Consider the two curves y = x^2 and y = 5 − x.

Question

Consider the two curves y = x² and y = 5 − x.

(a) Sketch the graphs of the two curves and shade the are between them in the first quadrant.

(b) Find the area of this region using integration with respect to x.

(c) Find the area of this region using integration with respect to y

Answer 1

find where they intersect

y=x^2 is an upward opening parabola with vertex=minimum point = the origin

y=5-x is a downward sloping straight line with intercepts (0,5) and (5,0)

x^2 = 5-x

x^2+x-5 =0

x=-1/2+/-(1/2)sqr(1+4(5))

x= -.5+.5sqr21,-.5-.5sqr21

use the larger for the upper limit of integration when integrating with respect to x

and use x=0 as the lower limit

find the y coordinate of the intersection points

5-x=y

x=5-y

x=+/-sqry=5-y

y = 25-10y+y^2

y^2-11y +25= 0

y=11/2 +/-(1/2)sqr(121-100)

y=5.5+.5sqr21, 5.5-.5sqr21 only the latter is in quadrant 1

use it as a limit of integration when integrating with respect to y

use y=5 as another limit where x=0 intersects y=5-x

and y=0 as another limit, the y coordinate of the lower bound of the region

part of the reigion has the line has the right boundary, from y=5.5-.5sqr21 to y=5

the other part has the parabolic curve for the right boudary from 0 to 5.5-5sqr21

the region is bounded on top by the line, and on the bottom by the parabola

integral of (5-x)-x^2 dx

= 5x- .5x^2 - (1/3)x^3

evaluated between zero and .5+.5sqr21

= 5(.5+.5sqr21) -.5(.5+.5sqr21)^2-(1/3)(.5+.5sqr21)^3 - zero

expand and simply, combine like terms

= 2.5+2.5sqr21- .125 -.25sqr21 - .125(21)-(1/3)(.25+.5sqr21+.25(21))(.5+.5sqr21)

= 2.375+2.25sqr21-2.625 - (1/3)(.125+.25sqr21+2.6375+ .125sqr21 +.25(21) +2.6375sqr21)

=-2.25+ 2.25sqr21-(1/3)(5.4+3.0125sqr21)

=

do a similar procedure for integration with respect to y, but break it into 2 definite integrals as you have 3 limits of integration

no guarantees the above calculations are error free, as it gets a little tedious