How do you integrate $int 6^x-2^xdx$ from $[1,e]$ ?

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Answer 1 · 2016-12-19T18:01:09

The answer is $=6^e/ln6-2 ^e/ln2-6/ln6+2/ln2=62.8129$

Let $u=6^x$

Then, $lnu=xln6$

$u=e^(xln6)$

Let $v=2^x$

$lnv=xln2$

$v=e^(xln2)$

Therefore

$int_1 ^e(6^x-2^x)dx$

$=int_1 ^e(e^(xln6)-e^(xln2))dx$

$= [e^(xln6)/ln6-e^(xln2)/ln2 ]_1^e$

$= [6^x/ln6-2^x/ln2 ]_1 ^e$

$=(6^e/ln6-2^e/ln2)-(6/ln6-2/ln2)$

$=6^e/ln6-2 ^e/ln2-6/ln6+2/ln2=62.8129$

How do you integrate int 6^x-2^xdxint 6^x-2^xdx from [1,e][1,e]?