Asked by John
How do you solve the definite integral of sqrt( (t+1)^2 ) from 0 to 1. I tried using u-substitution, but I am not sure if this is the way to go?
Answers
Answered by
oobleck
huh? for 0 <= t <= 1
√(t+1)^2 = t+1
∫[0,1] t+1 dt = 1/2 t^2 + t [0,1] = 3/2
You can use u-substitution, using u=t+1, so dt = du and
∫[0,1] √(t+1)^2 dt = ∫[1,2] √u^2 du = ∫[1,2] u du = 1/2 u^2 [1,2] = 3/2
what a strange integral, using √u^2. It might have been more interesting using a wider domain, since √u^2 = |u| so having u<0 would have required a little more manipulation.
√(t+1)^2 = t+1
∫[0,1] t+1 dt = 1/2 t^2 + t [0,1] = 3/2
You can use u-substitution, using u=t+1, so dt = du and
∫[0,1] √(t+1)^2 dt = ∫[1,2] √u^2 du = ∫[1,2] u du = 1/2 u^2 [1,2] = 3/2
what a strange integral, using √u^2. It might have been more interesting using a wider domain, since √u^2 = |u| so having u<0 would have required a little more manipulation.
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