To find area under curve using integral calculus

Question

• Area bounded by y2=x-1 and the line y=x-3

Answer 1

y^2 + 1 = x  

 

Substituting into the second equation:  y = y^2 + 1 - 3

 

                                                        0 = y^2 - y - 2

                                                         0 = ( y     -  2 )( y +    1    )

                                                              y-2 = 0 ---> y = 2  ---> x  = (2)^2 +1 = 4 +1 = 5 

 

                                                               y+1 = -1 --> x = (-1)^2 + 1 = 1+1 = 2

 

They meet at (5,2) and (2,-1)

Parabola is bounded on the right by the line.

 

 

integral    y+3 - (y^2+1)   dy   ,    y=[-1 , 2]

 

   = integral  y + 3 - y^2 - 1  dy 

 

   = integral -y^2 + y + 2 dy

 

     [ -y^3/3 + 1/2y^2 + 2y ] 

 

  upper limit: -(2)^3/3 + (1/2)(2^2) + 2(2)  = -8/3 + 2 + 4 = 6 - 8/3 = 18/3 - 8/3 = 10/3

 lower limit:  [ -(-1)^3/3 + (1/2)(-1)^2 + 2(-1) ] =  1/3 + 1/2 - 2 = 1/3-3/2 = 2/6-9/6 = -7/6

 

10/3 - (-7/6) = 10/3 + 7/6 = 20/6 + 7/6 = 27/6 = 17/3 = 9/2

 

  

Answer 2

To find the area between two curves, you subtract the integrals.  As these two lines intersect at two points, you first need to solve the system of equations to determine the bounds for the integrals.  

 

y² = x - 1

y = x - 3

 

can be solved by substitution.

Once you get the x values, you can plug them back into the second equation to show the lines intersect at (2,-1) and (5,2)

 

As the y has an exponent in the first equation, solve both equations for x to put the equations in terms of y to make integration easier.  You will be integrating in terms of y from the y values of the points above, from -1 to 2.  Be sure to subtract the integral of the right most function on the graph (x=y+3) from the integral of the leftmost function (x=y²+1) on that interval.