Question

Help with solving integrals

Answer 1

I basically would like to understand how my book solved thisintegral

Answer 2

Like they totally ignored a negative

Answer 3

du=-1

Answer 4

whoops d(theta)=-1

Answer 5

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

Answer 6

shldnt the answer be -(4-(theta))^-1

Answer 7

no the book is right

Answer 8

http://letmegooglethat.com/?q=How+to+solve+integrals

Answer 9

so like how did they ignore the negative

Answer 10

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

Answer 11

Th eintegral will be negative instead of poitive

Answer 12

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

Answer 13

wait i am just gonna process this

Answer 14

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

Answer 15

I would have done it differently

Answer 16

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 \[u=4-\theta\]

Answer 17

ok Thanks :D I got it

Answer 18

Yup i got it thanks for ur help

Answer 19

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!

Answer 20

ya that is the way i did it