## anonymous 4 years ago Help with solving integrals

1. anonymous

2. anonymous

I basically would like to understand how my book solved thisintegral

3. anonymous

Like they totally ignored a negative

4. anonymous

du=-1

5. anonymous

whoops d(theta)=-1

6. anonymous

Sorry there is some kind of bug. It is really all the same document

7. anonymous

8. anonymous

no the book is right

9. anonymous
10. anonymous

so like how did they ignore the negative

11. anonymous

like if u check the answer you wont get the same integral

12. anonymous

Th eintegral will be negative instead of poitive

13. anonymous

Let $u=4-\theta$ then $\frac{du}{d\theta} = -1$ so $d\theta = -du$ and hence our integral becomes, upon changing the limits, $\lim_{a\rightarrow4^{+}}\left(-\displaystyle\int_{4-a}^{-2}u^{-2}du \right) = \lim_{a\rightarrow4^{+}} \left[-(-u^{-1})\right]^{-2}_{4-a} = \lim_{a\rightarrow4^{+}}u^{-1} \vert_{4-a}^{-2} = \lim_{a\rightarrow4^{+}}\left(\frac{1}{-2} - \frac{1}{4-a}\right)$

14. anonymous

wait i am just gonna process this

15. anonymous

alright i get it but what i dont get is why u changed the bounds

16. anonymous

I would have done it differently

17. anonymous

I used the substitution $u=4-\theta$ and to find the new bounds, i.e. the ones for the integral in terms of u, one must substitute the old bounds into the theta position i.e. the first bound is $4-(6) = -2$ and the second is $4-(a) = 4-a$ in both cases, we use the equation $u=4-\theta$

18. anonymous

ok Thanks :D I got it

19. anonymous

Yup i got it thanks for ur help

20. anonymous

In fact, substitution doesn't change anything about the integral really, it's still the same integral, just expressed differently. If we rearrange to get $\theta = 4-u$ and then continue from here: $...=\lim_{a\rightarrow 4^{+}}u^{-1}\vert_{4-a}^{-2}$ we see that the bounds may be changed via:$4-(-2) = 6$ and $4-(4-a) = a$ now we simply sub back $u=4-\theta$ to arrive at the same thing your book had in the second to last step: $=\lim_{a\rightarrow 4^{+}}(4-\theta )^{-1}\vert^{6}_{a}$ so really, the true 'identity' of this integral is unchaged under the substitution - you can uncover it back!

21. anonymous

ya that is the way i did it