Skip to main content
Search IntMath

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.

Video transcript


Related lessons on IntMath

This 'Area under a curve' mini video lecture explains the processes in these two pages:

3. Area Under a Curve (in this chapter)

... and ...

Area Under a Curve by Integration (in the Applications of Integration chapter)

Area Between Two Curves by Integration (also in the Applications of Integration chapter)

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... 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

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. (Here's the calculation:)

`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.


Tips, tricks, lessons, and tutoring to help reduce test anxiety and move to the top of the class.