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Topic: Integral Math Homework Help?**Question:**
Find the area between curves : f(x)=42+x-x^2 and g(x)=x^2-7x over (1,4) , i already tried this problem and cant get the right answers when i look at the solution, apparently my g(x) prime is incorrect , help please

June 16, 2019 / By Charlotte

There is no need to involve g prime here. The curves f(x) and g(x) intersect at x = -3 and x = 7. To see this, note that setting them equal and solving for x gives 42 + x - x² = x² - 7x ==> 2x² - 8x - 42 = 2(x² - 4x - 21) = 2(x+3)(x-7) = 0 The important point is that everywhere between -3 and 7, the graph of f(x) is above the graph of g(x). Since the interval you are interested in---(1,4)---is inside the interval (-3,7), f will be your top function and g your bottom function. The area between them will be the integral from 1 to 4 of f(x) - g(x). The area A is A = int(x=1 to 4) (42 + x - x²) - (x² - 7x) dx =int(x=1 to 4) 42 + 8x - 2x² dx = 42x + 4x² - 2x^3/3 You have to evaluate at 4 and 1 and take the difference. A = 42(4) + 4(4)² - 2(4)^3/3 - (42(1) + 4(1)² - 2(1)^3/3) = 144

👍 282 | 👎 7

Did you like the answer? I = ∫ dx /√(600x + x^2) I = ∫ dx /√(600x + x^2 + 300^2 - 300^2) I = ∫ dx /√[(x + 300)^2 - 300^2] ----------------------------- let x+300 = 300 * sec p dx = 300 * sec p tan p dp I = ∫ 300 * sec p tan p dp /√(300^2)(sec^2 p-1) I = ∫ sec p tan p dp /sec p I = ∫ tan p dp I = - ln |cos p| + C I = ln |sex p| + C replace (p) to (x) I = ln [(x+300)/300] + C

Try completing the square on what is inside the radical. I get (x+300)^2 - 9000 inside the radical. Now you can use x = 300sec(theta) as a substitution.

perhaps it will help graphing the functions... but they are both parabolas which enclose one another.... do you know how to do double integrals? this would probably be easier to do, but we probably dont have to do that - just so you know, we can manipulate the equations to meet our demands as integrating often cuts our area off at the x-axis but we want more... we'll call the area we need our image, and g(x) tells us where the bottom of our image is...find g ' to find where it reaches a min: g' = 2x-7=0 ==> x = (7/2)...and g(7/2) = -(49/4) because we dont want interference with the x-axis, we can add this to both g(x) and f(x) as this doesnt change the enclosed area... so you have f(x) = 42+(49/4) + x -x^2 and g(x) = x^2 - 7x ==> set them equal to each other to find what x values they are both equal in and do the integral Sf(x) - g(x) with your limits being those two x values

👍 120 | 👎 0

First let's see where the curves intersect. f(x)=g(x) 42+x-x²=x²-7x 2x²-8x-42=0 x²-4x-21=0 (x-7)(x+3)=0 x=-3, x=7 So the curves do not intersect in (1,4). f(x)>g(x) for all x in (1,4). ∫ f(x)-g(x) dx = ∫ 42+8x-2x² dx = [ 42x+4x²-(2/3)x³ ] = (168+64-128/3) - (42+4-2/3) = 144 I hope this helps.

👍 114 | 👎 -7

Cost of an apple (y) in terms of cost of an orange (x)—2 equations: 4x + 5y = $3.56 y = ($3.56 - 4x)/5 3x + 4y = $2.76 y = ($2.76 - 3x)/4 Cost of an orange—x: 4($3.56 - 4x) = 5($2.76 - 3x) $14.24 - 16x = $13.80 - 15x x = $0.44 Cost of an apple—1st equation: = ($3.56 - 4[$0.44])/5 = ($3.56 - $1.76)/5 = $1.80/5 or $0.36 Answer: orange, $0.44; apple, $0.36 ----------- No. of student tickets sold—x: $4.50x + $9.00(200 - x) = $1,485.00 x + 2(200 - x) = 330 x + 400 - 2x = 330 x = 70 No. of adult tickets sold: = 200 - 70 = 130 Answer: 70 student tickets; 130 adult tickets ----------- No. of tickets of those who held activity cards—x: $2.50x + $4.00(203 - x) = $620.00 5x + 8(203 - x) = 1,240 5x + 1,624 - 8x = 1,240 3x = 384 x = 128 No. of tickets of those who didn't hold activity cards: = 203 - 128 = 75 Answer: with activity cards, 128; without activity cards, 75

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