Consider the area between the graphs x+y=25 and x+5=y2 . This area can be computed in two different ways using integrals

Question

Consider the area between the graphs x+y=25 and x+5=y2 . First of all it can be computed as a sum of two integrals ∫baf(x)dx+∫cbg(x)dx where a= , b= , c= and f(x)= g(x)= Alternatively this area can be computed as a single integral ∫βαh(y)dy where α= , β= and h(y)= Either way we find that the area is .

Answer 1

Let's rewrite each function:

x+y=25 --> x = 25-y

x+5=y² --> x = y²-5

Now we can determine the area between the two curves. Consider where they intersect:

25-y = y²-5

0 = y²+y-30

0 = (y+6)(y-5)

Thus, we can tell that our bounds are from α=-6 to β=5. Hence, our integral is:

∫[-6,5] [(25-y)-(y²-5)] dy

= ∫[-6,5] (-y²-y+30) dy

= -y³/3 - y²/2 + 30y [-6,5]

= [-5³/3 - 5²/2 + 30(5)] - [-(-6)³/3 - (-6)²/2 + 30(-6)]

= (-125/3 - 25/2 + 150) - (216/3 - 36/2 - 180)

= 1331/6

Hope this helped!

Answer 2

area between x+y = 25 and x+5=y^2

it helps to graph the line and parabola

the line has y intercept (25,0) and x intercept (25,0)

the parabola has vertex (-sqr5) and is rightward opening

it intersects the line at (20,5) and (31,-6)

y=-x+25

y=+/-sqr(x+5)

Break the area between them into 5 subareas

2 above the x axis and 3 below

above the x axis is the integral of sqr(x+5) evaluated from x=-5 to 20

then add 12 1/2, which is the integral of the line from x=20 to 25

below the x axis are 3 areas

integral of the parabola from x=-5 to 25

then add 3sqr30

and a 3rd area = 144 -20sqr30

the 5 areas sum to 239 5/6 + 3sqr30

= about 256.265

integrate from

Answer 3

curves intersect (20,5) and (31,-6)

a=-5

b=20

c=31

f(x)=2((x+5)^(1/2))

g(x)=(-x+25)+(x+5)^(1/2)

alpha=-6

beta=5

h(y)=(25-y)-(y²-5)=-y²-y+30

correction

Area=1431/6,,,,it is best to compute with dy