RI Applications of Integration Solns

RAFFLES INSTITUTION H2 Mathematics 9758 2023 Year 6 Term 3 Revision 11 (Summary and Tutorial) Topic: Applications of Integration Summary for Applications of Integration 1 Evaluating Areas Bounded by Curves and Axes 1.1 Area bounded by the curve and the x-axis In Fig. 1.1a, (i) f( ) d 0. b a xx >∫ Thus, area under f( )yx= from xa= to xb= is given by f( ) d b a xx∫ . (ii) f( ) d 0. c b xx <∫ Thus, area under f( )yx= from xb= to xc= is given by f( ) d c b xx−∫ or f( ) d c b xx∫ (see Fig. 1.1b). (iii) The area under f( )yx= from xa= to xc= is given by ( ) f( ) d f( ) d bc ab xx xx +−∫∫ or f( ) d c a xx∫ (see Fig. 1.1b). Fig 1.1a Fig 1.1b

1.2 Area bounded by the curve and the y-axis In Fig.1.2, (i) g( ) d 0. a b yy >∫ Thus, area bounded by g( )xy= and the y-axis from yb= to ya= is given by g( ) d a b yy∫ . (ii) g( ) d 0. b c yy <∫ Thus, area bounded by g( )xy= and the y-axis from yc= to yb= is given by g( ) d . b c yy−∫ (iii) The shaded area bounded by the curve g( )xy= and the y-axis from yc= to ya= is given by ( ) g( ) d g( ) d ab bc yy yy +−∫∫ . g( )xy= Fig 1.2 y x a b c

1.3 Area bounded between two curves In general, given 2 curves defined by the equations ( )1fyx= and ( )2fyx= , and an interval [ ],ab where ( ) ( )12ff xx ≥ within this interval, we can evaluate the area bounded between them from xa= to xb= by evaluating the definite integral ( ) ( )( ) 12ffd b a x xx−∫ . Similarly, given 2 curves defined by the equations ( )1gxy= and ( )2gxy= , and an interval [ ],cd where ( ) ( )12gg yy ≥ within this interval, we can evaluate the area bounded between them from yc= to yd= by evaluating the definite integral ( ) ( )( ) 12ggd d c y yy−∫ . xa b 1f( )yx= 2f()yx= x a b 1f( )yx= 2f()yx= x a b 1f( )yx= 2f()yx= x y c d 1g( )xy= 2g( )xy= x y c d 1g( )xy= 2g( )xy= x y c d 1g( )xy= 2g( )xy=

1.4 Evaluating areas involving curves which are defined pa rametrically Essentially, we write the required area as f( ) d b a xx∫ or d c g( ) dyy∫ like how we did in the earlier sections. However, since both y and x are now expressed in terms of a parameter t , the integration here is done with respect to the parameter t instead of the variable x. To do so, we apply the integration by substitution technique to change the integral to one involving the parameter t . Students are encouraged to sketch the curve with their GC in order to identify the required region. Example The parametric equations of a curve C is given by 3sin , cosx ty t= = , where 02 πt≤≤ . Find the area of bounded by the curve C, the lines x = 1 and x = 3. Solution: Required area 3 1 2 d yx= ∫ ( ) 2 1 1 3 2 1 1 3 sin sin d2 cos d d 2 cos (3cost) d xtt t tt π π − − =  = ∫ ∫ and we proceed to solve this integral. *note the substitution of dcos , d d d xy tx t t= = , and also the change of limits. 1 3

1.5 Area under a curve as the limiting sum of the areas of the rectangles Fig 1.3a Fig 1.3b Other than using integration, students are required to be able to estimate the area under a curve using the sum of the areas of the rectangles of equal width. For example, to obtain an estimate for the area under the curve f( )yx= between 0x = and 1,x = we can consider finding the sum of the areas of the rectangles as shown in Fig. 1.3a or Fig. 1.3b. 1 0 f ( ) d (0.2)(f (0) f (0.2) f (0.4) f (0.6) f (0.8))xx ≈ ++++∫ (in Fig. 1.3a) or 1 0 f ( ) d (0.2)(f (0.2) f (0.4) f (0.6) f (0.8) f (1))xx ≈ ++++∫ (in Fig. 1.3b). From the diagrams, we see that the former gives an approximation which is an underestimate while the latter provides an overestimate of the actual area under the curve. To obtain a better estimate, we can consider using more rectangles within the given interval. Let us consider the case in Fig. 1.3b, but instead of using just 5 rectangles, let’s use n rectangles. f( )yx= f( )yx= 1 width = ... ... ... ... Height = ... 1f n n −  1f n   2f n  

Thus 1 0 11 2 3f ( ) d f f f ... f nxx n nnn n     ≈ +++ +       ∫ . As we increase the number of rectangles n, we find that the sum of the areas of the n rectangles will approach the value of the integral 1 0 f( )dxx∫ . Thus, the limiting sum of the areas of the n rectangles can be used here to evaluate the area under the graph of f( )yx= from 0x = to 1x = . i.e. 1 0 1 1f ( ) d lim f n n r rxx nn→∞ = = ∑∫ . Note: This idea can also be similarly applied if we choose to consider the case in Fig. 1.3a.

2 Volume of Solid of Revolution 2.1 Rotation about the x-axis (y = 0) Fig. 2.1a In general, if the region R bounded by the curve ( )fyx= , the x -axis, the lines xa= and xb= , as shown in Fig.2.1a is rotated completely about the x -axis, the volume of solid of revolution formed is given by ( ) 22 d fd bb aa yx x xπ= π ∫∫ . If the region R is bounded by the curves ( )1fyx= , ( )2fyx= , the lines xa= and xb= as shown in Fig. 2.1b, Fig. 2.1b then the volume of solid of revolution formed when R is rotated completely about the x -axis is given by ( ) ( ) 22 12fd f d bb aa xx xxπ −π    ∫∫ ( ) ( ) 22 12f fd b a x xx= π−     ∫ y x a b R y x a b R

2.2 Rotation about the y-axis (x = 0) Fig. 2.2a In general, if the region S bounded by the curve ( )gxy= , the y -axis, the lines yc= and yd= , as shown in Fig. 2.2a is rotated completely about the y-axis, the volume of solid of revolution formed is given by ( ) 22 d gd dd cc xy y yπ= π ∫∫ If the region S is bounded by the curves ( )1gxy= , ( )2gxy= , the lines yc= and yd= as shown in Fig. 2.2b, Fig. 2.2b then the volume of solid of revolution formed when S is rotated completely about the y-axis is given by ( ) ( ) 22 12gd g d dd cc yy yyπ −π    ∫∫ ( ) ( ) 22 12g gd d c y yy= π−     ∫ S x x = g (y) y x d c S

2.3 Rotation about other lines (vertical or horizontal) When the required region is rotated through 2π about other vertical or horizontal line instead of one of the axes, we will need rewrite the problem (usually via translation of the graph) so that the line of rotation is one of the axis. This idea is illustrated in the following example. Example The region R is bounded by the curve 2yx= and the line 4y = . Find the volume of the solid formed when R is rotated completely about the line 4y = . Solution: Required volume 22 2 22 22 ( 4) d ( 4) dyx x xππ −− = −= −∫∫ In order to find the required volume, the region is translated 4 units in the negative y -direction so that the line of rotation is now the x-axis. Thus, the equation of the curve in consideration is now 2 4yx= − instead of 2yx= . 2yx= 2 4yx= − -2 -2 2 2 y =4 y =0

Revision Tutorial Questions Source of Question: IJC JC2 CT2 9758/2017/P1/Q3 1 The diagram shows a circle with equation 22( 1) 4xy−+= . (i) By using the substitution 1 2sinx θ= + , find the exact area bounded by the circle and the y-axis for 0x ≤ . [6] (ii) The circle cuts the axes at the points A and B as shown in the diagram. The region bounded by the minor arc AB and the line segment AB is rotated through 4 right angles about the x-axis. Find the exact volume of the solid of revolution. [4] Solution: (i) [6] 1 2sin d 2cos dxx θ θθ= + ⇒= When 22 2 0, ( 1) 4 ( 1) 4 1 or 3 yx y x x = −+= −= = − x