a to the u Intergal

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

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

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

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

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

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

\(\displaystyle \int_{9}^{59049} \dfrac{du}{2 \ln 3}\)..........Unnecessary - just leave the constant outside

\(\displaystyle \dfrac{1}{2\ln 3} |_{9}^{59049} \)................\(\displaystyle \int du = u\), ...Where is the "u" ?
\(\displaystyle \displaystyle \left.\dfrac{1}{2\ln 3} (u)\ \right| _9^{59049} = \dfrac{1}{2\ln 3} (59049 - 9) = \dfrac{29520}{\ln 3}\)
 
How about this? :)

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

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

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

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

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

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

\(\displaystyle \dfrac{59049}{2\ln 3} - \dfrac{9}{2\ln 3} = \dfrac{59040}{2\ln 3} = \dfrac{29520}{\ln 3}\).
 
How about this? :)

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

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

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

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

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

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

\(\displaystyle \dfrac{59049}{2\ln 3} - \dfrac{9}{2\ln 3} = \dfrac{59040}{2\ln 3} = \dfrac{29520}{\ln 3}\).
:)
The advantage of this substitution [instead of u=2x] is that you don't have to know a formula for \(\displaystyle \int a^u du\). Instead, you differentiate the function, using the chain rule (and knowing that \(\displaystyle a=\mathrm e^{\ln{a}}\)). Once the differentiation is done, the integral is trivial.
 
:)
The advantage of this substitution [instead of u=2x] is that you don't have to know a formula for \(\displaystyle \int a^u du\). Instead, you differentiate the function, using the chain rule (and knowing that \(\displaystyle a=\mathrm e^{\ln{a}}\)). Once the differentiation is done, the integral is trivial.

So was my logic correct?
 
How about this? :)

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

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

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

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

\(\displaystyle \dfrac{1}{2 \ln 3} \int_{9}^{59049} u(du)\). ← Incorrect - should be \(\displaystyle \displaystyle \frac{1}{2 \ln 3} \int_{9}^{59049} du\)

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

\(\displaystyle \dfrac{59049}{2\ln 3} - \dfrac{9}{2\ln 3} = \dfrac{59040}{2\ln 3} = \dfrac{29520}{\ln 3}\).
.
 
So was my logic correct?
Yes. But thanks to Subhotosh Kahn for pointing out an error that slipped past me. Logic correct, answer correct, though again one intermediate step was sloppy. As you can't appreciate Shakespeare without knowing grammar, you can't appreciate Calculus without Algebra.

I suppose I shouldn't be surprised that you ignore what I say when I tell you you are right, as much as when I tell you you are wrong. We would save about 2/3 of our postings if you would accept what you are told the first time, rather than the third or fourth..
 
So why is the u taken out of the last step of differentiation (before integration), with only du there?
 
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