Area between a curve and a straight line
Watch this video to learn about calculating the area between two curves or a curve and a straight line.
To find the limits of integration, we first need to:
- check if the curve and straight line intersect at all?
- find out at what point(s) they do intersect
To do this, we need to revise the discriminant and intersections.
Example
Find the area between the curve and straight line shown below.
Solution
Firstly, we need to prove whether or not the curve and straight line actually intersect at all by using the discriminant.
Remember to make the RHS of both equations equal to each other first.
\({x^2} = x\)
\({x^2} - x = 0\)
Now, we can see that \(a = 1,\,b = - 1\) and \(c = 0\).
\({b^2} - 4ac\)
\(= {( - 1)^2} - (4 \times 1 \times 0)\)
\(= 1 - 0\)
\(= 1\)
Therefore \({b^2} - 4ac\textgreater0\) so the roots are real and unequal.
Now we can find out these points by going back to:
\({x^2} - x = 0\)
\(x(x - 1) = 0\)
\(x = 0\) or \(x - 1 = 0\)
Therefore \(x = 0\) or \(x = 1\)
The curve and straight line intersect when \(x = 0\) and \(x = 1\).
0 and 1 are the limits of integration.
Therefore the area:
\(Area= \int_{0}^{1} (top-bottom)dx\)
\(Area = \int\limits_0^1 {(x - {x^2}} )dx\)
\(= \left[ {\frac{{{x^2}}}{2} - \frac{{{x^3}}}{3}} \right]_0^1\)
\(= \left( {\frac{1}{2} - \frac{1}{3}} \right) - (0)\)
\(= \frac{1}{6}unit{s^2}\)
