Area between two curves
The process for calculating the area between two curves is the same as finding the area between a curve and a straight line.
Example
Calculate the shaded area between the two curves below.
\(y=x^{2}\) and \(y=2x-x^{2}\)
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.
\({x^2} = 2x - {x^2}\)
\({x^2} - 2x + {x^2} = 0\)
\(2{x^2} - 2x = 0\)
Now we can see that \(a = 2,\,b = - 2\) and \(c = 0\).
\({b^2} - 4ac\)
\(= {( - 2)^2} - (4 \times 2 \times 0)\)
\(= 4 - 0\)
\(= 4\)
Therefore \({b^2} - 4ac\textgreater0\) so the roots are real and unequal.
Now we can find out these points by going back to:
\(2x(x - 1) = 0\)
\(2x = 0\) 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 equals:
\(Area=\int_{0}^{1}(top-bottom)dx\)
\(\int\limits_0^1 {(2x - {x^2}} ) - ({x^2})dx\)
\(= \int\limits_0^1 {(2x - 2{x^2}} )dx\)
\(= \left[ {{x^2} - 2\frac{{{x^3}}}{3}} \right]_0^1\)
\(= \left( {1 - \frac{2}{3}} \right) - \left( 0 \right)\)
\(= \frac{1}{3}uni{t^2}\)
