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Sketch the region bounded by the curves 5x^2 +y=33 and x^4-y−=3 , then find the area of the region.

User Panofish
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1 Answer

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The area bounded by the curves is 104.53 unit².

How to calculate area between curves.

Given

5x² + y = 33 and x⁴ - y = 3

Rearrange the second equation to y = x⁴ - 3 then substitute this expression into the first equation:

5x² + (x⁴ - 3) = 33

Combine like terms:

x⁴ + 5x² - 36 = 0

Now, factor the quadratic:

(x² - 4)(x² + 9) = 0

This gives two sets of solutions:

x² - 4 = 0, x = +-2

x² + 9 = 0 has no real solutions.

So, the points of intersection occur at x = -2 and x = 2.

The area A between the curves can be found using the integral:

A = ∫₋₂²[(33-5x²) - (x⁴- 3)]dx

A = ∫₋₂²[(33-5x²- x⁴ + 3)]dx

Simplify further:

A = ∫₋₂²[(36-5x² - x⁴)]dx

Now, integrate with respect to x

A = [(36x -5/3x³ - 1/5x⁵)]₋₂²

Evaluate the expression at the upper and lower limits:

A = [(36(2) -5/3(2)³ - 1/5(2)⁵) - [(36(-2) -5/3(-2)³ - 1/5(-2)⁵)]

A = [(72 -40/3 - 32/5) - [( -72 + 40/3 + 32/5)

= 144 - 80/3 - 64/5

= 104.53 unit²

Sketch the region bounded by the curves 5x^2 +y=33 and x^4-y−=3 , then find the area-example-1
User PEZ
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