## mariomintchev 3 years ago i need help with #17

1. mariomintchev

2. mariomintchev

i know the chain rule is used

3. TuringTest

not really, it should just be a u-sub$u=t+4$$du=dt$

4. mariomintchev

i need more help than that

5. TuringTest

well if you use that sub what does this integral become$\int_{0}^{x}\sqrt{t+4}dt$?write it in terms of u

6. mariomintchev

yeah im stuck at 1/2(t+4)^(-1/2)

7. TuringTest

you took the derivative or something...

8. TuringTest

we are integrating here, that's the opposite

9. mariomintchev

that response was to whoever said use the power rule... they deleted their comment

10. TuringTest

they just didn't quite understand your question I think $\int_{0}^{x}\sqrt{t+4}dt$$u=t+4\implies du=dt$now rewrite the integral in terms of u

11. mariomintchev

im not sure how to do that

12. TuringTest

$\int_{0}^{x}\sqrt{t+4}dt$$u=t+4\implies du=dt$so our integral (indefinite for the time being) is now$\int\sqrt udu$which you should be able to integrate

13. TuringTest

notice we put the $$u$$ where the $$t+4$$ was and the $$du$$0 where the $$dt$$ was. that's the essence of how a u-sub in integration works

14. mariomintchev

the only part im not sure how to get is the - 16 / 3

15. TuringTest

what do you get after integrating?

16. TuringTest

after integrating you get ________ evaluated from __ to __ fill in the blanks

17. mariomintchev

t+4 0 x

18. TuringTest

not quite... would it help you if I reminded you that we can rewrite this as$\int u^{1/2}du$??

19. TuringTest

then you can use the power rule for $$integration$$ (basically the opposite of the power rule for differentiation)

20. TuringTest

which is$\int x^ndx=\frac{x^{n+1}}{n+1}$

21. mariomintchev

can you solve the problem so i can see how its done? ill understand it better that way.

22. TuringTest

I will try to guide you a little farther than I normally would, but I hope you don't ask the exact same question again later!

23. mariomintchev

i try not to do that unless i forget a step or something

24. TuringTest

$\int_{0}^{x}\sqrt{t+4}dt$$u=t+4\implies du=dt$so our integral (indefinite for the time being) is now$\int\sqrt udu=\int u^{1/2}du=\frac23u^{3/2}$now put it back in terms of t and put in the evaluation$\int_{0}^{x}\sqrt{t+4}dt=\frac23(t+4)^{3/2}|_{0}^{x}$evaluate this $$carefully$$. Now do you see where the last term comes from?

25. mariomintchev

not really lol

26. TuringTest

ok full solution:

27. TuringTest

$\int_{0}^{x}\sqrt{t+4}dt$$u=t+4\implies du=dt$so our integral (indefinite for the time being) is now$\int\sqrt udu=\int u^{1/2}du=\frac23u^{3/2}$now put it back in terms of t and put in the evaluation$\int_{0}^{x}\sqrt{t+4}dt=\frac23(t+4)^{3/2}|_{0}^{x}=\frac23(x+4)-\frac23(0+4)^{3/2}$$=\frac23(x+4)-\frac23(4)^{3/2}=\frac23(x+4)-\frac23(8)=\frac23(x+4)-\frac{16}3$please don't ask me to give the final answer again, it's very buch not my style... I hope this makes sense to you after looking it over. Good luck!

28. TuringTest

very much*

29. mariomintchev

i wont. thanks for your time/help!

30. TuringTest

*typo above (I dropped the ^3/2 in the last few steps)$\int_{0}^{x}\sqrt{t+4}dt$$u=t+4\implies du=dt$so our integral (indefinite for the time being) is now$\int\sqrt udu=\int u^{1/2}du=\frac23u^{3/2}$now put it back in terms of t and put in the evaluation$\int_{0}^{x}\sqrt{t+4}dt=\frac23(t+4)^{3/2}|_{0}^{x}=\frac23(x+4)^{3/2}-\frac23(0+4)^{3/2}$$=\frac23(x+4)^{3/2}-\frac23(4)^{3/2}=\frac23(x+4)^{3/2}-\frac23(8)=\frac23(x+4)^{3/2}-\frac{16}3$

31. mariomintchev

ok i see what was done. thanks. :D

32. TuringTest

very welcome :)