Access to Science, Engineering and Agriculture: Mathematics 2 MATH00040 Chapter 4 Solutions

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1 Access to Science, Engineering and Agriculture: Mathematics MATH4 Chapter 4 Solutions In all these solutions, c will represent an arbitrary constant.. (a) Since f(x) 5 is a constant, 5dx 5x] 5. (b) Since f(x) πcos(e) is a constant, πcos(e)dx πcos(e)x+c. (c) Since f(x) x is of the form f(x) x n with n, x x + dx + ] x ]. (d) Since f(x) x 9 is of the form f(x) x n with n 9, x 9 dx 9/+ x9 + +c x +c. (e) Since f(x) x 5 is of the form f(x) x n with n 5, x 5 dx 5+ x 5+ ] ] 4 x ( 4 ) (f) Since f(x) x cos() is of the form f(x) x n with n cos(), x cos() dx cos()+ xcos()+ +c. (g) Since f(x) e 4x is of the form f(x) e ax with a 4, e 4x dx ] 4 e4x 4 e4() 4 e 4 (e8 ).

2 (h) Since f(x) e x is of the form f(x) e ax with a, e x dx / e x +c e x +c. (i) Since f(x) e 6x is of the form f(x) e ax with a 6, e 6x dx 6 e 6x ] ] 6 e 6x 6 e ( 6 ) e 6() 6 ( e 6 ). (j) Since f(x) e πx is of the form f(x) e ax with a π, e πx dx π eπx +c. (k) x dx ln(x)] ln() ln() ln() ln(). (l) Since f(x) sin(x) is of the form f(x) sin(ax) with a, sin(x)dx cos(x)+c. (m) Since f(x) sin( x) is of the form f(x) sin(ax) with a, π sin( x) dx cos( x) ]π cos( x) π ) cos ( ]π cos( π) cos() () (). cos( ()) (n) Since f(x) sin(ex) is of the form f(x) sin(ax) with a e, sin(ex)dx e cos(ex)+c.

3 (o) Since f(x) cos(x) is of the form f(x) cos(ax) with a, π π ]π cos(x)dx sin(x) π ( sin π ) sin(( π)) sin(π) sin( π) () (). (p) Since f(x) cos( πx) is of the form f(x) cos(ax) with a π, cos( πx)dx π sin( πx)+c π sin( πx)+c.. (a) We will first use the sum and multiple rules to find the corresponding definite integral. +x x +x 4x 4 dx dx+ xdx+ x dx+ x dx+ 4x 4 dx dx+ xdx x dx+ x dx 4 x 4 dx ( ) ( ) ( ) ( ) x+ x x + 4 x4 4 5 x5 +c x+ x x + 4 x4 4 5 x5 +c. Note that in your assignment or exam solutions you don t need to give as much detail as this. I am just setting out everything carefully until you get used to the ideas involved. +x x +x 4x 4 dx x+ x x + 4 x4 4 ] 5 x5 + ( ) ( )+ 4 (4 ) 4 5 (5 ) + () () + 4 ()4 45 ] ()

4 (b) Using the sum and multiple rules, x +sin4xdx x dx+ sin4xdx x dx+ sin4xdx x dx+ sin4xdx ln(x)+ ( 4 ) cos(4x) +c ln(x) cos(4x)+c. (c) We will first use the sum and multiple rules to find the corresponding definite integral. ( ) ( ) e x cos x dx e x dx+ cos x dx ( ) e x dx cos dx π e x cos x ( ( / e ) x / sin ( ) 6e x 4sin x. ( ) ( x dx 6e x 4sin x ( ) 6e π 4sin π )] π 6e π 4() 6() 4()] 6e π. ( x )) +c 6e () 4sin (d) Using the sum and multiple rules, 4cos( x) e x dx 4cos( x)dx+ e x dx 4 cos( x) dx e x dx ( ) ( 4 sin( x) / e )+c x 4 sin( x)+ e x +c. (e) We will first use the sum and multiple rules to find the corresponding definite ( )] () 4

5 integral. x +e cos()x dx x dx+ e cos()x dx x dx+ e cos()x dx ( ) x + cos() ecos()x +c x + cos() ecos()x +c. x +e cos()x dx x + ] cos() ecos()x ( )+ cos() ecos()() cos() ( e cos() e cos()) 6. (f) Using the sum and multiple rules, sin(x) sin(x)+cos(x) cos(x)dx sin(x)dx+ sin(x)dx+ cos(x)dx+ sin(x)dx sin(x)dx+ ( ) cos(x) ( ) cos(x) cos(x)+ cos(x)+ sin(x) sin(x)+c ( ) + ] cos() ecos()( ) cos(x) dx cos(x)dx cos(x) dx ( ) ( ) + sin(x) sin(x) +c (g) We will first use the sum rule to find the corresponding definite integral. e +e x 4dx e 4dx+ e x dx (e 4)x+ ex +c. Note that we didn t need the multiple rule here and also note that we could deal with e 4 all at once since e 4 is a constant. e +e x 4dx (e 4)x+ ] ex (e 4)()+ e() (e 4)()+ ] e() (e 4)+ e6 + e 8. ( e 6 e ) 5

6 (h) Using the sum and multiple rules, x +4x 4 +5x 5 +x dx x dx+ 4x 4 dx+ x dx+4 ( + x + x x5 5 4 x 4 +x+c. 5x 5 dx+ x dx x 4 dx+5 x 5 dx+ dx ) ( ) x4+ +5 ( 5+ x 5+ ) +x+c For the remainder of the questions we will still need the sum and multiple rules but I will not mention them explicitly.. (a) Here we use integration by substitution. Let u x+5, so that du dx. Then dx du du/dx du du. Also, when x, u 5 and when x, u 6. 6 ] 6 (x+5) 9 dx u 9 du 5 u (6 5 ). (b) Here we use integration by substitution. Let u x 5, so that du dx x. Then dx du du/dx du x. x (x 5) 5 dx x u 5 du x u5 du ( ) 6 u6 +c 8 u6 +c ( x 5 )6 +c. 8 (c) Here we use integration by substitution. Let u cos(x), so that du sin(x). Note we chose u cos(x) rather dx than u sin(x) since this ensures the sin(x) in the numerator cancels. This is part of the skill you will have to learn when performing integration; what 6

7 works and what doesn t. Then dx du du/dx du sin(x). Also, when x, u and when x π 4, u. π 4 sin(x) cos(x) dx sin(x) u u du du sin(x) ln(u)] ( ln ) ( ln()) ( ) ln + ( ) ln() ln(). (d) Here we use integration by substitution. Let u sin(x), so that du dx cos(x). Then dx du du/dx du cos(x). cos(x) cos(x) sin(x) dx u (e) Here we use integration by substitution. Let u du, so that x dx x. Then dx du du/dx du /x x du. Also, when x, u and when x, u. du cos(x) du ln(u)+c ln(sin(x))+c. u x e x dx e u ] x eu ( x )du e u du e ( e ) e e e e. 7

8 (f) Here we use integration by substitution. Let u x 4 x +x x, so that du dx 4x x +x. Then dx du du/dx du 4x x +x. x 9x +6x x x 4 x +x x dx 9x +6x u u du ln(u)+c ln(x 4 x +x x)+c. (g) Here we use integration by substitution. Let u x, so that du dx x. Then dx du du/dx du x. Also, when x, u and when x, u. x x dx x u du x u du ( u u ] )] ( du 4x x +x ) +. (h) Here we use integration by substitution. Let u x x, so that du dx 4x +4x. Then dx du du/dx du 4x +4x. 8

9 (x +x )sin (x ) x dx 4 sin(u)du (i) Here we use integration by substitution. Let u x +4, so that du dx x. Then dx du du/dx du x. Also, when x, u 4 and when x, u 5. (x +x )sin(u) du 4x +4x 4 ( cos(u))+c 4 cos ( x x )+c. x 5 x +4 dx x du 4 u x 5 du 4 u u ( u ] ) ] 5 (j) Here we use integration by substitution. Let u x +, so that du dx x. Then dx du du/dx du x. x ( x + ) dx x u du x u du ( ) 5 u5 +c 5 u5 +c ( x + )5. 5 (k) Here we use integration by substitution. Let u cos(x), so that du dx sin(x). 9 4

10 Then dx du du/dx du sin(x). Also, when x, u and when x π, u. π sin(x)e cos(x) dx sin(x)e u e u du e u ] e ( e ) +e e. du sin(x) (l) Here we use integration by substitution. Let u x 4, so that du dx 4x. Then dx du du/dx du 4x. x cos ( x 4) dx x cos(u) du 4x 4 cos(u)du 4 sin(u)+c 4 sin( x 4) +c. (m) Here we use integration by substitution. Let x sin(u), so that dx du cos(u). Then dx dx du cos(u)du. du Also, when x, u and when x, u π. x dx π π π π π sin (u)cos(u)du cos (u)cos(u)du cos(u) cos(u)du cos (u)du + cos(u)du

11 ( )]π sin(u) u+ u+ ]π 4 sin(u) π 4 + ( 4 sin(π) π 4 + (+) π sin() ) We don t cover this in MATH4 but the equation of a circle of radius one centred at the origin is x +y and on solving this for y we obtain y ± x. So what we have actually calculated here is the area inside this circle that lies in the first quadrant. The area inside a circle of radius r is πr π π and we have found a quarter of this area, that is π 4. Note in addition that if we replace x with r x and the upper limit with r then we can actually derive the formula πr for the area inside a circle. See if you can decide what the substitution should be in this case and then derive the formula. 4. (a) Here we use integration by parts. Let f(x) x and g (x) e x, so that f (x) and g(x) ex., using the integration by parts formula, xe x dx x ex ex dx xex ex dx xex ( ) ex +c (b) Here we use integration by parts. xex 4 ex +c Let f(x) 5x and g (x) cos(x), so that f (x) 5 and g(x) sin(x)., using the integration by parts formula, π ( )] π xcos(x)dx 5x sin(x) ] π 5 π xsin(x) π 5 sin(x)dx 5 sin(x)dx 5 (π)sin(π) 5 5 ()sin(()) ] π cos(x) 5 4 cos(π) 5 4 cos(()) ( cos(x) )] π

12 (c) Here we use integration by parts. Let f(x) ln(x) and g (x) x, so that f (x) x and g(x) x., using the integration by parts formula, ( ) ( ) xln(x)dx ln(x) x x x dx x ln(x) xdx (d) Here we use integration by parts. x ln(x) 4 x +c. Let f(x) ln(x) and g (x) x 4 +, so that f (x) x and g(x) 5 x5 +x., using the integration by parts formula, ( x 4 + ) ( )] ( ) ln(x)dx ln(x) 5 x5 +x x 5 x5 +x dx ( ) ] 5 x5 +x ln(x) 5 x4 +dx ( ) ( ) ] 5 5 +() ln() 5 5 +() ln() 5 x5 +x ( ) ( ) 5 +6 ln() 5 + ln() ( ) ( )] 5 5 +() 5 5 +() ln() 5 76 ] ln() 6 5. (e) Here we will have to integration by parts twice. First we let f(x) x and g (x) sin(x), so that f (x) x and g(x) cos(x)., using the integration by parts formula, x sin(x)dx x ( cos(x)) x( cos(x)) dx x cos(x)+ x cos(x) dx. () We will now find x cos(x) dx by integrating by parts again. So let f(x) x and g (x) cos(x), so that f (x) and g(x) sin(x). Then, using the integration by parts formula, xcos(x)dx xsin(x) sinxdx xsin(x) ( cos(x))+c xsin(x)+cos(x)+c. ()

13 If we now substitute () into (), we finally obtain x sin(x)dx x cos(x)+xsin(x)+cos(x)+c. (f) Again here we will have to integrate by parts twice. We will first find the associated definite integral x e x dx. Let f(x) x and g (x) e x, so that f (x) x and g(x) e x., using the integration by parts formula, x e x dx x e x xe x dx () We will now find xe x dx by integrating by parts again. Let f(x) x and g (x) e x, so that f (x) and g(x) e x., using the integration by parts formula, xe x dx xe x e x dx xe x e x. (4) If we now substitute (4) into (), we see that x e x dx x e x (xe x e x )+c x e x xe x +e x +c. x e x dx x e x xe x +e x] e ()e +e e ()e +e ] e e+e +] e. 5. (a) Here we use partial fractions. Since x 4 (x )(x+), we let x+ x 4 A x + B x+, (5) where A and B are constants we have to find. Multiplying both sides of (5) by x 4 we obtain x+ A(x+)+B(x ). (6) If we let x in (6), we obtain 4 4B, so that B. If we let x in (6), we obtain 8 4A, so that A. x+ x 4 dx x + x+ dx ln(x )+ln(x+)+c. Note that if you can t spot the integrals, then you can use the substitutions u x and u x+ for the two terms, respectively. Similar substitutions will work in all the examples below but I won t explicitly mention them.

14 (b) Again we use partial fractions. Since x +9x+ (x+4)(x+5), we let x +9x+ A x+4 + B x+5, (7) where A and B are constants we have to find. Multiplying both sides of (7) by x +9x+ we obtain A(x+5)+B(x+4). (8) If we let x 5 in (8), we obtain B, so that B. If we let x 4 in (8), we obtain A, so that A. x +9x+ dx (c) Again we use partial fractions. Since x +4x x(x+4), we let x+4 + x+5 dx ln(x+4)+ln(x+5)] ln(5)+ln(6) ( ln(4)+ln(5)) ln(6) ln(5)+ln(4) x 8 x +4x A x + B x+4, (9) where A and B are constants we have to find. Multiplying both sides of (9) by x +4x we obtain x 8 A(x+4)+Bx. () If we let x 4 in (), we obtain 4 4B, so that B. If we let x in (), we obtain 8 4A, so that A. x 8 x +4x dxx x + x+4 dx ln(x)+ln(x+4)+c ln(x+4) ln(x)+c. (d) Again we use partial fractions. Since (x )(x+) (x)(x+)(x+), we let x +4x (x )(x+) A x + B x+ + C x+, () where A, B and C are constants we have to find. Multiplying both sides of () by (x )(x+) we obtain x +4x A(x+)(x+)+B(x)(x+)+C(x)(x+). () 4

15 If we let x in (), we obtain C, so that C. If we let x in (), we obtain B, so that B. If we let x in (), we obtain 6 6A, so that A. x +4x (x )(x+) dx x + x+ + x+ dx ln(x)+ln(x+)+ln(x+)] ln()+ln(4)+ln(5) (ln()+ln()+ln(4)) ln()+ln(5) ln(). (e) Again we use partial fractions. Ths time we have a repeated root, so we let x (x ) A (x ) + B x, () where A and B are constants we have to find. Multiplying both sides of () by (x ) we obtain x A+B(x ) x Bx+(A B). Comparing the coefficients of x we obtain B and comparing the constant terms we obtain A B. Since B, this means that A as well. x (x ) dx (x ) + x dx x +ln(x )+c ln(x ) x +c. Note the substitution u x will also work for the term here. (x ) (f) Again we use partial fractions. Ths time we have a quadratic term in the denominator which doesn t factorise, so we let x +6x+4 (x +x+)(x+) Ax+B x +x+ + C x+, (4) where A, B and C are constants we have to find. Multiplying both sides of (4) by (x +x+)(x+) we obtain x +6x+4 (Ax+B)(x+)+C(x +x+) x +6x+4 Ax +Ax+Bx+B +Cx +Cx+C x +6x+4 (A+C)x +(A+B +C)+B +C. (5) Comparing coefficients of powers of x in (5), we get the simultaneous equations A+C, A+B +C 6 and B +C 4. 5

16 Subtracting the third of these from the second yields A. Then substituting A in to the first gives C and finally substitution C into the third gives B. Note if you can spot these manipuations, you can always do a formal row reduction. x +6x+4 (x +x+)(x+) dx x+ x +x+ + x+ dx ln ( x +x+ ) +ln(x+) ] ln(5)+ln() (ln()+ln()) ln(5). Note that if you can t spot the integral of the use the substitution u x +x+. x+ term then you can x +x+ 6. (a) The graph of f(x) x 7 lies below the x-axis between x and x and above the x-axis between x and x. Thus the required area is ] ] x 7 dx+ x 7 dx 8 x8 + 8 x8 ( 8 ) ( ) ()8 + 8 (8 ) (b) Thegraphoff(x) cos(x)lies belowthex-axis between the pointsx π and x π 6 and above the x-axis between the points π and. Thus the 6 required area is π 6 π cos(x)dx+ (. π 6 ] π 6 cos(x) dx ] sin(x) + sin(x) π π 6 ( sin π ) ) ( sin( π) + sin() ( sin π ) ) ( () ) ( () + () ) () (c) The graph of f(x) e x lies above the x-axis everyhere. Thus the required area is ] e x dx ex e e (e e ). 6

17 (d) Since f() and the graph of f(x) x x x + only crosses the x-axis at x between the points x and x, it follows that the graph of f(x) x x x + lies below the x-axis between the points x and x and above the x-axis between the points x and x (since f() > it is above the x-axis at x ). the required area is x x x+dx+ 4 x4 x x +x 4 +. x x x+dx ] ( x4 x x +x )] + b 7. In all these questions we will use the formula V π f(x) dx. a (a) The volume is (b) The volume is (c) The volume is π π π π (x ) dx π ] x 4 dx π 5 x5 5 π. ( ) π sin(x) dx π sin(x) dx π cos(x)] π π cos(π) ( cos())] π( () ()) π. ( ) e x dx π e x dx π ] e x ( π e ( )) e ( π ( e )) π ( ) e. ] ( 4 + )] 7

18 (d) The volume is π π π sin (x)dx π + cos(x)dx π x+ ] π 4 sin(x) ( π π + (+ 4 sin(π) 4 )) sin() ( π ) π + (+) π. 8