How do you find the definite integral for: #((2x^(2) +3)dx# for the intervals #[4, 6]#?
1 Answer
Explanation:
The definite integral of
#int_4^6(2x^2+3)dx#
To evaluate this, find the antiderivative
The antiderivative can be found using the rule:
#intx^ndx=x^(n+1)/(n+1)+C#
So, we can express the definite integral as:
#int_4^6(2x^2+3)dx=2int_4^6x^2+3int_4^6dx#
Note that
#2int_4^6x^2+3int_4^6dx=[2(x^(2+1)/(2+1))+3x]_4^6#
#=[2/3x^3+3x]_4^6#
Note that this is saying the same thing as
#[2/3x^3+3x]_4^6=(2/3(6)^3+6(3))-(2/3(4)^3+4(3))#
#=(2/3(216)+18)-(2/3(64)+12)#
#=(432/3+54/3)-(128/3+36/3)#
#=(432+54-128-36)/3#
#=322/3#
This can be interpreted to mean that the area bounded by the curve