Suppose f(x)=∫ on top x, on bottom 1, √t^2-6t+9dt.For which positive value of x does f'(x) equal 0 ?
And evaluate the integral ∫ on top 7, on bottom 0,√t+1dt
Please, please help? Im so confused
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And evaluate the integral ∫ on top 7, on bottom 0,√t+1dt
Please, please help? Im so confused
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for the first one,
sqrt(t^2-6t+9) can be rewritten as (t-3)^2, so the integral becomes ...
∫ (t-3) from 1 to x.
= t^2/2-3t from 1 to x
= ((x)^2/2 - 3(x)) - ( (1)^2/2 - 3(1))
= (x^2/2)-3x+(5/2)
=> f'(x) = x-3,
it only equals 0 when x = 3
∫ √t+1 dt
= ∫ (t+1)^(1/2) dt
= (2/3)(t+1)^(3/2) *1 from 0 to 7
= (32sqrt(2)/3) - (2/3)
= (-2+32sqrt(2))/3
= 14.4183