a to the u Intergal

[Jason76]

\(\displaystyle \int_{1}^{5} 3^{2x} dx\)

Using \(\displaystyle \int a^{u} du = \dfrac{a^{u}}{\ln a} + C\) and \(\displaystyle \dfrac{d}{dx} (a^{u}) = a^{u} * du * \ln a\)

\(\displaystyle u = 3^{2x}\)

\(\displaystyle du = 3^{2x} * 2 dx * \ln 3\)

Hint on next step? Do I put something to the left of the intergal sign at some point?

 

[Deleted member 4993]

\(\displaystyle \int_{1}^{5} 3^{2x} dx\)

Using \(\displaystyle \int a^{u} du = \dfrac{a^{u}}{\ln a} + C\) and \(\displaystyle \dfrac{d}{dx} (a^{u}) = a^{u} * du * \ln a\)

\(\displaystyle u = 3^{2x}\)

\(\displaystyle du = 3^{2x} * 2 dx * \ln 3\)

Hint on next step? Do I put something to the left of the intergal sign at some point?

a = 3

u = 2x

dx = du/2

\(\displaystyle \displaystyle \int_{1}^{5} 3^{2x} dx \ = \ \frac{1}{2}\int_{2}^{10} 3^{u} du \ \)... now continue...

 

[Jason76]

\(\displaystyle \int_{1}^{5} 3^{2x} dx\)

Using \(\displaystyle \int a^{u} du = \dfrac{a^{u}}{\ln a} + C\) and \(\displaystyle \dfrac{d}{dx} (a^{u}) = a^{u} * du * \ln a\)

\(\displaystyle u = 2x\)

\(\displaystyle du = 2\)

\(\displaystyle a = 3\)

\(\displaystyle \dfrac{1}{2} \int_{2}^{10} 3^{u} du\)

 

[DrPhil]

\(\displaystyle \int_{1}^{5} 3^{2x} dx\)

Using \(\displaystyle \int a^{u} du = \dfrac{a^{u}}{\ln a} + C\) and \(\displaystyle \dfrac{d}{dx} (a^{u}) = a^{u} * du * \ln a\)

\(\displaystyle u = 2x\)

\(\displaystyle du = 2\)...NO... du = 2 dx

\(\displaystyle a = 3\)...So what will you do with ln(a) ?

\(\displaystyle \dfrac{1}{2} \int_{2}^{10} 3^{u} du\) Good - do you know why the limits are (2,10)?

Carry on.

 

[Jason76]

\(\displaystyle \int_{1}^{5} 3^{2x} dx\)

Using \(\displaystyle \int a^{u} du = \dfrac{a^{u}}{\ln a} + C\) and \(\displaystyle \dfrac{d}{dx} (a^{u}) = a^{u} * du * \ln a\)

\(\displaystyle u = 2x\)

\(\displaystyle du = 2\)

\(\displaystyle a = 3\)

\(\displaystyle \dfrac{1}{2} \int_{1}^{5} 3^{u} du\) - decided to change back to \(\displaystyle 1\) and \(\displaystyle 5\). Will evaluate in next step.

\(\displaystyle \int_{1}^{5} \dfrac{3^{2x}}{2\ln 3}\) with \(\displaystyle u\) to be evaluated at \(\displaystyle 1\) and \(\displaystyle 5\)

\(\displaystyle \dfrac{3^{2(5)}}{2\ln 3}\) - \(\displaystyle \dfrac{3^{2(1)}}{2\ln 3}\)

\(\displaystyle \dfrac{3^{10}}{2\ln 3}\) - \(\displaystyle \dfrac{3^{2}}{2\ln 3}\)

\(\displaystyle \dfrac{3000}{2\ln 3}\) - \(\displaystyle \dfrac{9}{2\ln 3}\)

 

[DrPhil]

\(\displaystyle \int_{1}^{5} 3^{2x} dx\)

Using \(\displaystyle \int a^{u} du = \dfrac{a^{u}}{\ln a} + C\) and \(\displaystyle \dfrac{d}{dx} (a^{u}) = a^{u} * du * \ln a\)

\(\displaystyle u = 2x\)

\(\displaystyle du = 2\)...NO... du = 2 dx


\(\displaystyle a = 3\)

\(\displaystyle \dfrac{1}{2} \int_{1}^{5} 3^{u} du\) - decided to change back to \(\displaystyle 1\) and \(\displaystyle 5\). Will evaluate in next step....NO - if the integral is with respect to u, the limits have to be values of u
........\(\displaystyle x=(1,5) \implies u = \cdot \cdot \cdot\)

\(\displaystyle \int_{1}^{5} \dfrac{3^{2x}}{2\ln 3}\)du with \(\displaystyle u\) to be evaluated at \(\displaystyle 1\) and \(\displaystyle 5\).
......Now you really confused between x and u -- you can't substitute back unless you replace all.

\(\displaystyle \dfrac{3^{2(5)}}{2\ln 3}\) - \(\displaystyle \dfrac{3^{2(1)}}{2\ln 3}\)

\(\displaystyle \dfrac{3^{10}}{2\ln 3}\) - \(\displaystyle \dfrac{3^{2}}{2\ln 3}\)

\(\displaystyle \dfrac{3000}{2\ln 3}\) - \(\displaystyle \dfrac{9}{2\ln 3}\)....3^10 is something completely different from 3×10^3

sigh

 

[Jason76]

\(\displaystyle \int_{1}^{5} 3^{2x} dx\)

Using \(\displaystyle \int a^{u} du = \dfrac{a^{u}}{\ln a} + C\) and \(\displaystyle \dfrac{d}{dx} (a^{u}) = a^{u} * du * \ln a\)

\(\displaystyle u = 2x\)

\(\displaystyle du = 2\)

\(\displaystyle a = 3\)

\(\displaystyle \dfrac{1}{2} \int_{1}^{5} 3^{u} du\) - decided to change back to \(\displaystyle 1\) and \(\displaystyle 5\). Will evaluate in next step.

\(\displaystyle \int_{1}^{5} \dfrac{3^{2x}}{2\ln 3}\) with \(\displaystyle u\) to be evaluated at \(\displaystyle 1\) and \(\displaystyle 5\)

\(\displaystyle \dfrac{3^{2(5)}}{2\ln 3}\) - \(\displaystyle \dfrac{3^{2(1)}}{2\ln 3}\)

\(\displaystyle \dfrac{3^{10}}{2\ln 3}\) - \(\displaystyle \dfrac{3^{2}}{2\ln 3}\)

\(\displaystyle \dfrac{59049}{2\ln 3}\) - \(\displaystyle \dfrac{9}{2\ln 3} = \dfrac{59040}{2\ln 3} = \dfrac{29520}{\ln 3}\) correction

 

[Jason76]

Does look right?

\(\displaystyle \int_{1}^{5} 3^{2x} dx\).

Using \(\displaystyle \int a^{u} du = \dfrac{a^{u}}{\ln a} + C\) and \(\displaystyle \dfrac{d}{dx} (a^{u}) = a^{u} * du * \ln a\).

\(\displaystyle u = 2x\).

\(\displaystyle du = 2\).

\(\displaystyle a = 3\).

\(\displaystyle \dfrac{1}{2} \int_{1}^{5} 3^{u} du\).

\(\displaystyle \int_{1}^{5} \dfrac{1}{2} 3^{u} du \).

\(\displaystyle \int_{1}^{5} \dfrac{3^{u}}{2} du \).

\(\displaystyle \dfrac{3^{2(u)}}{2\ln 3} |_{1}^{5} \).

\(\displaystyle \dfrac{3^{2(5)}}{2\ln 3}\) - \(\displaystyle \dfrac{3^{2(1)}}{2\ln 3}\).

\(\displaystyle \dfrac{3^{10}}{2\ln 3}\) - \(\displaystyle \dfrac{3^{2}}{2\ln 3}\).

\(\displaystyle \dfrac{59049}{2\ln 3}\) - \(\displaystyle \dfrac{9}{2\ln 3} = \dfrac{59040}{2\ln 3} = \dfrac{29520}{\ln 3}\)

 

[Jason76]

How about going this route? Making u to equal the whole thing: \(\displaystyle 3^{2x}\)

\(\displaystyle \int_{5}^{1} 3^{2x} dx \).

\(\displaystyle u = 3^{2x}\).

\(\displaystyle du = 3^{2x} 2 \ln 3dx \rightarrow\).

 

[Deleted member 4993]

Now, it's all coming together (sorry to bump thread again).

Pretty sure, this is right:

\(\displaystyle \int_{5}^{1} 3^{2x} dx \).

\(\displaystyle u = 2x\).

\(\displaystyle du = 2 dx \rightarrow du(\dfrac{1}{2}) = dx \rightarrow \dfrac{du}{2} = dx\).

\(\displaystyle \dfrac{1}{2} \int_{5}^{1} 3^{u} du \rightarrow \int_{5}^{1} (\dfrac{1}{2}) 3^{u} du \rightarrow \dfrac{3^{u}}{2}\).

\(\displaystyle = \dfrac{3^{u}}{2\ln 3} |_{5}^{1} \rightarrow \dfrac{3^{2x}}{2\ln 3} |_{5}^{1}\).

\(\displaystyle \dfrac{3^{2(5)}}{2\ln 3} - \dfrac{3^{2(1)}}{2\ln 3} \rightarrow \dfrac{3^{10}}{2\ln 3} - \dfrac{3^{2}}{2\ln 3}\).

\(\displaystyle = \dfrac{59049}{2\ln 3} \rightarrow \dfrac{29520}{\ln 3}\).

Those statements are mathematically incorrect - reasons have been discussed by Dr. Phil in previous posts.

 

[Jason76]

So when du is substituted back, then what do you put there? Hint

 

[DrPhil]

So when du is substituted back, then what do you put there? Hint

You don't have to backsubstitute. IF you want to do that extra step AFTER integrating, then the exponent would become 2x and the limits would be x=1 to 5.

\(\displaystyle \displaystyle \left. \dfrac{3^u}{2\ \ln{3}} \right|_2^{10} = \left. \dfrac{3^{2x}}{2\ \ln{3}} \right|_1^5\)

Either of those representations gives the same result. You got there, but your intermediate steps were not written correctly.

\(\displaystyle \displaystyle \dfrac{3^{10}-3^2}{2\ \ln{3}} = \dfrac{25920}{\ln{3}}\)

 

[Jason76]

How about if

\(\displaystyle \int_{5}^{1} 3^{2x} dx\)

\(\displaystyle u = 3^{2x}\)

\(\displaystyle du = 3^{2x} * \ln 3 * 2 dx\)

 

[DrPhil]

How about if

\(\displaystyle \displaystyle \int_{1}^{5} 3^{2x} dx\)

\(\displaystyle u = 3^{2x}\)

\(\displaystyle du = 3^{2x} * \ln 3 * 2 dx\)

Have you attempted to substitute that into the integral? It just might work - it would totally bypass the form you were given for \(\displaystyle \int a^u\ du\) ... instead, you may get the simplest of all possible integrals. TRY IT!

Remember that you must change the limits of integration:
\(\displaystyle x=1 \implies u=...\),........\(\displaystyle x=5 \implies u=...\)

If it works, it will be far more clever than the method you were "supposed" to use. But I want to see YOU work it out completely, without algebraic snafus.

 

[Jason76]

\(\displaystyle \int_{1}^{5} 3^{2x} dx\)

\(\displaystyle u = 3^{2x}\)

\(\displaystyle du = 3^{2x} * \ln 3 * 2 dx\)

\(\displaystyle \dfrac{du}{ \ln 3 * 2} = 3^{2x} dx\)

\(\displaystyle \dfrac{1}{2} \int_{u}^{u} \dfrac{du}{\ln 3}\)

\(\displaystyle \int_{u}^{u} \dfrac{du}{2 \ln 3}\)

 

[DrPhil]

\(\displaystyle \int_{1}^{5} 3^{2x} dx\)...lLimits are 1 to 5, not 5 to 1

\(\displaystyle u = 3^{2x}\)....This is the "new" substitution.

\(\displaystyle du = 3^{2x} * \ln 3 * 2 dx\)

\(\displaystyle \dfrac{du}{ \ln 3 * 2} = 3^{2x} dx\)....Good

YOU MUST CHANGE LIMITS WHEN SUBSTITUTING

x=1 --> u = ... .......x=5 --> u= ...

\(\displaystyle \dfrac{1}{2} \int_{5}^{1} \dfrac{du}{\ln 3}\)......NO .. Limits?

\(\displaystyle \int_{5}^{1} \dfrac{du}{2 \ln 3}\)......NO

\(\displaystyle \dfrac{du}{2 \ln 3} |_{5}^{1}\)......NONSENSE .. du??

\(\displaystyle \dfrac{3^{2(5)}}{2\ln 3}\) - \(\displaystyle \dfrac{3^{2(1)}}{2\ln 3}\).

\(\displaystyle \dfrac{3^{10}}{2\ln 3}\) - \(\displaystyle \dfrac{3^{2}}{2\ln 3}\).

\(\displaystyle \dfrac{59049}{2\ln 3}\) - \(\displaystyle \dfrac{9}{2\ln 3} = \dfrac{59040}{2\ln 3} = \dfrac{29520}{\ln 3}\)

I would rather see you carry through with the substitution u=3^(2x) rather than revert back to u=2x. Don't try to mix two different approaches.

 

[Jason76]

\(\displaystyle \int_{1}^{5} 3^{2x} dx\)

\(\displaystyle u = 3^{2x}\)

\(\displaystyle du = 3^{2x} * \ln 3 * 2 dx\)

\(\displaystyle \dfrac{du}{ \ln 3 * 2} = 3^{2x} dx\)

\(\displaystyle \dfrac{1}{2} \int_{u}^{u} \dfrac{du}{\ln 3}\)

\(\displaystyle \int_{u}^{u} \dfrac{du}{2 \ln 3}\)

 

[Jason76]

:cool: - ok might be getting the hang of this.

\(\displaystyle \int_{1}^{5} 3^{2x} dx\)

\(\displaystyle u = 3^{2x}\)

\(\displaystyle du = 3^{2x} * \ln 3 * 2 dx\)

\(\displaystyle \dfrac{du}{ \ln 3 * 2} = 3^{2x} dx\)

\(\displaystyle \dfrac{1}{2} \int_{9}^{59049} \dfrac{du}{\ln 3}\)

\(\displaystyle \int_{9}^{59049} \dfrac{du}{2 \ln 3}\)

\(\displaystyle \dfrac{1}{2\ln 3} |_{9}^{59049} \).

\(\displaystyle \dfrac{59049}{2\ln 3}\) - \(\displaystyle \dfrac{9}{2\ln 3} = \dfrac{59040}{2\ln 3} = \dfrac{29520}{\ln 3}\)

 

[DrPhil]

\(\displaystyle \int_{1}^{5} 3^{2x} dx\)

Using \(\displaystyle \int a^{u} du = \dfrac{a^{u}}{\ln a} + C\) and \(\displaystyle \dfrac{d}{dx} (a^{u}) = a^{u} * du * \ln a\)

\(\displaystyle u = 3^{2x}\)

\(\displaystyle du = 3^{2x} * 2 dx * \ln 3\)

Hint on next step? Do I put something to the left of the intergal sign at some point?

You had previously written

\(\displaystyle \dfrac{du}{ \ln 3 * 2} = 3^{2x} dx\)

which I had said was Good

You can take constants outside the integral sign: \(\displaystyle \displaystyle \dfrac{1}{2\ \ln{3}}\int \cdot \cdot \cdot \)

What is the integrand after the substitution? [should be very simple!]

What are the limits of integration with respect to u?

 

[Jason76]

You had previously written

\(\displaystyle \dfrac{du}{ \ln 3 * 2} = 3^{2x} dx\)

which I had said was Good

You can take constants outside the integral sign: \(\displaystyle \displaystyle \dfrac{1}{2\ \ln{3}}\int \cdot \cdot \cdot \)

What is the integrand after the substitution? [should be very simple!]

What are the limits of integration with respect to u?

The integrand after substitution (after constants taken out) should be \(\displaystyle 1\). Right,

\(\displaystyle \int_{1}^{5} 3^{2x} dx\)

\(\displaystyle u = 3^{2x}\)

\(\displaystyle du = 3^{2x} * \ln 3 * 2 dx\)

\(\displaystyle \dfrac{du}{ \ln 3 * 2} = 3^{2x} dx\)

\(\displaystyle \dfrac{1}{2 \ln 3} \int_{9}^{59049} du\)

\(\displaystyle \int_{9}^{59049} \dfrac{du}{2 \ln 3}\)

\(\displaystyle \dfrac{1}{2\ln 3} |_{9}^{59049} \).

\(\displaystyle \dfrac{59049}{2\ln 3}\) - \(\displaystyle \dfrac{9}{2\ln 3} = \dfrac{59040}{2\ln 3} = \dfrac{29520}{\ln 3}\)