Find the average value of the function over the indicated interval
f(x)=x^2-3 on [-1,1]
f(x)=x^2-3 on [-1,1]
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The average value = 1/(b-a) * INT f(x)dx on (a,b)
1/(2) INT [ x^2-3] dx on (-1,1)
=(1/2) [ x^3/3- 3x] |(-1,1)
= (1/2) [ (1/3 -3)- (-1/3 +3]
= (1/2) [ 2/3 -6]
= -8/3
Hoping this helps!
1/(2) INT [ x^2-3] dx on (-1,1)
=(1/2) [ x^3/3- 3x] |(-1,1)
= (1/2) [ (1/3 -3)- (-1/3 +3]
= (1/2) [ 2/3 -6]
= -8/3
Hoping this helps!
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Average =∫f(x)dx / ∫dx
Area ∫f(x)dx=∫(x^2-3)dx between x= -1 and +1
=[x^3/3-3x+C] between -1 and +1
=[1/3-3+C]-[-1/3+3+C]=-5⅓
∫dx=[x] -1 to +1 = [1]-[-1]=2
Average = -5⅓/2= -2.66'
Area ∫f(x)dx=∫(x^2-3)dx between x= -1 and +1
=[x^3/3-3x+C] between -1 and +1
=[1/3-3+C]-[-1/3+3+C]=-5⅓
∫dx=[x] -1 to +1 = [1]-[-1]=2
Average = -5⅓/2= -2.66'
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[ f(1) - f(-1) ] / [ 1 - (-1) ] = slope of secant line = answer