Problem 1 (i) Determine the (x, y) coordinates of the points of intersection for the curves described by the following equations.

wstecznyg5 2022-07-25 Answered
Problem 1
i) Determine the (x, y) coordinates of the points of intersection for the curves described by the following equations.
y 1 = x 2 4 x + 3 y 2 = x 2 + 2 x + 3
ii) Set up a definite integral to give the area of the region bounded by the two curves
iii) Evaluate the definite integral to give the area
Proble 2
Same as i, ii, iii in problem 1 above for following equations:
y 1 = x 2 y 2 = x 3
Problem 3
Same as i, ii, iii in problem 1 above for following equations:
x = 4 y 2 x = y 2
Problem 4
Same as i, ii, iii in problem 3 above for the following equations:
y = x 2 y = 6 x
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Answers (2)

eyiliweyouc
Answered 2022-07-26 Author has 15 answers
Step 1
y 1 = x 2 4 x + 3 y 2 = x 2 + 2 x + 3 y 1 = y 2 x 2 4 x + 3 = x 2 + 2 x + 3 2 x 2 6 x = 0 x ( 2 x 6 ) = 0 x = 0 ,   x = 3 x = 0 y = 0 + 0 + 3 y = 3 ( 0 ,   3 ) x = 3 y = 0 ( 3 ,   0 )
i) ( x ,   y ) coordints   ( 0 ,   3 )   ( 3 ,   0 ) y = x 2 4 x + 3 y = 3 x = 1 ,   y = 0 x = 2 ,   y = 1 y = x 2 + 2 x + 3 x = 0 ,   x = 1 y = 3 ,   y = 4 x = 2 A = a b f ( x ) g ( x ) d x f ( x ) = x 2 + 2 x + 3 g ( x ) = x 2 4 x + 3
ii) f ( x ) g ( x ) x 2 + 2 x + 3 x 2 + 4 x 3 = 2 x 2 + 6 x
iii) A = 0 3 2 x 2 + b × d x 2 0 3 x 2 d x + 6 0 3 x d x 2 | x 3 3 | 0 3 + 6 | x 2 2 | 0 3 2 3 [ 27 ] + 6 2 [ 9 ] = 2 3 [ 27 9 ] + 3 [ 9 ] = 18 + 27 9 m 2
Step 2
y 1 = x 2 y 2 = x 3 y 1 = y 2 x 2 = x 3 x 2 x = x 2 x = 1 y = 1 i )   x , y   coordinats   ( x ,   y ) = ( 1 , 1 ) i i ) f ( x ) g ( x ) ? f ( x ) = x 2 g ( x ) = x 3 x 2 x 3 i i i ) 0 1 x 2 x 3 d x | x 3 3 | 0 1 | x 4 4 | 0 1 1 3 1 4 4 3 12 1 2 1 12 m 2

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Matias Aguirre
Answered 2022-07-27 Author has 3 answers
Step 1
Problem 3
When x = 4 y 2 ,   x = y 2 intersect
4 y 2 = y 2 y 2 4 + y 2 = 0 y 2 + y 6 = 0 ( y + 3 ) ( y 2 ) = 0 y = 3 ,   y = 2 y = 3 x = 3 2 x = 5 y = 2 x = 2 2 x = 0
points of intersection are (-5, -3), (0, 2)
ii) are of the region = 3 2 ( ( 4 y 2 ) ( y 2 ) ) d y
iii) are of the region:
⇒= 3 2 ( 4 y 2 y + 2 ) d y ⇒= 3 2 ( 6 y 2 y ) d y ⇒= 3 2 ( 6 y ( 1 3 ) y 3 ( 1 2 ) y 2 ) ⇒= ( 6 ( 2 ) ( 1 3 ) ( 2 ) 3 ( 1 2 ) ( 2 ) 2 ) ( 6 ( 3 ) ( 1 3 ) ( 3 ) 3 ( 1 2 ) ( 3 ) 2 ) ⇒= ( 12 ( 8 3 ) 2 ) ( 18 + 9 ( 9 2 ) ) ⇒= 12 ( 8 3 ) 2 + 18 9 + ( 9 2 ) ⇒= 19 ( 8 3 ) + ( 9 2 ) ⇒= 125 6 square units
Step 2
Problem 4
y = x 2 ,   y = 6 x i )   equating   x 2 = 6 x x 2 + x 6 = 0 x 2 + 3 x 2 x 6 = 0 x 2 + 3 x 2 x 6 = 0 x 2 + 3 x 2 x 6 = 0 x ( x + 3 ) 2 ( x + 3 ) = 0 ( x + 3 ) ( x 2 ) = 0 x = 3 ,   2 ( 3 ,   9 ) ,   ( 2 ,   4 ) i i )   Area   a b ( f ( x ) g ( x ) ) d x = 3 2 ( 6 x x 2 ) d x i i i )   A = [ 6 x x 2 2 x 3 3 ] 3 2 = ( 12 4 2 8 3 ) ( 28 9 2 + 27 3 ) = 125 6

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