Find the area of the region enclosed by y=2x and y=x^2-15.
2007-11-26 01:21:48 · 5 answers · asked by pearlnruby89 1 in Science & Mathematics ➔ Mathematics
Intersect when:-
2x = x² - 15
x² - 2x - 15 = 0
(x - 5)(x + 3) = 0
x = 5 , x = - 3
A = ∫ 2x - (x² - 15) dx between lims of - 3 and 5
A = x² - x³/3 + 15x between - 3 and 5
A = [ 25 - 125/3 + 75 ] - [ 9 + 9 - 45 ]
A = 100 - 125/3 + 27
A = 127 - 125/3
A = 381/3 - 125/3
A = 256 / 3
Area is 85 / 3 units ²
2007-11-28 19:39:43 · answer #1 · answered by Como 7 · 1⤊ 1⤋
You do know how to solve this type of question right ?
OK. First we find where the two curves meet.
2x = x^2 - 15
=> x = 5, -3.
Thus area lies between x = -3 and x = 5. Taking x = 0, we find that y = 2x lies above y = x^2 - 15.
Now integrate.
Area = Integration of (2x - x^2 + 15) from -3 to +5.
= x^2 -x^3 / 3 + 15x from -3 to+5
= 25 - 125/3 + 75 - 9 - 9 + 45
= 256 / 3
2007-11-26 01:34:50 · answer #2 · answered by Anonymous · 2⤊ 1⤋
Integrate them to give you y=x^2 and y=(1/3)x^3-15x and then work out the area for each by making x the two places where the curves meet and then subtract one from the other.
2007-11-26 01:38:21 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Solve 2x = x^2 - 15 for the limits of integration x = -3 and x = 5
Now integrate [2x - (x^2 -15) ] with these limits
2007-11-26 01:32:21 · answer #4 · answered by lienad14 6 · 1⤊ 1⤋
im am so happy i dont have your class
2007-11-26 01:25:05 · answer #5 · answered by Anonymous · 1⤊ 0⤋