## anonymous one year ago Find....

?

2. anonymous

|dw:1437384213608:dw|

3. anonymous

I have tried diffrent approaches but end up with the same result that doesnt match the correct answer.

4. amoodarya

$t^2-3=u\\t=3 \rightarrow 9-3=u \rightarrow u=6\\t=4\rightarrow 16-3=u \rightarrow u=13\\t^2-3=u \rightarrow 2tdt=du \\\frac{tdt}{\sqrt{t^2-3}}=\frac{1}{2} \frac{du}{\sqrt{u}}$

5. UsukiDoll

u subsitution like what @amoodarya did

6. anonymous

|dw:1437384771141:dw|

7. UsukiDoll

and then take the antiderivative after u = ... du = dx steps

8. amoodarya

or see this $\frac{t}{\sqrt{t^2-3}}=\frac{2t}{2\sqrt{t^2-3}}=\frac{(t^-3)'}{2\sqrt{t^2-3}}\\=\frac{d}{dt}\sqrt{t^2-3}\\ \int\limits_{3}^{4}\frac{d}{dt}\sqrt{t^2-3} =\sqrt{t^2-3}|_{3,4}=\sqrt{3^2-3}-\sqrt{4^2-3}$

9. UsukiDoll

then plug u back in and evaluate f(4)-f(3)

10. anonymous

why doesnt mine work?

11. UsukiDoll

|dw:1437385040459:dw| find the derivative of u

12. anonymous

2t and then?

13. anonymous

ohhhh Now I get it...

14. UsukiDoll

|dw:1437385150396:dw| so we just need to divide 2 on both sides

15. UsukiDoll

then that tdt should be 1/2

16. UsukiDoll

|dw:1437385229449:dw|

17. UsukiDoll

|dw:1437385312000:dw|

18. UsukiDoll

|dw:1437385348540:dw|

19. UsukiDoll

|dw:1437385408712:dw|

20. UsukiDoll

now sub u back u = t^2 -3 |dw:1437385433088:dw|

21. UsukiDoll

|dw:1437385518425:dw|

22. UsukiDoll

|dw:1437385586233:dw| you should be getting this after evaluation

23. anonymous

Yeah, this was way easier. THanks!

24. UsukiDoll

I was double checking with wolfram we got the right answer, but they took it a step further by multiplying 4 throughout the answer |dw:1437385928137:dw|

25. UsukiDoll

wouldn't matter...it's the same thing :P

26. anonymous

Appreciate ur time and patience. Thank you:)

27. UsukiDoll

you're welcome :)