How do you evaluate the definite integral #int -8/3*2^xdx# from -2 to 9?
1 Answer
Explanation:
If you already know the formula for integrals involving
We can use the formula:
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
#-8/3int_(-2)^9 2^xdx=-8/3int_(-2)^9e^(xln2)dx#
Here, let
#=-8/(3ln2)int_(-2)^9 ln2*e^(xln2)dx#
Before we make the
#=-8/(3ln2)int_(-2ln2)^(9ln2)e^udu=-8/(3ln2)[e^u]_(-2ln2)^(9ln2)#
From here, the