B1 = 2
b2 = (b1)^2 + 1 = 2^2 + 1 = 5
b3 = (b2)^2 + 1 = 5^2 + 1 = 26
b4 = (b3)^2 + 1 = 26^2 + 1 = 676+1=<span>677</span>
Because of the symmetry, we can just go from x=0 to x=2 to find the area between
<span>y = x^2 and y = 4 </span>
<span>that area = ∫4-x^2 dx from 0 to 2 </span>
<span>= [4x - (1/3)x^3] from 0 to 2 </span>
<span>= 8 - 8/3 - 0 </span>
<span>= 16/3 </span>
<span>so when y = b </span>
<span>x= √b </span>
<span>and we have the area as </span>
<span>∫(b - x^2) dx from 0 to √b </span>
<span>= [b x - (1/3)x^3] from 0 to √b </span>
<span>= b√b - (1/3)b√b - 0 </span>
<span>(2/3)b√b = 8/3 </span>
<span>b√b =4 </span>
<span>square both sides </span>
<span>b^3 = 16 </span>
<span>b = 16^(1/3) = 2 cuberoot(2) </span>
<span>or appr 2.52</span>
Answer:
x = 0
Step-by-step explanation:
if 2y + x = 4
we can assume that y = 2
if y - 3x = 2
it means 2 - 0 = 2
therefore x = 0
