calculus problem

[Ashford]

hey,
I've got another calculus problem. The text that I'm using gives an example but the solution to the problem seems a bit incorrect. So I just want to know if it's a mistake by the author or is it actually correct and if so can you explain it to me? Here's the problem





I think that the solution should actually be:



since the anti-derivative of 30 is suppose to be 30t

 

[Ratchit]

Ashford,

What is the variable of integration? Is it "t"? If so, there should be a dt under the ∫ sign? If the variable is "t", then the answer is 20(a^(3/2) - b^(3/2))/(b-a)

Ratch

 

[tapeworm]

Ratchit is correct, so long as you are integrating with respect to t. (dt)
You should get:
20/(b-a) * [a^(3/2) - b^(3/2)]


Mistake:
30 t^(1/2) is together, you don't split it up and integrate each individual part. That is your mistake.

30 t^(1/2) becomes 20t^(3/2) and not 30t [2/3t^(3/2)]

 

[MrAl]

Hi,

Yes normally the "dt" would also be shown within the integrand so we are sure of what the variable of integration is.
For this it is most likely 't' and so we would first move the constant '30' outside of the integral, then treat sqrt(t) as t^(1/2), then follow the rule of exponents for integrals which is to increase the exponent by 1 and then divide by that result:

Integral(x^n)=(x^(n+1))/(n+1)