OpenStudy (anonymous):

can we solve this integration? some help please.

4 years ago
OpenStudy (anonymous):
4 years ago

OpenStudy (zehanz):

First write as two separate fractions. Then apply the rule:\[\frac{ a^x }{ b^x }=\left( \frac{ a }{ b }\right)^x\] Now you can integrate them both...

4 years ago
OpenStudy (zehanz):

That is, if you know how to integrate an exponential function. Do you need more help?

4 years ago
OpenStudy (zehanz):

You integrate exponential function zis way ;) \[\int\limits_{}^{}a^xdx=\frac{ 1 }{ lna }a^x+C\]

4 years ago
OpenStudy (zehanz):

@tomiko: do you know how to do the next step now?

4 years ago
OpenStudy (anonymous):

i'm thinking now...just a minute

4 years ago
OpenStudy (anonymous):

is it correct for me to continue like this: \[\int\limits_{}^{}\left[ \left( \frac{ 1 }{ 5 } \right)^{x} - \left( \frac{ 1 }{ 2 } \right)^{x} \right] dx\]

4 years ago
OpenStudy (zehanz):

Yes, it is

4 years ago
OpenStudy (anonymous):

then the above is equal to: \[\int\limits_{}^{}\left( \frac{ 1 }{ 5 } \right)^{x}dx - \int\limits_{}^{}\left( \frac{ 1 }{ 2 } \right)^{x}dx + C\]

4 years ago
OpenStudy (anonymous):

\[\frac{ 1 }{ \ln \frac{ 1 }{ 5} }\frac{ 1 }{ 5 }^{2} - \frac{ 1 }{ \ln \frac{ 1 }{ 2} }\frac{ 1 }{ 2 }^{2} + C\]

4 years ago
OpenStudy (anonymous):

right @ZeHanz

4 years ago
OpenStudy (zehanz):

Change the exponents to x's

4 years ago
OpenStudy (zehanz):

Also \[\frac{ 1 }{ \ln \frac{ 1 }{ 5 } }=\frac{ 1 }{ \ln1-\ln5 }=\frac{ 1 }{ -\ln5 }=-\frac{ 1 }{ \ln5 }\]So you could simplify a little more, but that's only a minor improvement imo ;)

4 years ago
OpenStudy (anonymous):

thank you very much. you've been great help!! i have a test tomorrow! hopefully i can solve a question like this when i see one. thanks.

4 years ago
OpenStudy (zehanz):

You can do it! Lots of success!

4 years ago