How to integrate #int_(1"/"2)^1 4^x "d"x#?

2 Answers
Jun 21, 2018

#int_(1/2)^1 4^xdx = 2/ln(4)#

Explanation:

The trick part about this integral is that we don't know the exponential function with base 4 very well.

Rewrite #4^x# with base #e# as follows:

#color(red)4^x = color(red)(e^ln(4))^x = e^(ln(4)*x)#

And now, it's easier to find an antiderivative.

#int_(1/2)^1 e^(ln(4)*x)dx#

Although you could make a substitution, it can be quicker to do this in your head. Consider the derivative of #e^(ln(4)*x)#:

#d/dx(e^(ln(4)*x)) = ln(4)e^(ln(4)*x)#

We can see that if we divide both sides by the constant #color(blue)ln(4)#, we find an expression for the antiderivative of #e^(ln(4)*x)#.

#d/dx(1/color(blue)ln(4)e^(ln(4)*x))=e^(ln(4)*x)#

We can simplify the antiderivative. Let #F(x) = 1/ln(4)e^(ln(4)*x)#.

#F(x) = 4^x/ln(4)#

Now that we have the antiderivative, can substitute the upper and lower bounds as follows.

#int_(color(red)(1/2))^color(red)1 e^(ln(4)*x)dx#
#= F(color(red)1) - F(color(red)(1/2))#
#= 4^color(red)1/ln(4) - 4^color(red)(1/2)/ln(4)#
#= (4^color(red)1- 4^color(red)(1/2))/ln(4)#
#= (4 - 2)/ln(4)#
#= 2/ln(4)#

Jun 21, 2018

#int_(1/2)^1 4^x "d"x=2/ln4#

Explanation:

Let #u=4^x# so #lnu=xln4#.

Then #1/u"d"u=ln4"d"x# so #1/ln4"d"u=u"d"x=4^x"d"x#

Also, #u(1"/"2)=4^(1/2)=2# and #u(1)=4^1=4#

Thus

#int_(1/2)^1 4^x "d"x=int_(2)^4 1/ln4"d"u=[u/ln4]_2^4=((4-2)/ln4)=2/ln4#