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

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)`

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`