (21,32) or (20,32) or (21,31), etc.Wait, let's compute x + y at the intersection points.At (20,32.5): x + y=52.5At (21.6667,30): x + y≈51.6667So, the maximum x + y is at (20,32.5), which is 52.5But since y must be integer, we can have y=32 or 33.At y=32, x=20: x + y=52At y=33, x would have to be less.Wait, if y=33, then 3x +2*33=3x +66=125 =>3x=59 =>x≈19.6667, which is less than 20, which violates the x≥20 constraint.So, y cannot be 33 if x has to be at least 20.Therefore, the maximum x + y is 52, achieved at (20,32).Wait, but earlier, when x=20, y=32, total cost is 24,000 +25,600=49,600, which is under budget.Alternatively, if we set x=21, y=31:Total cost: 1200*21=25,200; 800*31=24,800. Total=50,000.So, that's exactly the budget.x + y=52.Similarly, x=21, y=31: total cost=50,000.So, that's another point where x + y=52.So, both (20,32) and (21,31) give x + y=52.But (21,31) uses the entire budget, while (20,32) leaves some money unused.So, perhaps the maximum total number is 52, achieved at either (20,32) or (21,31).But the question says "the maximum number of football matches and educational workshops they can organize while staying within the budget."Hmm, so it might be referring to the maximum total, which is 52.But earlier, when maximizing x and y individually, we found x=21 and y=32.But if we are to maximize both, it's a bit ambiguous.Wait, let me re-read the question:"Additionally, determine the maximum number of football matches and educational workshops they can organize while staying within the budget."Hmm, so it's asking for the maximum number of each, but not necessarily together. Or maybe the maximum total.But the wording is a bit unclear. It could be interpreted as the maximum number of each, given the constraints, which would be x=21 and y=32. Or it could be the maximum total number, which is 52.But since in the first part, they asked for the system of inequalities, and then in the second part, they ask for the maximum number, perhaps it's the maximum of each individually.But in the second part, they talk about the attendance function, which depends on both x and y, so maybe they want the maximum total.Wait, no, in the second part, it's given the constraints from the first sub-problem, so it's likely that the first part is just setting up the constraints, and the second part is about optimizing attendance.But the first part's second question is about the maximum number of football matches and workshops. So, perhaps it's the maximum of each individually.So, to answer the first part: the system of inequalities is:1200x + 800y ≤50,000x ≥20y ≥30And the maximum number of football matches is 21, and the maximum number of workshops is 32.But let me confirm.If we set y=30, x can be up to 21.If we set x=20, y can be up to 32.So, those are the maximums for each individually.But if we consider the total number, it's 52.But since the question says "the maximum number of football matches and educational workshops," it might be referring to each separately.So, I think the answer is x=21 and y=32.But to be thorough, let's see.If we set x=21, y=30: total cost=25,200 +24,000=49,200, which leaves 800 unused.If we set x=20, y=32: total cost=24,000 +25,600=49,600, which leaves 400 unused.Alternatively, if we set x=21, y=31: total cost=25,200 +24,800=50,000, which uses the entire budget.So, in this case, x=21, y=31 is another feasible point.So, depending on whether we want to maximize x, y, or x + y, the answers differ.But since the question is about the maximum number of each, I think it's referring to each individually, so x=21 and y=32.But let me check the exact wording:"determine the maximum number of football matches and educational workshops they can organize while staying within the budget."Hmm, it's a bit ambiguous. It could be interpreted as the maximum total number, or the maximum of each.But in the context of the problem, since they have separate costs, and the question is about the number of each, I think it's asking for the maximum number of each, given the constraints.So, x=21 and y=32.But just to be safe, I'll note both interpretations.Now, moving on to the second part.The attendance increase is modeled by A(x, y)=500 +10x +15y.We need to find the maximum possible increase in attendance given the constraints from the first part.So, the constraints are:1200x +800y ≤50,000x ≥20y ≥30x,y integersWe need to maximize A(x,y)=500 +10x +15y.Since A(x,y) is a linear function, its maximum will occur at one of the corner points of the feasible region.So, we need to identify the corner points of the feasible region.From the first part, we have the budget line 3x +2y=125.The feasible region is bounded by:x=20, y=30, and 3x +2y=125.So, the corner points are:1. Intersection of x=20 and y=30: (20,30)2. Intersection of x=20 and 3x +2y=125: (20,32.5)3. Intersection of y=30 and 3x +2y=125: (21.6667,30)But since x and y must be integers, the actual corner points are the integer points near these intersections.So, the feasible integer points are:(20,30), (20,31), (20,32), (21,30), (21,31), (22,30) is not feasible because x=22 would require y= (125 -3*22)/2=(125-66)/2=59/2=29.5, which is less than 30, so not feasible.Similarly, y=33 would require x=(125 -2*33)/3=(125-66)/3=59/3≈19.6667, which is less than 20, so not feasible.So, the feasible integer points are:(20,30), (20,31), (20,32), (21,30), (21,31)Now, let's compute A(x,y) at each of these points.1. (20,30):A=500 +10*20 +15*30=500 +200 +450=11502. (20,31):A=500 +200 +15*31=500 +200 +465=11653. (20,32):A=500 +200 +15*32=500 +200 +480=11804. (21,30):A=500 +10*21 +15*30=500 +210 +450=11605. (21,31):A=500 +210 +15*31=500 +210 +465=1175So, the maximum A occurs at (20,32) with A=1180.But wait, let's check if (21,31) is feasible.At (21,31):Total cost=1200*21 +800*31=25,200 +24,800=50,000, which is exactly the budget.So, that's feasible.Similarly, (20,32):Total cost=24,000 +25,600=49,600, which is under budget.So, both are feasible.But A(x,y) is higher at (20,32):1180 vs (21,31):1175.So, the maximum increase in attendance is 1180.But wait, the function is A(x,y)=500 +10x +15y.So, the increase is A(x,y) -500=10x +15y.So, the increase is 10x +15y.So, at (20,32):10*20 +15*32=200 +480=680At (21,31):10*21 +15*31=210 +465=675So, the maximum increase is 680.But the question says "the maximum possible increase in attendance," so it's 680.But let me confirm:At (20,32): increase=680At (21,31): increase=675So, 680 is higher.Therefore, the maximum increase is 680.But wait, let me check if there are other points.Wait, is (20,33) feasible? Let's see:x=20, y=33:Total cost=24,000 +800*33=24,000 +26,400=50,400>50,000. Not feasible.Similarly, (22,30):x=22, y=30:Total cost=26,400 +24,000=50,400>50,000. Not feasible.So, the only feasible points are the ones I listed.Therefore, the maximum increase is 680.But wait, let me check if (20,32) is indeed feasible.Yes, 20 matches and 32 workshops cost 24,000 +25,600=49,600, which is under 50,000.So, that's fine.Alternatively, if we use the entire budget at (21,31), we get a slightly lower increase.So, the maximum increase is 680.But let me also check if there's a point beyond (20,32), but since y can't go beyond 32 without violating the budget when x=20, that's the maximum.So, I think that's the answer.**Final Answer**1. The system of inequalities is: [ begin{cases} 1200x + 800y leq 50000 x geq 20 y geq 30 end{cases} ] The maximum number of football matches is boxed{21} and the maximum number of educational workshops is boxed{32}.2. The maximum possible increase in attendance is boxed{680}.

answer：Okay, so I have this problem where I need to find the closed-form expression for the integral ( I(t) = int_{0}^{t} e^{-kx} cos(mx) , dx ). Hmm, I remember that integrals involving exponentials and trigonometric functions can be solved using integration techniques, maybe integration by parts or some formula. Let me think.I think the standard integral for something like ( int e^{ax} cos(bx) , dx ) is known. Maybe I can recall the formula or derive it. Let me try to recall. I think it involves integrating by parts twice and then solving for the integral.Let me set up the integral:Let ( I = int e^{-kx} cos(mx) , dx ).Let me use integration by parts. Let me set ( u = cos(mx) ) and ( dv = e^{-kx} dx ). Then, ( du = -m sin(mx) dx ) and ( v = -frac{1}{k} e^{-kx} ).So, integration by parts gives:( I = uv - int v du )Plugging in:( I = -frac{1}{k} e^{-kx} cos(mx) - int left( -frac{1}{k} e^{-kx} right) (-m sin(mx)) dx )Simplify:( I = -frac{1}{k} e^{-kx} cos(mx) - frac{m}{k} int e^{-kx} sin(mx) dx )Now, let me call the remaining integral ( J = int e^{-kx} sin(mx) dx ). I need to compute ( J ) using integration by parts again.So, set ( u = sin(mx) ), ( dv = e^{-kx} dx ). Then, ( du = m cos(mx) dx ), ( v = -frac{1}{k} e^{-kx} ).Thus,( J = uv - int v du = -frac{1}{k} e^{-kx} sin(mx) - int left( -frac{1}{k} e^{-kx} right) (m cos(mx)) dx )Simplify:( J = -frac{1}{k} e^{-kx} sin(mx) + frac{m}{k} int e^{-kx} cos(mx) dx )But notice that ( int e^{-kx} cos(mx) dx ) is our original integral ( I ). So,( J = -frac{1}{k} e^{-kx} sin(mx) + frac{m}{k} I )Now, substitute back into the expression for ( I ):( I = -frac{1}{k} e^{-kx} cos(mx) - frac{m}{k} J )But ( J = -frac{1}{k} e^{-kx} sin(mx) + frac{m}{k} I ), so plug that in:( I = -frac{1}{k} e^{-kx} cos(mx) - frac{m}{k} left( -frac{1}{k} e^{-kx} sin(mx) + frac{m}{k} I right ) )Simplify term by term:First term: ( -frac{1}{k} e^{-kx} cos(mx) )Second term: ( - frac{m}{k} times -frac{1}{k} e^{-kx} sin(mx) = frac{m}{k^2} e^{-kx} sin(mx) )Third term: ( - frac{m}{k} times frac{m}{k} I = - frac{m^2}{k^2} I )So, putting it all together:( I = -frac{1}{k} e^{-kx} cos(mx) + frac{m}{k^2} e^{-kx} sin(mx) - frac{m^2}{k^2} I )Now, bring the last term to the left side:( I + frac{m^2}{k^2} I = -frac{1}{k} e^{-kx} cos(mx) + frac{m}{k^2} e^{-kx} sin(mx) )Factor out ( I ):( I left( 1 + frac{m^2}{k^2} right ) = -frac{1}{k} e^{-kx} cos(mx) + frac{m}{k^2} e^{-kx} sin(mx) )Simplify the coefficient:( 1 + frac{m^2}{k^2} = frac{k^2 + m^2}{k^2} )So,( I = frac{ -frac{1}{k} e^{-kx} cos(mx) + frac{m}{k^2} e^{-kx} sin(mx) }{ frac{k^2 + m^2}{k^2} } )Multiply numerator and denominator:( I = frac{ -frac{1}{k} e^{-kx} cos(mx) + frac{m}{k^2} e^{-kx} sin(mx) }{ frac{k^2 + m^2}{k^2} } = frac{ -k e^{-kx} cos(mx) + m e^{-kx} sin(mx) }{ k^2 + m^2 } )So, the integral ( I ) is:( I = frac{ e^{-kx} ( -k cos(mx) + m sin(mx) ) }{ k^2 + m^2 } + C )But since we are dealing with a definite integral from 0 to t, we need to evaluate this expression from 0 to t.So, the definite integral ( I(t) ) is:( I(t) = left[ frac{ e^{-kx} ( -k cos(mx) + m sin(mx) ) }{ k^2 + m^2 } right ]_{0}^{t} )Compute this at t and at 0:At x = t:( frac{ e^{-kt} ( -k cos(mt) + m sin(mt) ) }{ k^2 + m^2 } )At x = 0:( frac{ e^{0} ( -k cos(0) + m sin(0) ) }{ k^2 + m^2 } = frac{1 ( -k cdot 1 + m cdot 0 ) }{ k^2 + m^2 } = frac{ -k }{ k^2 + m^2 } )So, subtracting the lower limit from the upper limit:( I(t) = frac{ e^{-kt} ( -k cos(mt) + m sin(mt) ) }{ k^2 + m^2 } - left( frac{ -k }{ k^2 + m^2 } right ) )Simplify:( I(t) = frac{ -k e^{-kt} cos(mt) + m e^{-kt} sin(mt) + k }{ k^2 + m^2 } )We can factor out the negative sign in the first term:( I(t) = frac{ k (1 - e^{-kt} cos(mt) ) + m e^{-kt} sin(mt) }{ k^2 + m^2 } )Alternatively, we can write it as:( I(t) = frac{ e^{-kt} ( -k cos(mt) + m sin(mt) ) + k }{ k^2 + m^2 } )Either form is acceptable, but perhaps the first one is more compact.So, that's the closed-form expression for ( I(t) ).Now, moving on to part 2: Given ( k = 0.5 ) and ( m = pi ), compute ( I(t) ) at ( t = 5 ) years.So, plug in k = 0.5, m = π, t = 5.First, let's compute each part step by step.Compute ( e^{-kt} = e^{-0.5 times 5} = e^{-2.5} ). Let me compute that value.( e^{-2.5} ) is approximately equal to... since e^2 ≈ 7.389, so e^2.5 ≈ e^2 * e^0.5 ≈ 7.389 * 1.6487 ≈ 12.182. So, e^{-2.5} ≈ 1 / 12.182 ≈ 0.0821.Next, compute ( cos(mt) = cos(pi times 5) = cos(5π) ). Since cosine has a period of 2π, 5π is equivalent to π (since 5π = 2π*2 + π). So, cos(5π) = cos(π) = -1.Similarly, compute ( sin(mt) = sin(5π) ). Sin(5π) = sin(π) = 0.So, let's plug these into the expression:( I(5) = frac{ e^{-2.5} ( -0.5 times (-1) + pi times 0 ) + 0.5 }{ (0.5)^2 + (pi)^2 } )Simplify term by term:First, compute the numerator:Inside the exponential term:-0.5 * (-1) = 0.5π * 0 = 0So, the exponential term is e^{-2.5} * (0.5 + 0) = e^{-2.5} * 0.5 ≈ 0.0821 * 0.5 ≈ 0.04105Then, add 0.5: 0.04105 + 0.5 = 0.54105Now, compute the denominator:(0.5)^2 = 0.25π^2 ≈ 9.8696So, denominator = 0.25 + 9.8696 ≈ 10.1196Therefore, I(5) ≈ 0.54105 / 10.1196 ≈ 0.05346So, approximately 0.0535.Let me verify the steps again to make sure I didn't make a mistake.Wait, let me re-examine the expression:I(t) = [ e^{-kt} (-k cos(mt) + m sin(mt)) + k ] / (k² + m²)So, plugging in:e^{-2.5} ≈ 0.0821- k cos(mt) = -0.5 * (-1) = 0.5m sin(mt) = π * 0 = 0So, e^{-2.5}*(0.5 + 0) ≈ 0.0821 * 0.5 ≈ 0.04105Then, add k = 0.5: 0.04105 + 0.5 = 0.54105Denominator: 0.25 + π² ≈ 0.25 + 9.8696 ≈ 10.1196So, 0.54105 / 10.1196 ≈ 0.05346, which is approximately 0.0535.Alternatively, using more precise calculations:Compute e^{-2.5} more accurately:e^{-2.5} ≈ 0.082085So, 0.082085 * 0.5 = 0.0410425Add 0.5: 0.0410425 + 0.5 = 0.5410425Denominator: 0.25 + π² ≈ 0.25 + 9.8696044 ≈ 10.1196044So, 0.5410425 / 10.1196044 ≈ Let's compute this division.10.1196044 * 0.0535 ≈ 10.1196 * 0.05 = 0.50598, 10.1196 * 0.0035 ≈ 0.03542, total ≈ 0.5414.So, 0.5410425 / 10.1196044 ≈ approximately 0.0535.So, I(t) at t=5 is approximately 0.0535.Alternatively, if I use more precise calculation:Compute 0.5410425 / 10.1196044.Let me do this division step by step.10.1196044 goes into 0.5410425 how many times?Well, 10.1196044 * 0.05 = 0.50598022Subtract that from 0.5410425: 0.5410425 - 0.50598022 ≈ 0.03506228Now, 10.1196044 goes into 0.03506228 approximately 0.00346 times (since 10.1196044 * 0.00346 ≈ 0.03506)So, total is approximately 0.05 + 0.00346 ≈ 0.05346, which is about 0.0535.So, I(t) ≈ 0.0535.Alternatively, if I use a calculator for more precision:Compute numerator: 0.5410425Denominator: 10.1196044So, 0.5410425 / 10.1196044 ≈ 0.05346So, approximately 0.0535.Therefore, the value of I(t) at t=5 is approximately 0.0535.I think that's the answer. Let me just recap:1. Found the closed-form integral using integration by parts twice, leading to an expression involving exponentials, sines, and cosines.2. Plugged in the given values k=0.5, m=π, t=5, computed each term step by step, and arrived at approximately 0.0535.I think that's correct. I don't see any mistakes in the steps.**Final Answer**1. The closed-form expression for ( I(t) ) is ( boxed{dfrac{e^{-kt}(-k cos(mt) + m sin(mt)) + k}{k^2 + m^2}} ).2. The value of ( I(5) ) is approximately ( boxed{0.0535} ).

answer：Okay, so I have this problem about Mullins, an introverted college student who likes to study in secluded spots in the library. The library is laid out as a 10x10 grid, and each spot has a probability associated with it based on how likely Mullins is to choose that spot. The problem has two parts: first, calculating the normalization constant N, and second, adjusting the probabilities considering that Mullins avoids spots within a Manhattan distance of 3 from the center (spot (5,5)). Let me start with the first part. The probability P(i, j) is given by (11 - i)(11 - j) divided by N. So, N is the normalization constant that makes sure the sum of all probabilities equals 1. Since it's a 10x10 grid, i and j both range from 1 to 10. So, to find N, I need to compute the sum over all i and j of (11 - i)(11 - j). That is, N = sum_{i=1 to 10} sum_{j=1 to 10} (11 - i)(11 - j). Hmm, I notice that the expression (11 - i)(11 - j) can be separated into two separate sums because it's a product of functions each depending only on i or j. So, I can rewrite N as [sum_{i=1 to 10} (11 - i)] multiplied by [sum_{j=1 to 10} (11 - j)]. Since the sums over i and j are the same, this simplifies to [sum_{k=1 to 10} (11 - k)] squared.Let me compute sum_{k=1 to 10} (11 - k). That's the same as sum_{m=1 to 10} m, where m = 11 - k. So, when k=1, m=10; when k=10, m=1. So, it's the sum from 1 to 10. The sum of the first n integers is n(n + 1)/2, so here n=10, so sum is 10*11/2 = 55.Therefore, N = 55 * 55 = 3025. So, the normalization constant N is 3025. That makes sense because each spot's probability is proportional to (11 - i)(11 - j), which is higher for spots closer to the edges, as the problem states.Now, moving on to the second part. Mullins avoids spots within a Manhattan distance of 3 from the center (5,5). So, I need to adjust the probabilities by excluding those spots and then re-normalizing the probabilities.First, let's recall what Manhattan distance is. The Manhattan distance between two points (i, j) and (5,5) is |i - 5| + |j - 5|. So, spots where |i - 5| + |j - 5| <= 3 are the ones Mullins avoids. Therefore, the adjusted probability P'(i, j) will be equal to P(i, j) if the spot is outside this distance, and 0 otherwise. But since we have to re-normalize, we need to compute the sum of P(i, j) over all spots that are outside the forbidden area and then set P'(i, j) = P(i, j) / S, where S is this new sum.So, the general expression for P'(i, j) is:P'(i, j) = { P(i, j) / S, if |i - 5| + |j - 5| > 3, 0, otherwise}Where S is the sum of P(i, j) for all (i, j) such that |i - 5| + |j - 5| > 3.But since P(i, j) is (11 - i)(11 - j)/N, and N is 3025, we can write P'(i, j) as:P'(i, j) = { (11 - i)(11 - j) / (N * S'), if |i - 5| + |j - 5| > 3, 0, otherwise}Wait, actually, S is the sum of (11 - i)(11 - j) over the allowed spots. Since N was the total sum, S = sum_{allowed spots} (11 - i)(11 - j). So, the new normalization constant would be S, and P'(i, j) = (11 - i)(11 - j) / S for allowed spots.But let me clarify: originally, P(i, j) = (11 - i)(11 - j)/N, with N=3025. After excluding certain spots, the new probability is P'(i, j) = P(i, j) / S, where S is the sum of P(i, j) over allowed spots. So, S = sum_{allowed} P(i, j) = sum_{allowed} (11 - i)(11 - j)/N. Therefore, S = (1/N) * sum_{allowed} (11 - i)(11 - j). So, P'(i, j) = (11 - i)(11 - j)/N divided by S, which is (11 - i)(11 - j)/N divided by (sum_{allowed} (11 - i)(11 - j)/N) = (11 - i)(11 - j) / sum_{allowed} (11 - i)(11 - j). Therefore, the new normalization constant is sum_{allowed} (11 - i)(11 - j). Let's denote this sum as S'. So, S' = sum_{allowed} (11 - i)(11 - j). Then, P'(i, j) = (11 - i)(11 - j) / S' for allowed spots, and 0 otherwise.So, the general expression is:P'(i, j) = { (11 - i)(11 - j) / S', if |i - 5| + |j - 5| > 3, 0, otherwise}Where S' is the sum of (11 - i)(11 - j) over all (i, j) where |i - 5| + |j - 5| > 3.Therefore, the normalization constant changes from N=3025 to S', which is the sum over the allowed spots.To compute S', I would need to calculate the sum of (11 - i)(11 - j) for all (i, j) not within Manhattan distance 3 from (5,5). Alternatively, since the original sum N is 3025, S' = N - sum_{forbidden} (11 - i)(11 - j). So, if I can compute the sum over forbidden spots, subtract that from 3025, and that will give me S'.But the problem doesn't ask me to compute S' numerically, just to provide the general expression and describe how the normalization constant changes. So, I think that's sufficient.In summary, the original normalization constant N is 3025. After excluding the spots within Manhattan distance 3 from the center, the new normalization constant becomes S', which is the sum of (11 - i)(11 - j) over all allowed spots, or equivalently, N minus the sum over forbidden spots. The adjusted probability P'(i, j) is zero for forbidden spots and (11 - i)(11 - j)/S' for allowed spots.**Final Answer**1. The normalization constant ( N ) is boxed{3025}.2. The adjusted probability ( P'(i, j) ) is given by: [ P'(i, j) = begin{cases} dfrac{(11 - i)(11 - j)}{S'}, & text{if } |i - 5| + |j - 5| > 3, 0, & text{otherwise} end{cases} ] where ( S' ) is the sum of ( (11 - i)(11 - j) ) over all allowed spots, serving as the new normalization constant.

answer：First, I need to determine how to distribute the total 1200 proportionally based on the number of students in each school.I'll start by calculating the total number of students across all three schools. School A has 30 students, School B has 45 students, and School C has 25 students. Adding these together gives a total of 100 students.Next, I'll find the amount of money each student effectively "earns" by dividing the total amount by the total number of students. So, 1200 divided by 100 students equals 12 per student.Now, I'll calculate the allocation for each school by multiplying the number of students in each school by the per-student amount. For School A, 30 students multiplied by 12 equals 360. For School B, 45 students multiplied by 12 equals 540. For School C, 25 students multiplied by 12 equals 300.This ensures that each school receives a fair share of the funds based on the number of students attending the trip.For the second part, the parent decides to add an additional 200 bonus to the school with the highest number of students, which is School B. This means School B will receive an extra 200 on top of its proportional share.After allocating the bonus, the remaining amount to be distributed is 1000. I'll redistribute this remaining amount proportionally among the three schools based on the number of students, following the same per-student allocation method as before.Finally, I'll add the bonus to School B's share to determine the new total each school receives.

answer：Alright, so I have this problem about Dr. Eleanor analyzing ancient trade routes. It's divided into two parts. Let me try to tackle each part step by step.Starting with part 1: She wants the shortest time from city A to city B using Dijkstra's algorithm. Hmm, okay, I remember Dijkstra's algorithm is used for finding the shortest path in a graph with non-negative weights. The adjacency matrix is given as W, where W_ij is the travel time between city i and j, and infinity if there's no direct route.So, the general formula and steps for Dijkstra's algorithm. Let me recall. The algorithm maintains a priority queue where each node has a tentative distance from the start node. Initially, the start node has a distance of zero, and all others are set to infinity. Then, it repeatedly extracts the node with the smallest tentative distance, updates the distances of its neighbors, and continues until the destination node is reached or all nodes are processed.Breaking it down:1. Initialize the distance array: distance[A] = 0, distance[others] = infinity.2. Create a priority queue and add all nodes with their tentative distances.3. While the queue is not empty: a. Extract the node u with the smallest tentative distance. b. For each neighbor v of u: i. Calculate the tentative distance through u: temp_dist = distance[u] + W_uv. ii. If temp_dist < distance[v], update distance[v] to temp_dist. iii. Reinsert v into the priority queue with the new distance.4. The algorithm stops when the destination node B is extracted from the queue or all nodes are processed.So, the formula for the tentative distance is straightforward: it's the sum of the current node's distance and the edge weight. The key steps involve the priority queue and relaxing the edges.Now, part 2 is a bit trickier. It mentions that trade routes are subject to periodic flooding, which increases travel time by a factor of f (where f > 1). Flooding occurs every k months and lasts for l months. We need to find the expected travel time from A to B over a long-term period, assuming the flooding effect is uniformly distributed.Okay, so first, I need to model the travel time considering the flooding. Let's think about the time intervals. Flooding happens every k months and lasts l months each time. So, the cycle is k months, with l months of flooding and (k - l) months of normal conditions.The travel time during normal months is W_ij, and during flooding, it's f * W_ij. Since the flooding is periodic, the expected travel time would be the average over the cycle.So, for each edge, the expected travel time E_ij would be:E_ij = (probability of flooding) * (f * W_ij) + (probability of no flooding) * W_ijThe probability of flooding in a given month is l/k, and no flooding is (k - l)/k.Therefore, E_ij = (l/k) * f * W_ij + ((k - l)/k) * W_ijSimplify that:E_ij = W_ij * [ (l f + (k - l)) / k ] = W_ij * (l f + k - l)/k = W_ij * (k + l(f - 1))/kSo, the expected weight for each edge is scaled by (k + l(f - 1))/k.Therefore, to find the expected shortest path from A to B, we can construct a new graph where each edge weight is replaced by its expected value E_ij, and then apply Dijkstra's algorithm on this new graph.Alternatively, since the expected value is linear, we can compute the expected shortest path as the shortest path in the original graph scaled by the same factor. Wait, is that correct?Wait, no, because the shortest path might change depending on the expected weights. So, we can't just scale the original shortest path; we need to compute the shortest path in the expected graph.Therefore, the steps would be:1. For each edge (i, j), compute E_ij = W_ij * (k + l(f - 1))/k2. Use Dijkstra's algorithm on this new expected adjacency matrix to find the shortest path from A to B.But wait, is the expected value additive? Because the total travel time is the sum of the expected times for each edge on the path. Since expectation is linear, the expected total time is the sum of the expected times for each edge. So, yes, we can compute the expected shortest path by replacing each edge with its expected value and then running Dijkstra's.Therefore, the expected travel time is the shortest path in the graph where each edge weight is E_ij as defined above.So, putting it all together, for part 2, we first compute the expected weight for each edge, then apply Dijkstra's algorithm on this expected graph to get the expected minimum travel time.Wait, but is there another way? Maybe considering the periodicity and integrating over time? Hmm, but since the flooding is periodic and uniformly distributed, the expected value approach should suffice.Alternatively, if we model the travel time as a random variable with two possible outcomes: W_ij with probability (k - l)/k and f W_ij with probability l/k. Then, the expected value is as we computed.So, I think the approach is correct.Therefore, the expected travel time is the shortest path in the graph with edge weights E_ij = W_ij * (k + l(f - 1))/k.So, summarizing:1. For each edge, compute E_ij = W_ij * (k + l(f - 1))/k2. Apply Dijkstra's algorithm on the graph with these expected weights to find the shortest path from A to B.Hence, the expected travel time is the result of this algorithm.I think that's the solution. Let me just check if there's any alternative interpretation. Maybe the flooding affects the entire journey, but since the journey can take multiple months, perhaps the flooding affects each month of travel? But the problem says the flooding occurs every k months and lasts l months, and the effect is uniformly distributed. So, each month, the probability of being in a flood month is l/k, so each edge's travel time is multiplied by f with probability l/k. Hence, the expected value per edge is as computed.Yes, that seems correct.