Home > Calculus > Integrals of Exponential Functions

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

Answer 1

Ezra Adams

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#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Adam Adkins

To integrate ( \int_{1}^{e} (6^x - 2^x) , dx ):

Integrate ( 6^x ): [ \int 6^x , dx = \frac{6^x}{\ln(6)} ]
Integrate ( 2^x ): [ \int 2^x , dx = \frac{2^x}{\ln(2)} ]

Now, compute each integral separately over the interval [1,e] and subtract the results:

[ \int_{1}^{e} 6^x , dx = \frac{6^e - 6^1}{\ln(6)} ]

[ \int_{1}^{e} 2^x , dx = \frac{2^e - 2^1}{\ln(2)} ]

Finally, subtract the second integral from the first:

[ \int_{1}^{e} (6^x - 2^x) , dx = \frac{6^e - 6^1}{\ln(6)} - \frac{2^e - 2^1}{\ln(2)} ]

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.