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

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Answer 1 · 2016-12-19T18:01:09

The answer is $= \frac{6^{e}}{ln 6} - \frac{2^{e}}{ln 2} - \frac{6}{ln 6} + \frac{2}{ln 2} = 62.8129$ $=6^e/ln6-2 ^e/ln2-6/ln6+2/ln2=62.8129$

Let $u = 6^{x}$ $u=6^x$

Then, $ln u = x ln 6$ $lnu=xln6$

$u = e^{x ln 6}$ $u=e^(xln6)$

Let $v = 2^{x}$ $v=2^x$

$ln v = x ln 2$ $lnv=xln2$

$v = e^{x ln 2}$ $v=e^(xln2)$

Therefore

$\int_{1}^{e} (6^{x} - 2^{x}) d x$ $int_1 ^e(6^x-2^x)dx$

$= \int_{1}^{e} (e^{x ln 6} - e^{x ln 2}) d x$ $=int_1 ^e(e^(xln6)-e^(xln2))dx$

$= {[\frac{e^{x ln 6}}{ln 6} - \frac{e^{x ln 2}}{ln 2}]}_{1}^{e}$ $= [e^(xln6)/ln6-e^(xln2)/ln2 ]_1^e$

$= {[\frac{6^{x}}{ln 6} - \frac{2^{x}}{ln 2}]}_{1}^{e}$ $= [6^x/ln6-2^x/ln2 ]_1 ^e$

$= (\frac{6^{e}}{ln 6} - \frac{2^{e}}{ln 2}) - (\frac{6}{ln 6} - \frac{2}{ln 2})$ $=(6^e/ln6-2^e/ln2)-(6/ln6-2/ln2)$

$= \frac{6^{e}}{ln 6} - \frac{2^{e}}{ln 2} - \frac{6}{ln 6} + \frac{2}{ln 2} = 62.8129$ $=6^e/ln6-2 ^e/ln2-6/ln6+2/ln2=62.8129$

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