Asked by Leah
Calculate d/dt when a=t^2, b=2 and sqrt(x+1)dx.
My answer turns out to be incorrect despite my checking it several times. I got -2/3t* sqrt(t^2+1)
Can someone explain to me what is wrong with my answer and how to go about obtaining it the right way? Thank you everyone. :-)
My answer turns out to be incorrect despite my checking it several times. I got -2/3t* sqrt(t^2+1)
Can someone explain to me what is wrong with my answer and how to go about obtaining it the right way? Thank you everyone. :-)
Answers
Answered by
Steve
f(t) = ∫[2,t^2] √(x+1) dx
df/dt = √(t^2+1) (2t)
Recall that d/dt ∫[a,b(t)] f(x) dx = f(b(t)) db/dt
see wikipedia's article on differentiating under the integral sign.
df/dt = √(t^2+1) (2t)
Recall that d/dt ∫[a,b(t)] f(x) dx = f(b(t)) db/dt
see wikipedia's article on differentiating under the integral sign.
Answered by
Leah
Thank you for the equation and the useful article. :)
So this is what I get.
f(t) dt= 2(0)- [sqrt(t^2+1) * 2t]
f(t) dt= -sqrt(t^2+1) *2t
Is this correct? Thank you so much. :-)
So this is what I get.
f(t) dt= 2(0)- [sqrt(t^2+1) * 2t]
f(t) dt= -sqrt(t^2+1) *2t
Is this correct? Thank you so much. :-)