The Area of inclosed region between two curves , f(x) = x² + 2 and g(x) = 10x² is 8√2/9.
Let y = f(x) and y = g(x) be functions, such that f (x) ≥ g(x). Define the interval of x to be [ a,b ]. The area A of the region formed by the two curves at the given interval is then computed by
A = ₐ∫ᵇ[ f(x) - g(x) ] dx
the integrand is the lower function subtracted from the upper function.
We have given that, Y = 10 x² and Y=x²+2
First we find the point of intersection of the curves, 10x² = x² +2
=> 10x² - x² = 2
=> 9x² = 2
=> x² = 2/9
=> x = ± √2/9 = ± √2/3
so, - √2/3 ≤ x ≤√2/3
Now, we determined the area of enclosed region between two curves.
A = ₐ∫ᵇ[ f(x) - g(x) ] dx = ∫( x² + 2 - 10x²)dx here, b=√2/3 ≤ x ≤ a =-√2/3
∫ ( -9x² + 2)dx = [-9x³/3 + 2x] , now put the upper limit is √2/3 and lower limit - √2/3
A= [- 3(√2/3)³+2√2/3 + 3(-√2/3)³- (-2√2/3 )]
A = - 3×2×2√2/27 + 4√2/3
= - 4√2/9 +4√2/3
= 8√2/9
Hence, area of enclosed region is 8√2/9.
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