Question

How do you evaluate the definite integral `int -8/3*2^xdx` from -2 to 9?

Answer 1

`-4094/(3ln2)`

Explanation:

`" "`

If you already know the formula for integrals involving `bb(a^x)`:

We can use the formula: `inta^xdx=1/lnaa^x+C`

So, we see that:

`int_(-2)^9-8/3*2^xdx=-8/3int_(-2)^9 2^xdx=-8/(3ln2)[2^x]_(-2)^9`

Evaluating, this becomes:

`-8/(3ln2)[2^x]_(-2)^9=-8/(3ln2)(2^9-2^(-2))=-8/(3ln2)(512-1/4)`

`=-8/(3ln2)(2047/4)=-4094/(3ln2)`

`" "`

Without knowing the formula:

You should know that `inte^udu=e^u+C`. We can use this to our advantage here, since `2^x=e^(ln(2^x))=e^(xln2)`. Thus:

`-8/3int_(-2)^9 2^xdx=-8/3int_(-2)^9e^(xln2)dx`

Here, let `u=xln2`. This implies that `du=ln2*dx`. So, multiply the integrand by `ln2` and the exterior by `1/ln2`.

`=-8/(3ln2)int_(-2)^9 ln2*e^(xln2)dx`

Before we make the `u` and `du` substitutions, recall that we will have to switch our bounds by plugging them in for `x` in `u=xln2`. Thus `-2->-2ln2` and `9rarr9ln2`.

`=-8/(3ln2)int_(-2ln2)^(9ln2)e^udu=-8/(3ln2)[e^u]_(-2ln2)^(9ln2)`

From here, the `e` and `ln` will cancel out (remember to write `-2ln2` as `ln(2^(-2))=ln(1/4)` and `9ln2` as `ln(2^9)=ln512`), and then the work becomes the same as the method outlined above.