SOLUTION :
Average value of the given function in interval 0 to 2 :
2
= ∫ f(x) dx / (2 - 0)
0
2
= 1/2 ∫ 5x^2 sqrt(1 + x^3) dx
0
Let 1 + x^3 = t
=> 3x^2 dx = dt
=> 5 x^2 dx = 5dt/3
At x = 0, t = 1 and at x = 2, t = 9
So, average value :
9 9
= 1/2 ∫ 5/3 sqrt(t) dt = 5/6 * 2/3 [ t^(3/2) ]
1 1
= 5/9 [9^(3/2) - 1^(3/2)]
= 5/9 * 26
= 130/9 = 14.44 (ANSWER)
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