How do you integrate this problem?

[Ghost3k]

Evaluate the following definite integral:

ok so I tried using u substitution.

u = t^1/2
du = 1/(2sqrt(t)) dt
but then we need to have t-1 dt so does it become
(t-1)(2sqrt(t))du = t-1 dt?
If so, what do I do after that.

 

[Deleted member 4993]

Evaluate the following definite integral:

ok so I tried using u substitution.

u = t^1/2
du = 1/(2sqrt(t)) dt
but then we need to have t-1 dt so does it become
(t-1)(2sqrt(t))du = t-1 dt?
If so, what do I do after that.

Rather than substituting ....

\(\displaystyle \int_4^9\dfrac{t-1}{\sqrt{t}}dt\)

\(\displaystyle = \int_4^9\ t^{\frac{1}{2}} dt - \int_4^9 t^{-\frac{1}{2}}dt\)

can you continue now.....

 

[Ghost3k]

Rather than substituting ....

\(\displaystyle \int_4^9\dfrac{t-1}{\sqrt{t}}dt\)

\(\displaystyle = \int_4^9\ t^{\frac{1}{2}} dt - \int_4^9 t^{-\frac{1}{2}}dt\)

can you continue now.....

thanks, I was able to follow your way and I was able to get the answer. I got the answer to be 32/3.
However could you explain to me how you got from \(\displaystyle \int_4^9\dfrac{t-1}{\sqrt{t}}dt\) to \(\displaystyle \int_4^9\ t^{\frac{1}{2}} dt - \int_4^9 t^{-\frac{1}{2}}dt\)

 

[Deleted member 4993]

thanks, I was able to follow your way and I was able to get the answer. I got the answer to be 32/3.
However could you explain to me how you got from \(\displaystyle \int_4^9\dfrac{t-1}{\sqrt{t}}dt\) to \(\displaystyle \int_4^9\ t^{\frac{1}{2}} dt - \int_4^9 t^{-\frac{1}{2}}dt\)

By "division"!!!

Think about it.

\(\displaystyle \dfrac{t}{t^{\frac{1}{2}}} \ = \ ???\)

 

[Ghost3k]

By "division"!!!

Think about it.

\(\displaystyle \dfrac{t}{t^{\frac{1}{2}}} \ = \ ???\)

Oh I see!!! Such simpleness seemed so complex!! Thanks