RI 2025 H2 C8B Apps of Integration Tutorial Qns

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 6 ________________________________ Tutorial 8B: Applications of Integration Page 1 of 3 Tutorial 8B: Applications of Integration 1 Evaluate 2 1ed a x a x x   where 0a  , giving your answer in terms of a . [3] [ 2e e aa  ] 2 (i) Sketch, on the same diagram, the graphs of 21 422 9 and e x xy y  . [2] (ii) The finite region in the first quadrant bounded by the curves 21 422 9, e x xy y  , the xaxis and the y axis is denoted by R . (a) Shade the region R . [1] (b) Find the volume of the solid of revolution formed when R is rotated through 2 radians about the x -axis. [4] [(ii)(b) 22.7] 3 It is given that 27f o r 02 ,f( ) 21 f o r 2 4 , xxx xx      and that f( ) f( 4 )xx for all real values of x. (i) Evaluate f(27) + f(45). [2] (ii) Sketch the graph of f( )yx for 71 0x . [3] (iii) Find 3 4 f( )dx x  . [3] [(i) 11 (iii) 110 3 ] 4 With reference to the given figure, (i) find the exact area of the shaded region. (ii) find the exact volume generated by the shaded region, when it is rotated through 4 right angles about the y-axis. [(i) 13 12 (ii) 23 30 ] 1 1 2 2 y x 3yx 21( 1 )yx  0

Raffles Institution H2 Mathematics 2025 Year 6 __________________________________________________________________________________________ ________________________________ Tutorial 8B: Applications of Integration Page 2 of 3 5 It is given that 64f( ) 3 7 .xx x  The diagram shows the curve with equation f( )yx and the line with equation 7y , for 0x . The curve crosses the positive x axis at x , and the curve and the line meet where 0x and x . (i) Find the value of  , giving your answer correct to 3 decimal places, and find the exact value of  . [2] (ii) Evaluate f( )dx x   , giving your answer correct to 3 decimal places. [2] (iii) Find, in terms of 3 , the area of the finite re gion bounded by the curve and the line, for 0x . [3] (iv) Show that f ( ) f ( )x x . What can be said about the six roots of the equation f( ) 0?x  [4] [(i) 1.885  , 3  (ii) 0.597 (iii) 54 335 ] 6 O is the origin and A is the point on the curve tanyx where 3x  . (i) Calculate the area of the region R enclosed by the arc OA, the x −axis, and the line 3x  , giving your answer in an exact form. [3] (ii) The region S is enclosed by the arc OA, the y -axis, and the line 3y . Find the volume of the solid of revolution formed where S is rotated 360 o about the x −axis, giving your answer in an exact form. [6] (iii) Find 3 1 0 tan d yy  , giving your answer in an exact form. [3] [(i) ln2 (ii) 24 33   (iii) 3 ln 23   ]

Raffles Institution H2 Mathematics 2025 Year 6 __________________________________________________________________________________________ ________________________________ Tutorial 8B: Applications of Integration Page 3 of 3 7 The curve C has equation y = f(x), where f(x) = 2 e xx  . (i) Sketch the curve C. [2] (ii) Find the exact coordinates of the turning points on the curve. [4] (iii) Use the substitution 2ux to find  0 fd , n x x for 0.n Hence find the area of the region between the curve and the positive x-axis. [4] (iv) Find the exact value of  2 2 fd x x  . [2] (v) Find the volume of revolution when the region bounded by the curve, the lines 0x , 1x and the x -axis is rotated completely about the x -axis. Give your answer correct to 3 significant figures. [2] [(ii) 11 1 1,, , 22 e 2 2 e         (iii) 211(1 e );22  n (iv) 41e  (v) 0.363] 8 The region R is bounded by the curve C with equation x xy 2 and the line 3y . (i) Calculate the exact area of R . [4] (ii) Write down the equation of the curve obtained when C is translated 3 units in the negative y−direction. [2] (iii) Hence, show that the volume of the solid formed when R is rotated completely about the line 3y is given by 4 1 12 461 3 dx xx xx      . [4] (iv) Determine the exact volume of the solid. [2] [(i) 1 3 (iv) 1(8ln 2 5 ) 2  ] 9 (i) Given that f is a continuous function, explain, with the aid of a sketch, why the value of 11 2lim f f ... f n n nn n n         is 1 0 f( ) dx x . [2] (ii) Hence evaluate 33 3 3 1 1 2 ...lim n n n n    . [3] [(ii) 3 4 ]