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Find the area of the region included between the parabolas y? = 4( 3+1)(x + 3 + 1), and y2 = 4(72 +1)32 +1 – x) 1+ 1 3

User Bourneli
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1 Answer

3 votes

Answer: look at explanation

Explanation:

To find the area between the two parabolas, we need to first find their intersection points. Let's start with the first parabola:

y² = 4(3+1)(x + 3 + 1) = 16(x + 4)

y = ± 4√(x + 4)

Now, let's work on the second parabola:

y² = 4(72 +1)32 +1 – x) 1+ 1/3

y² = 4(73)^(1/3) - 4x^(1/3) + 1

y = ± √(4(73)^(1/3) - 4x^(1/3) + 1)

We can set the two equations equal to each other to find the intersection points:

4√(x + 4) = ± √(4(73)^(1/3) - 4x^(1/3) + 1)

Squaring both sides and simplifying, we get:

16(x + 4) = 4(73)^(1/3) - 4x^(1/3) + 1

20x^(1/3) + 16x - 292 = 0

Let u = x^(1/3), then we have:

20u³ + 16u³ - 292 = 0

u³ = 73/5

u = (73/5)^(1/3)

x = u³ = 73/5

Now we can integrate to find the area between the parabolas:

A = ∫(from x=0 to x=73/5) [(4(73)^(1/3) - 4x^(1/3) + 1) - (4(x + 4))] dx

A = ∫(from x=0 to x=73/5) [4(73)^(1/3) - 4x^(1/3) - 4x - 15] dx

A = [4(73)^(1/3)x - 4(3/4)x^(4/3) - 2x² - 15x] (from x=0 to x=73/5)

A = 84(73)^(1/3)/5 - 292/5 - 438(73/5)^(4/3)/25 - 621/2

Therefore, the area between the parabolas is approximately 449.428 square units.

User Theutz
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