Question

How to evaluate the definite integral using substitution?

Answer 1

\[\int\limits_{-2/9}^{1/3} \frac{ -4 }{ \sqrt{7-9x}} dx\]

Answer 2

u-substitution.

Answer 3

Have you worked u=substitution problems previously?

Answer 4

yes but im a bit confused. would u=7-9x and du=-9dx?

Answer 5

Yes, that is exactly it!

Answer 6

Alternatively, you can multiply your integral by (-9)/(-9).
\(\color{#000000}{ \displaystyle \int\limits_{1/3}^{-2/9}\frac{-4\color{red}{\times (-9)}}{\sqrt{7-9x}\color{red}{\times (-9)}}dx }\)

and then the top -9 with dx goes away, being replaced by dx.
(and the bottom -9 remains.)

Answer 7

u = 7-9x definitely works,
can I also suggest another u substitution, which you can try after you're done with the first?
Its always good to know alternative ways to solve a problem :)
try \(u = \sqrt{7-9x}\) too.

Answer 8

i would like to use the second option dividing both sides by 9.

Answer 9

So, you want to solve for dx?
Ok, we can do this.

Answer 10

\[dx = -\frac{ 1 }{9 } du\]

Answer 11

Yes, and now you substitute:

u=7-9x (that hasn't changed, of course)
(-1/9) du = dx

Answer 12

before you go ahead and make the substitution, its a good practice to calculate new limits first.

Answer 13

\[\int\limits_{-2/9}^{1/3} \frac{ -4 }{ \sqrt{u} } (\frac{ -1 }{ 9 }du)\]

Answer 14

Yes, you can go ahead and change the limits of integration, but then, don't substitute the x back.

Or, you can integrate without changing the limits of integration, which would require you to sub back the x.

Answer 15

how do i calculate new limits?

Answer 16

What is your "u" ?

Answer 17

7-9x

Answer 18

Plug in the upper limit for x, x=1/3 into "7-9x".
Plug in the lower limit for x, x=-2/9 into "7-9x".

This will give you the new upper and lower limits for u, respectively.

Answer 19

7-9(1/3)=4
7-9(-2/9)=9

Answer 20

Yes, very good!

Answer 21

So, your new limits of integration are 4 and 9 respectively.

Answer 22

\(\displaystyle\int\limits_{9}^{4} \frac{ -4 }{ \sqrt{u} } (\frac{ -1 }{ 9 }du)\)

Answer 23

Well, you can simplify this to,

\(\displaystyle\frac{4}{9}\int\limits_{9}^{4} u^{-1/2}~du\)

And integrate that.

Answer 24

(Without having to substitute back the x, once you changed the limits of integration to fit the new variable u)

Answer 25

That integral is just a power rule.

Answer 26

\[\frac{ 4 }{ 9 } 2\sqrt{u} +c \]

Answer 27

Yes, that would be the indefinite integral,

(Which alternatively is written as \(\color{#000000}{ \displaystyle \frac{8}{9}\sqrt{u} \color{gray}{+C} }\))

Answer 28

so i plug 7-9x back in and then plug in 9 and 4?

Answer 29

But, with your upper and lower limits of 4 and 9, respectively - for the definite integral, you have,

\(\color{#000000}{ \displaystyle \left.\frac{8}{9}\sqrt{u} \right|_{\large ~u=9}^{\large ~u=4} }\)

Answer 30

why is that?

Answer 31

If you didn't change the limits of integration, then you would need to plug in the variable x back. Since, you did change the limits of integration, you mdon't need to plug the x back in.

Answer 32

(Plugging the x back in, once you have changed the limits of integration to fit "u", would give you the wrong result in 99.9999999 percent of the time.)

Answer 33

ok. why isnt the integral from 4 to 9? why is it 9 to 4?

Answer 34

Well, how did we obtain these limits of integration? You plugged your old limits into the expression for u, or into the u=7-9x. (Correct?)

the lower limit "x=-2/9" gave you a new lower limit "u=9"
the upper limit "x=1/3" gave you a new upper limit "u=4"

Answer 35

if you really really wish, you may also apply,

\(\color{#000000}{ \displaystyle \int\limits_a^b f(x)~dx:=-\int\limits_b^a f(x)~dx }\)

but that is really unnecessary.

Answer 36

oh im sorry i wrote it wrong. \[\int\limits_{1/3}^{-2/9}\]

Answer 37

Oh:)

Answer 38

So, then, you are just having:

\(\color{#000000}{ \displaystyle \int\limits^{-2/9}_{1/3}\frac{-4}{\sqrt{7-9x}}dx= \frac{4}{9}\int\limits^{9}_{4}\frac{1}{\sqrt{u~}}du=\left.\frac{8}{9}\sqrt{u} \right|_{\large ~u=4}^{\large ~u=9} }\)

Answer 39

If -2/9 is the upper limit and 1/3 is the lower limit, respectively,
then, just swipe your limits (don't need to work everything over).

Answer 40

ok. and whats my next step?

Answer 41

Then, ust plug in your limits of integration and evaluate.

Answer 42

\(\color{#000000}{ \displaystyle \left.\frac{8}{9}\sqrt{u} \right|_{\large ~u=\color{red}{4}}^{\large ~u=\color{blue}{9}} =\left[\frac{8}{9}\sqrt{\color{blue}{9}}\right]-\left[\frac{8}{9}\sqrt{\color{red}{4}}\right] }\)

Answer 43

This notation does through a lot of people off ....
if you have any questions, then you know ...

Answer 44

oh ok. if its a definite integral and the limits are given, then u will always be my new limits? I dont plug in u and then plug x?

Answer 45

If you are given the limits of integration at the beginning, and then you make a u-substitution, you can do one of the two things:)

way 1:
a) Make your u-substitution.
b) Change limits of integration by plugging them into the expression for "u".
c) Integrate (with variable u).
d) Do NOT substite the x back into the equation, (rather, just,)
e) Evaluate your integrand by plugging in the new limits of integration.

(This is what we done)

Answer 46

way 2:
a) Make your u-substitution.
b) Do NOT change the limits of integration ((leave the x-limits)).
c) Integrate (with variable u; BUT with limits that correspond to x).
d) You MUST substitute the x back into the equation. (and then,)
e) Having substituted the x back, evaluate your integrand under the original limits of integration.

Answer 47

(I am kind of lagging -:( )

Answer 48

((If you want an example of way 1 and way 2, let me know.))

Answer 49

an example would be great.

Answer 50

ok,

Answer 51

\(\color{#000000}{ \displaystyle \int\limits_2^6 \frac{1}{\sqrt{4x+1}}dx }\)
---------------------------------------------------------------
Way 1:

\(\color{#000000}{ \displaystyle u=4x+1 }\)
\(\color{#000000}{ \displaystyle du=4{\tiny~}dx \quad \Longrightarrow \quad (1/4)du=dx }\)

Upper limit: \(\color{#000000}{ \displaystyle x=6\quad \Longrightarrow \quad u=4(6)+1 \quad \Longrightarrow \quad u=25 }\)
Lower limit: \(\color{#000000}{ \displaystyle x=2\quad \Longrightarrow \quad u=4(2)+1 \quad \Longrightarrow \quad u=9 }\)

\(\color{#000000}{ \displaystyle \int\limits_{9}^{25} \frac{1}{\sqrt{u}}[(1/4)du]=\int\limits_{9}^{25} \frac{1}{4}u^{-1/2}du =\left.\frac{1}{4}(2u^{1/2}) \right|_{u=9}^{u=25} =\left.\frac{1}{2}\sqrt{u~} \right|_{u=9}^{u=25} }\)

\(\color{#000000}{ \displaystyle = \frac{1}{2}\sqrt{25}-\frac{1}{2}\sqrt{9}= \frac{5}{2}-\frac{3}{2}=1 }\)

Answer 52

\(\color{#000000}{ \displaystyle \int\limits_2^6 \frac{1}{\sqrt{4x+1}}dx }\)
---------------------------------------------------------------
Way 2:

\(\color{#000000}{ \displaystyle u=4x+1 }\)
\(\color{#000000}{ \displaystyle du=4{\tiny~}dx \quad \Longrightarrow \quad (1/4)du=dx }\)

\(\color{#000000}{ \displaystyle \int\limits_{2}^{6} \frac{1}{\sqrt{u}}[(1/4)du]=\int\limits_{2}^{6} \frac{1}{4}u^{-1/2}du =\left.\frac{1}{4}(2u^{1/2}) \right|_{x=2}^{x=6}=\left.\frac{1}{2}\sqrt{u~} \right|_{x=2}^{x=6} }\)

Here, (ot at any time after integrating, and before plugging in the limits of inetgration, you should substitute the x back: (We had that u=2x+1, so let's plug that)


\(\color{#000000}{ \displaystyle \left.\frac{1}{2}\sqrt{4x+1} \right|_{x=2}^{x=6} = \frac{1}{2}\sqrt{4(6)+1}-\frac{1}{2}\sqrt{4(2)+1}={\small \frac{1}{2}\sqrt{25}-\frac{1}{2}\sqrt{9}= \frac{5}{2}-\frac{3}{2}=1} }\)

Answer 53

And the result is the same, the difference is:

Changing the limits of integration, versus having to substitute the x back in.

Answer 54

Not to get too much off track, in your particular case we have left off at:

\(\color{#000000}{ \displaystyle \left.\frac{8}{9}\sqrt{u} \right|_{\large ~u=\color{red}{4}}^{\large ~u=\color{blue}{9}} =\left[\frac{8}{9}\sqrt{\color{blue}{9}}\right]-\left[\frac{8}{9}\sqrt{\color{red}{4}}\right] }\)

which I know isn't a problem for you, from then and on....

Answer 55

got it, thank you very much!

Answer 56

YW