I need to find the average value of the function f(x)=√(2x+1) ; [0,4]
The back of the book is saying the answer is 13/6 but i keep getting 4.333 (which is double the answer of the book). Can someone please help me I don't see where I'm going wrong and i want to understand how to do these problems.
The back of the book is saying the answer is 13/6 but i keep getting 4.333 (which is double the answer of the book). Can someone please help me I don't see where I'm going wrong and i want to understand how to do these problems.
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y_avg = (∫(from 0 to 4)(√(2x+1)dx))/(4-0)
∫√(2x+1)dx
Set u = √(2x+1)
u^2 = 2x + 1
2udu = 2dx
udu = dx
∫√(2x+1)dx = ∫u^2dx = (u^3)/3 = ((√(2x+1))^3)/3
Evaluate this from (0,4) we get 26/3
So y_avg = (26/3)/4 = 26/12 = 13/6
∫√(2x+1)dx
Set u = √(2x+1)
u^2 = 2x + 1
2udu = 2dx
udu = dx
∫√(2x+1)dx = ∫u^2dx = (u^3)/3 = ((√(2x+1))^3)/3
Evaluate this from (0,4) we get 26/3
So y_avg = (26/3)/4 = 26/12 = 13/6