Problem 1i) 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 + 3ii) Set up a definite integral to give the area of the region bounded by the two curvesiii) Evaluate the definite integral to give the areaProble 2Same as i, ii, iii in problem 1 above for following equations: y 1 = x 2 y 2 = x 3 Problem 3Same as i, ii, iii in problem 1 above for following equations: x = 4 − y 2 x = y − 2Problem 4Same as i, ii, iii in problem 3 above for the following equations: y = x 2 y = 6 − x

Question

Problem 1i) 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  +  3ii) Set up a definite integral to give the area of the region bounded by the two curvesiii) Evaluate the definite integral to give the areaProble 2Same as i, ii, iii in problem 1 above for following equations:      y    1    =      x    2          y    2    =      x    3  Problem 3Same as i, ii, iii in problem 1 above for following equations:  x  =  4  −      y    2      x  =  y  −  2Problem 4Same as i, ii, iii in problem 3 above for the following equations:  y  =      x    2      y  =  6  −  x

eyiliweyouc · Accepted Answer

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  +  3ii)   f  (  x  )  −  g  (  x  )  ⇒  −      x    2    +  2  x  +            ⧸            3    −      x    2    +  4  x  −            ⧸            3      =  −  2      x    2    +  6  xiii)   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

Matias Aguirre · Accepted Answer

Step 1Problem 3When   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  =  0points of intersection are (-5, -3), (0, 2)ii) are of the region   =      ∫          −      3        2    (  (  4  −      y    2    )  −  (  y  −  2  )  )  d  yiii) 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 unitsStep 2Problem 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

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

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