can we solve this integration? some help please.

Question

anonymous · Answer

attachment

zehanz · Answer

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...

zehanz · Answer

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

zehanz · Answer

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

zehanz · Answer

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

anonymous · Answer

i'm thinking now...just a minute

anonymous · Answer

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

zehanz · Answer

Yes, it is

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

right @ZeHanz

zehanz · Answer

Change the exponents to x's

zehanz · Answer

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 ;)

anonymous · Answer

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.

zehanz · Answer

You can do it! Lots of success!