## mariomintchev Group Title i need help with #17 2 years ago 2 years ago

1. mariomintchev Group Title

2. mariomintchev Group Title

i know the chain rule is used

3. TuringTest Group Title

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

4. mariomintchev Group Title

i need more help than that

5. TuringTest Group Title

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 Group Title

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

7. TuringTest Group Title

you took the derivative or something...

8. TuringTest Group Title

we are integrating here, that's the opposite

9. mariomintchev Group Title

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

10. TuringTest Group Title

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 Group Title

im not sure how to do that

12. TuringTest Group Title

$\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 Group Title

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 Group Title

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

15. TuringTest Group Title

what do you get after integrating?

16. TuringTest Group Title

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

17. mariomintchev Group Title

t+4 0 x

18. TuringTest Group Title

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

19. TuringTest Group Title

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

20. TuringTest Group Title

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

21. mariomintchev Group Title

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

22. TuringTest Group Title

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 Group Title

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

24. TuringTest Group Title

$\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 Group Title

not really lol

26. TuringTest Group Title

ok full solution:

27. TuringTest Group Title

$\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 Group Title

very much*

29. mariomintchev Group Title

i wont. thanks for your time/help!

30. TuringTest Group Title

*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 Group Title

ok i see what was done. thanks. :D

32. TuringTest Group Title

very welcome :)