7f. Area Under a Curve - Definite Integration
Integration Mini Video Lecture
Historically, areas between curves were a hot problem and inpsired the development of integral calculus. This video gives an overview on how to use integration to find an area under a curve.
Area under curve definite integral... a mini lecture.
One of the first applications of integration was to find the area under a curve.
Though there were approximate ways of finding this, nobody had come up with an accurate way of finding an answer [until Newton and Leibniz developed integral calculus].
So let's have a look at this example.
Find the area under the curve y equals 2x3 + 5 between x equals 1 and x equals 2.
Let's see what we are talking about here.
OK, here' the curve y equals 2x3 + 5 and here is x equals 1 and x equals 2.
So what we are doing is to find this area in here.
Now the formula for area is integral a to b of function x [f(x)] dx.
Well in this particular example, our function is y equals 2x3 + 5, so I copy this...
And.so we can say that the area equals:
The integral from 1 to 2 ... of 2x3 + 5 dx.
Thats the... 2x3 + 5 in brackets. Don't forget the brackets, that's always important.
Now, how do we do this problem?
Well what I must do is integrate this thing, and substitute 2 in and then minus substitute 1 in. That's how we do a definite integral.
So, first step, put square brackets [ ]. And then integrate... [x4] plus 5 times x, lower value is 1, upper value is 2.
It's going to be 2 and then its going to be x to the power of 4 because we are addding one to the index and then dividing by the new number.
Plus 5 times x because the integral of 5 is 5x.
Lower value is 1 and upper value is 2.
Then this equals... The 2 and the 4 cancel..so
Let's cheat a little. Copy and paste this. And cancel out the 2 and 4 [to become 2 in the denominator].
So great! I have done the integration and now what I need to do is to do the substitution steps. So it's going to look like this.
I am going to copy this bit, and paste it, minus, paste it.
And substituing in the 2 [in the power 4 expression] ,and then [plus] 5 times 2 ... minus substituting in the 1 [in the power 4 expression] and [Minus] 5 times 1.
So, just to tell you what I did again, I integrate this expression 2x3 + 5 and I get x to the power 4 on 2 plus 5x. `(int (2x^3+5) dx = x^4/2 + 5x)`.
Then for definite integral we substitute 2 into this, and then 1 into this to get this and then using Scientific Notebook to give me the answer.
I get 25 over 2, or 12.5. `(int_1^2 (2x^3+5) dx = [x^4/2 + 5x]_1^2 = 25/2 = 12.5)`.
And of course we don't know what the units are. But we need to put something like "units squared" to indicate we know it's an area.