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Travka [436]
3 years ago
12

Find the number b such that the line y = b divides the region bounded by the curves y = 25x2 and y = 1 into two regions with equ

al area. (Round your answer to two decimal places.)
Mathematics
1 answer:
Dennis_Churaev [7]3 years ago
4 0

Answer:

y=1/∛4 divides the area in half

Step-by-step explanation:

since the minimum value of x² is 0 (for x=0 ) and for y=1

1 = 25*x² → x= ±√(1/25) = ±1/5

then the total area between y=1  and y = 25*x²  is bounded to  x=±1/5 and y=0 . Since there is a direct relationship between x and y , we can find the value of x=a that divides the region in 2 of the same area. thus

Area below x=C = Area above x=C

Area below x=C = Total area -  Area below x=C

2*Area below x=C = Total area

Area below x=C = Total area /2

∫ 25*x² dx from x=c to x=-c =  1/2 ∫ 25*x² dx from x=1/5  to x=-1/5

25*[c³/3 - (-c)³/3] = 25/2 * [(1/5)³/3 - (-1/5)³/3]

2*c³/3 = (1/5)³/3

c = 1/(5*∛2)

thus

y=25* x²  = 25*[1/(5*∛2)]² = 1/∛4

thus the line y=1/∛4 divides the area in half

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Check the picture below.


based on the equation, if we set y = 0, we'd end up with 0 = 0.5(x-3)(x-k).

and that will give us two x-intercepts, at x = 3 and x = k.

since the triangle is made by the x-intercepts and y-intercepts, then the parabola most likely has another x-intercept on the negative side of the x-axis, as you see in the picture, so chances are "k" is a negative value.

now, notice the picture, those intercepts make a triangle with a base = 3 + k, and height = y, where "y" is on the negative side.

let's find the y-intercept by setting x = 0 now,


\bf y=0.5(x-3)(x+k)\implies y=\cfrac{1}{2}(x-3)(x+k)\implies \stackrel{\textit{setting x = 0}}{y=\cfrac{1}{2}(0-3)(0+k)} \\\\\\ y=\cfrac{1}{2}(-3)(k)\implies \boxed{y=-\cfrac{3k}{2}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of a triangle}}{A=\cfrac{1}{2}bh}~~ \begin{cases} b=3+k\\ h=y\\ \quad -\frac{3k}{2}\\ A=1.5\\ \qquad \frac{3}{2} \end{cases}\implies \cfrac{3}{2}=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)


\bf \cfrac{3}{2}=\cfrac{3+k}{2}\left( -\cfrac{3k}{2} \right)\implies \stackrel{\textit{multiplying by }\stackrel{LCD}{2}}{3=\cfrac{(3+k)(-3k)}{2}}\implies 6=-9k-3k^2 \\\\\\ 6=-3(3k+k^2)\implies \cfrac{6}{-3}=3k+k^2\implies -2=3k+k^2 \\\\\\ 0=k^2+3k+2\implies 0=(k+2)(k+1)\implies k= \begin{cases} -2\\ -1 \end{cases}


now, we can plug those values on A = (1/2)bh,


\bf \stackrel{\textit{using k = -2}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-2)\left(-\cfrac{3(-2)}{2} \right)\implies A=\cfrac{1}{2}(1)(3) \\\\\\ A=\cfrac{3}{2}\implies A=1.5 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{using k = -1}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-1)\left(-\cfrac{3(-1)}{2} \right) \\\\\\ A=\cfrac{1}{2}(2)\left( \cfrac{3}{2} \right)\implies A=\cfrac{3}{2}\implies A=1.5

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