AREAS USING INTEGRATION

We shall use the result that the area, A, bounded by a curve, y = f(x), the x axis and the lines x = a, and x = b, is given by:

Hence, the shaded area = (units squared) Example 1:Find the area shown enclosed by the curve y = x 2, the x-axis, and the lines x = 1 and x = 2. I = 2 1 x 2 dx x 33x 33       2 1 =       =       – 7373 = – 7373 =

Example 2: Find the area enclosed by the curve y = 3x 2 – 6x and the x axis. Firstly we need to find where the curve crosses the x-axis. 0 = 3x 2 – 6x x = 0 or 2 The graph looks like this: = [ 8 – 12 ] – [ 0 ] = – 4 n.b. The integral is negative, this is because the area lies below the x axis. The actual area is 4 (units squared) i.e. When y = 0; 0 = 3x(x – 2 ) Now, I =3x 2 – 6x dx 2 0 =       2 0 3x 3 3 6x 2 2 – =       2 0 – x3x3 3x23x2

Since an area below the x axis gives a negative value, care must be taken when part of the required region is above the x axis and part is below. e.g.

Example 3: Find the area shown enclosed by the curve y = x ( x – 1), the x-axis, and the line x = 2. We need to calculate 2 separate integrals: For the actual area, we add the 2 answers, ignoring the negative: I 1 = 1 0 x 2 – x dx I 2 = 2 1 x 2 – x dx and,

i.e. Simultaneous equations are used to find the points of intersection. y = f(x) y = g(x) y x To find the area bounded by two curves, ( or a curve and a line ), consider the graph shown: R Then R = f(x) dx b a g(x) dx b a – a b Area bounded by two curves: The required area, R, is the area under y = f(x) between x = a and x = b, minus the area under y = g(x) between x = a and x = b. f(x) – g(x) dx b a = R Since the limits are the same for both integrals, they can be combined to a single integral:

Example 4: R x y The curve y = x 2 – 4x + 5 and the line y = 2x intersect as shown. Find the area of the region R enclosed by the two graphs. Firstly we need the two points of intersection, A and B. A B x 2 – 4x + 5 = 2x x 2 – 6x + 5 = 0 ( x – 1 )( x – 5 ) = 0 The required area, R, is the area under the line, between x = 1 and x = 5, minus the area under the curve between x = 1 and x = 5. i.e. R = –x 2 – 4x + 5 dx 5 1 2x dx x = 1 or 5 Note: The 1 st part of the integral 2x dx 5 1 i.e. ( the area under the line) could be found by calculating the area of a trapezium.

6x6x 5 1 – x 2 – 5 dx R = 3x23x2 x 33x 33 –    – 5x 5 1 = ( 75 – – 25 ) –( 3 – 1313 – 5 ) –2x dx 5 1 x 2 – 4x + 5 dx 5 1 R = We now have: R =    32 3 = Since the limits are the same for both integrals, they can be combined to a single integral: f(x) – g(x) dx b a = R i.e. –2x2x 5 1 ( x 2 – 4x + 5 ) dx R =

Summary of key points: 1.The area, A, bounded by a curve, y = f(x),the x axis and the lines x = a, and x = b, is given by: 2.A region which lies below the x axis gives a negative result for the integral. 3.Care must be taken when part of the required region is above the x axis, and part is below. This PowerPoint produced by R.Collins ; Updated Apr The area, R bounded by two functions which intersect at x = a, and x = b is given by: f(x) – g(x) dx b a = R

We shall now use the result that the area, A, bounded by a curve, x = f(y), the y axis and the lines y = a, and y = b, is given by: n.b. This is no longer on the syllabus.

Example 4:Find the area shown, bounded by the curve the y axis, and the lines y = 1, and y = 4. Firstly we need the equation in the form x = f(y): Square both sides: So x = 3y 2 = [ 64 ] – [ 1 ] = 63