Questions and Answers #4 Integrals

Questions and Answers Sheet 4 Integrals Question #1 The integral represents the volume of a solid. Describe the solid. 1 π ∫0 (𝑦 4 − 𝑦 8 )𝑑𝑦 a) The integral describes the volume of the solid obtained by rotating the region 𝑦 4 ≤ 𝑥 ≤ 𝑦 2 } of the xy-plane about the x-axis. b) The integral describes the volume of the solid obtained by rotating the region 𝑦 2 ≤ 𝑥 ≤ 𝑦 4 } of the xy-plane about the x-axis. c) The integral describes the volume of the solid obtained by rotating the region 𝑦 4 ≤ 𝑥 ≤ 𝑦 2 } of the xy-plane about the y-axis. d) The integral describes the volume of the solid obtained by rotating the region 𝑦 2 ≤ 𝑥 ≤ 𝑦 4 } of the xy-plane about the y-axis. e) The integral describes the volume of the solid obtained by rotating the region 𝑦 4 ≤ 𝑥 ≤ 𝑦 8 } of the xy-plane about the y-axis. 𝑅 = {{𝑥, 𝑦}|0 ≤ 𝑦 ≤ 1, 𝑅 = {{𝑥, 𝑦}|0 ≤ 𝑦 ≤ 1, 𝑅 = {{𝑥, 𝑦}|0 ≤ 𝑦 ≤ 1, 𝑅 = {{𝑥, 𝑦}|0 ≤ 𝑦 ≤ 1, 𝑅 = {{𝑥, 𝑦}|0 ≤ 𝑦 ≤ 1, Answer: 1 π ∫0 (𝑦 4 − 𝑦 8 )𝑑𝑦 = 1 𝑑1 = π[(𝑦 2 )2 − (𝑦 4 )2 ]𝑑𝑦 Axis of solution =y-axis upper boundary 𝑥 = 𝑦 2 lower boundary 𝑥 = 𝑦 4 region in x-y plane 𝑦4 < 𝑥 < 𝑦2 0≤𝑦≤1 Answer: (c) Question #2 Evaluate the iterated integral by converting to polar coordinates. √(2𝑥−𝑥 2 ) 2 ∫ ∫ 0 𝑥𝑦𝑑𝑦𝑑𝑥 0 Answer: Step1 Integrating the given equation after converting to polar coordinates: 2 √(2𝑥−𝑥 2 ) 𝑥𝑦𝑑𝑦𝑑𝑥 ∫0 ∫0 Let is assume 𝑥 = 𝑟 cos(θ) and 𝑦 = 𝑟 sin(θ) after putting in above expression and corresponding to that changing the limiting value: } Expression will change into π 2 cos θ 3 𝐼 = ∫02 ∫0 𝑟 cos(θ) sin(θ) 𝑑𝑟 𝑑 π 2 2 cos θ 3 = ∫0 cos(θ) sin θ ⋅ [∫0 π 2 𝑟 𝑑𝑟] 𝑑 θ = ∫0 cos 5(θ) sin(θ) 𝑑 θ Now assuming = cos(θ) so 𝑑𝑢 = − sin(θ) Putting this to solve further π 2 𝐼 = − ∫ 𝑢5 𝑑𝑢 0 π −4 cos 6(θ) 2 =[ ] 6 0 = 2 3 Result: 2 = 3 Question #3 1 Evaluate the double integral 𝑥𝑦 2 𝑑𝐴, D is enclosed by x=0 and 𝑥 = (1 − 𝑦 2 )2 Answer: The important thing to note here are the boundaries of the integral. y varies between -1 and 1 and x is given as a function of y from one side and as 0 from the other. We set up the integral and calculate it: 1 √1−𝑦 2 ∫−1 ∫0 1 𝑦 2 −𝑦 4 𝑥𝑦 2 𝑑𝑥𝑑𝑦 = ∫−1 2 𝑑𝑦 = 2 15 Result: 2 15 Question #4 User polar coordinates to find the volume of the solid above the cone 𝑧 = √(𝑥 2 + 𝑦 2 ) and below the sphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 1 Answer: Step 1 Remember that: Volume under the surface 𝑧 = 𝑓(𝑥, 𝑦) and above the region D in xy-plane is ∫ ∫𝐷 𝑓(𝑥, 𝑦)𝑑𝐴 Step 2 In the given problem 𝑧 = √𝑟22 − 𝑥 2 − 𝑦 2 The following integral gives the required volume which is above 𝑧-axis ∫ ∫𝐷 √𝑟2 − 𝑥 2 − 𝑦 2 𝑑𝐴 Where D is the region between the circles of radius 𝑟1 and 𝑟2 But this is only half the volume, therefore volume of the solid which remains is 𝑉 = 2 ∫ ∫𝐷 √𝑟2 − 𝑥 2 − 𝑦 2 𝑑𝐴 Step 3 In polar coordinates, we can define the region D as {(r , )  D | r1  r  r2 , 0    2 } Therefore V = 2 2 0  r2 r1 r2 r22 − r2 (r )drdV = 4  r r22 − r2 dr r1 Substitute r − r = u And − 2rdr = du 2 2 2 Limits of inregration will change from _ r| r2 to 0 1 2 V = −2  2 2 u du 0 r2 − r1  r22 −r12 2 3 V = −2 [ u 2 ]0r 2 − r 2 2 1 3 4 32 0 V = [− u ]r 2 − r 2 2 1 3 4 2 2 32 V= (r2 − r1 ) 3 Result 𝑉= 4π 3 3 (𝑟22 − 𝑟12 )2 Question #5 Use a double integral to find the volume of the indicated solid. 𝑧 = 6 − 2𝑦, 0 ≤ 𝑥 ≤ 4, 0 ≤ 𝑦 ≤ 2 Answer: Step 1 The integral implied by the volume given 4 2 ∫0 ∫0 −2𝑦𝑑𝑦𝑑𝑥 Step 2 Integration with respect to y 4 ∫ (6𝑦 − 𝑦 2 )𝑑𝑥 0 Step 3 Evaluation of integral over y at end points (2 and 0) 4 ∫0 8𝑑𝑥 Step 4 Integration with respect to x 8x Step 5 Evaluation of integral over x at end points 32 Result 32 Question #6 Let X be a random variable with probability density function. c(1 − x 2 ) − 1  x  1 f ( x) =  otherwise 0 (a) What is the value of c? (b) What is the cumulative distribution function of X? Answer: Step 1 Given: Density function of a random variable X as c(1 − x 2 ) − 1  x  1 f ( x) =  otherwise 0 To find: (a) value of (c) cummulative distribution function of X Solution (a) We know that pdf of any random variable integrates to 1. Therefore, we have ∞ ∫−∞ 𝑓(𝑥)𝑑𝑥 = 1 −1 1 ∞ ∫−∞ 0𝑑𝑥 + ∫−1 𝑐(1 − 𝑥 2 )𝑑𝑥 + ∫1 0𝑑𝑥 = 1 1 ∫−1 𝑐(1 − 𝑥 2 )𝑑𝑥 = 1 𝑐 [𝑥 − 𝑥3 1 ] 3 −1 1 =1 𝑐 [1 − − (−1) + −1 3 3 ]=1 4 𝑐=1 3 𝑐= 4 Thus, the density function is 3 3 2  (1 − x ) −1  x  1 f ( x) =  4 0 otherwise Step 2 (b) Cummulative distribution function of X is given as 𝐹𝑋 (𝑥) as π π 3 𝐹𝑥 (𝑥) = 𝑃[𝑋 ≤ 𝑥] = ∫−∞ 0𝑑𝑡 + ∫−1 (1 − 𝑡 2 )𝑑𝑡 4 𝑥 3 𝑡3 4 3 3 −1 𝑥3 4 3 3 𝑥3 4 3 3 1 4 4 = [𝑡 − ] = [𝑥 − = [𝑥 − − (−1) + −1 3 ] 2 + ] 3 3 = 𝑥− 𝑥 + 1 2 Thus, CDF of X is 0 3 1 1  FX ( x) =  x − x 3 + 4 2 4 1 x  −1 −1  x  1 x 1 Result 3 𝑐 = and CDF of X is 4 0 3 1 1  FX ( x) =  x − x 3 + 4 2 4 1 x  −1 −1  x  1 x 1 Question #7 1 1 1 Rewrite the integral as an equivalent iterated integral in the five other orders. ∫0 ∫0 −𝑦 2 ∫0 −𝑦 𝑑𝑧 𝑑𝑥 𝑑𝑦𝑥 ≥ 0, 𝑦 ≥ 0, 𝑧 ≥ 0. Answer: Step 1 we are given this iterated integral, we need to find the other equivalent orders 1−𝑦 2 1 ∫0 ∫0 1−𝑦 ∫0dz dx dy Step 2 dz stays the same because we didnt change the order. we find ''dy'' by solving for y in the equation 1-y^2 you can do the same for the rest of these 1 1−𝑦 √1 − 𝑥0 ∫0dz dx dy ∫0 ∫ Step 3 1 1 1−𝑧 1 1−√1−𝑥 ∫0 ∫√1−𝑥 ∫0dz dx dy ∫0 ∫0 √1 − 𝑥0dz dx dy ∫ Step 4 1 2𝑧−𝑧 2 1 1 ∫0 ∫0 1−𝑧 ∫0dz dx dy ∫0 ∫2𝑧−𝑧 2 ∫ √1 − 𝑥 2 0dz dx dy Step 5 1 1−𝑦 2 1−𝑧 ∫ ∫ 0 ∫ 0 0dz dx dy Step 6 1 1−𝑦 2 1−𝑦 ∫ ∫ 0 ∫ 0 0dz dx dy Question #8 Calculate the double integral. ∫∫ 2(1+𝑥 2 ) 1+𝑦 2 𝑑𝐴, 𝑤ℎ𝑒𝑟𝑒 𝑅 = {(𝑥, 𝑦)|0 ≤ 𝑥 ≤ 2,0 ≤ 𝑦 ≤ 1} Answer: "{𝐼𝑡 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑡ℎ𝑎𝑡 }𝑅 = {(𝑥, 𝑦)|0 ≤ 𝑥 ≤ 2,0 ≤ 𝑦 ≤ 1}. "{𝑊𝑒 ℎ𝑎𝑣𝑒 𝑡𝑜 𝑓𝑖𝑛𝑑} 2(1 + 𝑥 2 ) ∫∫ 𝑑𝐴 2 𝑅 1+𝑦 Notice that 1 1 2(1 + 𝑥 2 ) 2(1 + 𝑥 2 ) ∫∫ 𝑑𝐴 = ∫ ∫ 𝑑𝑥𝑑𝑦 2 2 𝑅 1+𝑦 𝑦=0 𝑥=0 1 + 𝑦 1 2 2 =∫ [ ⋅ ∫ 1 + 𝑥 2 𝑑𝑥] 𝑑𝑦 2 1 + 𝑦 0 0 1 2 8 =∫ [ (2 + )] 𝑑𝑦 2 3 0 1+𝑦 1 = ∫0 ( = 28 3 ⋅ 28 ) 𝑑𝑦 3(1+𝑦 2 ) 1 1 ∫0 1+𝑦2 𝑑𝑦 = 28 𝑦= = [arctan(𝑦)]10 3 [∵ ∫ 1 1+𝑦 2 𝑑𝑦 = arctan(𝑦)] 28 π ⋅ 3 4 7π 3 Result 2 2(1+𝑥 2 ) 1 ∫0 ∫0 1+𝑦 2 𝑑𝑥𝑑𝑦 = 7π 3 Question #9 A variable force of 5𝑥 −2 pounds moves an object along a straight line when it is feet from the origin. Calculate the work done in moving the object from x = 1 ft to x = 10 ft Answer: Equation #4 from this section: If an object moves along x-axis in the positive direction from direction from x=a to x=b and f(x) is a force applied on that object. Then the work done by this force f(x) on the object is given by 𝑏 𝑊 = ∫𝑎 𝑓(𝑥)𝑑𝑥 In the given case 𝑓(𝑥) = 5𝑥 −2 and [a,b]=[1,10] Therefore 10 𝑊 = ∫1 5𝑥 −2 𝑑𝑥 =[ =[ 5𝑥 −2+1 10 ] −2+1 1 10 5𝑥 −1 −1 ] 1 5 10 = [− ] 𝑥 1 5 5 10 1 = [− ] − [− ] = 0.5 + 5 = 4.5𝑙𝑏 ⋅ 𝑓𝑡 Result The work done by the given variable force on the object is 4.5𝑙𝑏 ⋅ 𝑓𝑡 Question #10 Evaluate the following problem. 1 ∫ 𝑥 3 (𝑎2 + 𝑥 2 )2 𝑑𝑥 Answer: Our Aim is to evaluate the integral given below: ∫ 𝑥 3 (√𝑎2 + 𝑥 2 )𝑑𝑥 − (𝑖) Considering the integral given below: ∫ 𝑥 3 (√𝑎2 + 𝑥 2 )𝑑𝑥 − (𝑖) For the integrand 𝑥 2√𝑎 1 = 𝑢√𝑎2 + 𝑢𝑑𝑢 2 +𝑥 2 , substitute 𝑢 = 𝑥 2 and 𝑑𝑢 = 2𝑥𝑑𝑥 2 For, the integrand 𝑢√𝑎2 + 𝑢, substitute 𝑠 = 𝑎2 + 𝑢 and ds=du 1 ∫ √𝑠(𝑠 − 𝑎2 )𝑑𝑠 2 3 Expanding the integral √𝑠(𝑠 − 𝑎2 ) given 𝑠 2 − 𝑎2√𝑠 3 1 = ∫ (𝑠 2 − 𝑎2√𝑠 ) 𝑑𝑠 2 Integrate the sum term by term and taking out constants: 1 3 = ∫ 𝑠 2 𝑑𝑠 − 2 𝑎2 2 ∫ √𝑠𝑑𝑠 5 3 2𝑠 2 The integral of 𝑠 2 is 5 = 𝑠2 − 5 𝑎2 2 5 √𝑠𝑑𝑠 ∫ 3 The integral of \sqrt{s} is 5 = 𝑠2 2𝑠 2 3 3 2 2 1 − 𝑎 𝑠 +𝐶 5 3 Substitute back for 𝑠 = 𝑎2 + 𝑢 5 1 3 1 = (𝑎2 + 𝑢2 )2 − 𝑎2 (𝑎2 + 𝑢)2 + 𝐶 5 3 Substitute back for 𝑢 = 𝑥 2 5 1 3 1 = (𝑎2 + 𝑥 2 )2 − 𝑎2 (𝑎2 + 𝑥 2 )2 + 𝐶 5 3 Answer: ⇒ 1 15 3 (𝑎2 + 𝑥 2 )2 (3𝑥 2 − 2𝑎2 ) + 𝐶 Question #11 Evaluate integration 1 ∫ 3 𝑑𝑥 (𝑥 2 −9)2 using trigonometry substitution Answer: Consider the given integral as ∫ 1 3 (𝑥 2 −9)2 𝑑𝑥 We use the trigonometric substitution 𝑥 = 3 𝑠𝑒𝑐 to evaluate the above integral. 3 When 𝑥 = 3 sec , the expression (𝑥 2 − 9)2 becomes 3 3 (𝑥 2 − 9)2 = ((3 sec 𝑡)2 − 9)2 3 = (9 sec 2 𝑡 − 9)2 3 = (9(sec 2 𝑡 − 1))2 3 3 = 92 (sec 2 𝑡 − 1)2 1 3 3 = (92 ) (sec 2 𝑡 − 1)2 3 = 33 (tan2 𝑡) 2 1 3 = 27 ((tan2 𝑡)2 ) = 27(tan 𝑡)3 27 tan3 𝑡 Further, 𝑥 = 3 𝑠𝑒𝑐 ⇒ 𝑑𝑥 = 𝑑(sec 𝑡) ⇒ 𝑑𝑥 = 3 sec 𝑡 tan 𝑡 𝑑𝑡 3 Substituting (𝑥 2 − 9)2 = 27 tan3 𝑡 and 𝑑𝑥 = 3 sec 𝑡 tan 𝑡 𝑑𝑡 in the integral ∫ ∫ 1 1 3 (𝑥 2 −9)2 =∫ =∫ =∫ 1 𝑑𝑥 = ∫ × 3 sec 𝑡 tan 𝑡 𝑑𝑡 27 tan3 𝑡 sec 𝑡 9 tan2 𝑡 1 ( ) 𝑑𝑡 cos 𝑡 sin 𝑡 2 ) 9( cos 𝑡 cos 𝑡 9 sin2 𝑡 cos 𝑡 𝑑𝑡 𝑑𝑡 = ∫ (sin 2 𝑑𝑡 9 𝑡) 1 (sin 𝑡)′ = ∫ 𝑑𝑡 9 (sin 𝑡)2 1 3 (𝑥 2 −9)2 𝑑𝑥 we get 1 = ∫(sin 𝑡)′ (sin 𝑡)−2 𝑑𝑡 9 1 (sin 𝑡)−2+1 = [ 9 −2+1 1 (sin 𝑡)−1 = [ 9 =− −1 1 ]+𝐶 + 𝐶] +𝐶 9 sin 𝑡 From 𝑥 = 3 𝑠𝑒𝑐 , we have 𝑥 = 3 sec 3 ⇒𝑥= cos 𝑡 3 ⇒ cos 𝑡 = 𝑥 3 2 2 ⇒ (cos 𝑡) = ( ) 𝑥 9 ⇒ cos 2 𝑡 = 𝑥2 ⇒ 1 − sin2 𝑡 = 9 𝑥2 1 ⇒ sin2 𝑡 = (𝑥 2 −9)2 𝑥 1 x 2 − 9) 2 Substituting sin t = in x ∫ 1 3 (𝑥 2 −9)2 1 =− 9( =− 𝑑𝑥 = − 1 (𝑥2−9)2 𝑥 𝑥 1 9(𝑥 2 −9)2 1 9 sin 𝑡  1 ( x 2 − 9) 3 2 dx = − 1 + C we get 9sin t +𝐶 +𝐶 ) +𝐶 Hence, the given integral is ∫ 1 3 (𝑥 2 −9)2 𝑑𝑥 = − 𝑥 1 9(𝑥 2 −9)2 + 𝐶, where C is the constant of integration. Question #12 Evaluate the line integral ∫𝑐 𝑥𝑦𝑑𝑠 Where c is the given curve 𝑐: 𝑥 = 𝑡 2 , 𝑦 = 2𝑡, 0 ≤ 𝑡 ≤ 5 Answer: Explanation Line integrals can be used to evaluate integrals of two or three dimensional curves The curve c is expressed by the following parametric equations given by 𝑥 = 𝑡2 and y=2t Thus, to evaluate the following integral given by 𝐼 = ∫𝑐 𝑥𝑦𝑑𝑠 we use the following line integral formula given by 𝑏 ∫𝑐 𝑓(𝑥, 𝑦)𝑑𝑠 = ∫𝑎 𝑓(𝑥(𝑡), 𝑦(𝑡))|𝑟 ′ |𝑑𝑡 b dr | dt dt dx dy f ( x(t ), y (t )) ( )2 + ( ) 2 dt dt dt =  f ( x(t ), y (t )) | a = b a Since 𝑥(𝑡) = 𝑡 2 and y(t)=2t We want to evaluate ∫𝑐 𝑥𝑦𝑑𝑠 as follows Firstly, we must find the integrand f(x(t),y(t)) as follows f(x(t),y(t))=xy = 𝑡 2 (2𝑡) = 2𝑡 3 Also, the derivatives of x(f) and y(t) must be 𝑑 𝑑 𝑥(𝑡) = (𝑡 2 ) = 2𝑡 𝑑𝑡 𝑑𝑡 and 𝑑 𝑑 𝑦(𝑡) = (2𝑡) = 2 𝑑𝑡 𝑑𝑡 Since 0 ≤ 𝑡 ≤ 5 , so that the upper and lower limits of the integral must be a=0 and b=5 Thus, using the line integral formula, we get 𝑑𝑥 2 𝑏 𝑑𝑦 2 ∫𝑐 𝑥𝑦𝑑𝑠 = ∫𝑎 𝑓(𝑥(𝑡))√( 𝑑𝑡 ) + ( 𝑑𝑡 ) 𝑑𝑡 5 = ∫0 (2𝑡 3 )√(2𝑡)2 + (2)2 𝑑𝑡 5 = 2 ∫0 𝑡 3√4𝑡 = = 2 +4 5 𝑑𝑡 = 2 ∫0 𝑡 3√4(𝑡 2 +1) 𝑑𝑡 5 2 2 ∫0 2𝑡 3√(𝑡 +1) 𝑑𝑡 5 2 4 ∫0 𝑡 3√(𝑡 +1) 𝑑𝑡 5 2 To evaluate the following integral ∫0 𝑡 3√(𝑡 +1) 𝑑𝑡, we use the method of substitution as follows Let 𝑢 = 𝑡 2 + 1, so that Differentiate both sides of the following equation 𝑢 = 𝑡 2 + 1 implies that du=2tdt and thus 𝑑𝑢 𝑡𝑑𝑡 = 2 Also, we have 𝑡2 = 𝑢 − 1 Also, we find that the upper and lower limits of the integral must be 𝑡 = 0 ⇒ 𝑢 = 02 + 1 = 1 and 𝑡 = 5 ⇒ 𝑢 = 52 + 1 = 26 Thus, the line integral becomes 5 √ ∫𝑐 𝑥𝑦𝑑𝑠 = 4 ∫𝑡=0 𝑡 2 𝑡 = = = 2 +1 𝑡𝑑𝑡 26 𝑑𝑢 4 ∫𝑢=1(𝑢 − 1)√𝑢 ( ) 2 4 26 (𝑢 − 1) 𝑢 𝑑𝑢 ∫ √ 2 𝑢=1 3 1 26 2 ∫𝑢=1 (𝑢2 − 𝑢2 ) 𝑑𝑢 Thus, evaluating the following line integral implies that 3 26 5 𝑢2 𝑢2 ∫𝑥𝑦𝑑𝑠 = 2 [ − ] 3 5 𝑐 2 𝑢=1 2 5 = 2[ 2𝑢2 5 = 2 [( = 8 15 3 − 2𝑢2 3 ] 5 3 5 3 2(26)2 2(26)2 2(1)2 2(1)2 5 − 3 )−( 5 − 3 )] (949√26 + 1) Result: We have shown that the following line integral implies that 8 ∫𝑐 𝑥𝑦𝑑𝑠 = (949√26 + 1) 15 for the curve c expressed by the following parametric equations given by 𝑥 = 𝑡2 and y=2t Question #13 Determine whether the following integrals are convergen: ∞ arctan 𝑥 𝑑𝑥 ∫5 2 𝑥 +3𝑥+5 Answer: Solution: ∞ arctan 𝑥 ∫5 𝑥 2 +3𝑥+5 𝑑𝑥 Let 𝑎𝑛 and 𝑏𝑛 be two sequence such that far n, 𝑎𝑛 ≤ 𝑏𝑛 π π − ≤ arctan(𝑥) ≤ 2 ⇒ 2 ∞ arctan 𝑛 ∫5 𝑛2+3𝑛+5 Now, ∞ 𝑎𝑛 = ∫5 ≤ π 2 𝑛2 +3𝑛+5 1 𝑛2 +3𝑛+5 ∞ 1 ∫5 𝑛2 ∞ 1 ∫5 𝑛2 - Converges 𝑎𝑛 = 𝑎𝑛 = Question #14 Solved by the radical integration. ∫ √1−𝑥 2 𝑥4 𝑑𝑥 Answer: Solution: ∫ √1−𝑥 2 𝑥4 𝑑𝑥 x = sin   dx = cos   d =  cos 2  Let cos  d =  sin 4  d =  cot 2  csc2  d Let 𝑢 = cot θ ⇒ 𝑑𝑢 = − csc 2 θ𝑑θ. = − ∫ 𝑢2 𝑑𝑢 =− =− =− ∫ 𝑢3 +𝑐 3 cot3 θ 3 +𝑐 (1−𝑥 2 )√1−𝑥 2 3×3 √1−𝑥 2 𝑥4 𝑑𝑥 = − +𝑐 (1−𝑥 2 )√1−𝑥 2 3×3 Question #15 Evaluate iterated integral. 2 9 3+5𝑦 ∫1 ∫4 √𝑥 𝑑𝑥𝑑𝑦 Answer: +𝑐 1 − sin 2  sin 4  We can evaluate the double integral by integrating with respect to each variable one by one according to their order in the integration term. After integrating with respect to one variable , we have to apply its limit and then we can perform integration using next variable . The following integration formulas can be used here, ∫ 𝑥 𝑛 𝑑𝑥 = 𝑥 𝑛+1 𝑛+1 𝑥2 ∫ 𝑘𝑥𝑑𝑥 = 𝑘 ∫ 1 √𝑥 +𝐶 +𝐶 2 1 − 𝑑𝑥 = ∫ 𝑥 𝑑𝑥 2 2 9 3+5𝑦 Consider the double integral ∫1 ∫4 Where √𝑥 can be written as 𝑥 1 2 √𝑥 𝑑𝑥𝑑𝑦 𝑚 1 1 The expression can be written as (3 + 5𝑦)𝑥 −2 by taking 𝑥 2 to the numerator. 2 1 9 Thus we can write the integral as ∫1 ∫4 (3 + 5𝑦)𝑥 −2 𝑑𝑥𝑑𝑦 First we have to integrate with respect to x by taking y terms outside as a constant. Use the formula ∫ 𝑥 𝑛 𝑑𝑥 = 2 ∫1 (3 + 1 9 5𝑦) [∫4 𝑥 −2 𝑑𝑥] 𝑑𝑦 2 = ∫1 (3 + 5𝑦) [( = = 2 ∫1 (3 2 ∫1 (3 1 𝑥2 1 2 𝑥 𝑛+1 𝑛+1 +𝐶 1 = 2 ∫1 (3 − +1 𝑥 2 + 5𝑦) [( −1 2 +1 9 ) ] 𝑑𝑦 4 9 ) ] 𝑑𝑦 4 1 2 1 + 5𝑦) [2 (9 − 42 )] 𝑑𝑦 + 5𝑦)2(3 − 2)𝑑𝑦 2 ∫1 2(3 + 5𝑦)𝑑𝑦 Now we can integrate the last term with respect to y . 2 2 ∫1 (3 + 5𝑦)𝑑𝑦 = 2 [(3𝑦 + 5 = 2 [( = 2( = 2( 6𝑦+5𝑦 2 2 12+5×4 2 32−11 2 2 𝑦2 2 2 ) ] 1 ) ] 1 6+5 − 2 ) ) 21 = 2( ) 2 =21 thus we obtained required integration value . 2 9 3+5𝑦 Hence we have ∫1 ∫4 √𝑥 𝑑𝑥𝑑𝑦 = 21 Question #16 Solve the equation: ∫ 2𝑒 −5𝑥 𝑑𝑥 Answer: We have to find integration: 𝐼 = ∫ 2𝑒 −5𝑥 𝑑𝑥 -5x=t -5dx=dt −1 𝐼 = 2 ( ) ∫ 𝑒 𝑡 𝑑𝑡 5 𝐼= 𝐼= 𝐼= −2 5 −2 5 −2 5 ∫ 𝑒 𝑡 𝑑𝑡 𝑒𝑡 + 𝐶 𝑒 −5𝑥 + 𝐶 𝑛 using the conversion formula of radical to exponent √𝑎𝑛 = 𝑎𝑚 Question #17 Trigonometric substitutions Evaluate the following integrals using trigonometric substitution. ∫ 𝑑𝑥 ,𝑥 > 1 2 𝑥 3√𝑥 −1 Answer: Consider the provided integral, ∫ 𝑑𝑥 ,𝑥 > 1 2 𝑥 3√𝑥 −1 Evaluate the following integrals using trigonometric substitution. Apply the trigonometric substitution let 𝑥 = sec(𝑢) ⇒ 𝑑𝑥 = sec(𝑢) tan(𝑢) 𝑑𝑢 So, ∫ =∫ =∫ 𝑑𝑥 𝑥 3 (√𝑥 2 −1) =∫ tan(𝑢) sec 2 (𝑢) tan(𝑢) 1 sec2 (𝑢) sec(𝑢) tan(𝑢)𝑑𝑢 sec3 (𝑢)√sec2 (𝑢)−1 𝑑𝑢 𝑑𝑢 Using trigonometric identity simplifying further, 𝑑𝑥 ∫ = ∫ cos 2(𝑢) 𝑑𝑢 √𝑥 2 −1 3 𝑥 1+cos(𝑢) =∫ 𝑑𝑢 2 1 = (∫ 1𝑑𝑢 + ∫ 𝑐𝑜𝑠(2𝑢) 𝑑𝑢) 2 1 1 = (𝑢 + sin(2𝑢)) + 𝐶 2 2 Substitute back 𝑥 = sec(𝑢). 𝑑𝑥 1 1 ∫ = (𝑎𝑟𝑐𝑠𝑒𝑐(𝑥) + 𝑎𝑟𝑐𝑠𝑒𝑐(𝑥)) + 𝐶 2 √ 2 𝑥 3 𝑥 −1 2 Hence. Question #18 Integrals Evaluate the following integrals. Include absolute values only when needed. 𝑒 2𝑥 ∫ 4+𝑒 2𝑥 𝑑𝑥 Answer: Given, ∫ 𝑒 2𝑥 4+𝑒 2𝑥 𝑑𝑥 Put, 4 + 𝑒 2𝑥 = 𝑡 ⇒ 0 + 2𝑒 2𝑥 = 𝑑𝑡 𝑑𝑥 ⇒ 2𝑒 2𝑥 𝑑𝑥 = 𝑑𝑡 ⇒ 𝑒 2𝑥 𝑑𝑥 = 𝑑𝑡 2 Therefore integral becomes, 𝑒 2𝑥 ∫ 4+𝑒 2𝑥 𝑑𝑥 = ∫ 1 = ∫ 2 1 𝑑𝑡/2 𝑡 𝑑𝑡 𝑡 = ln|𝑡| + 𝐶 2 1 = ln|4 + 𝑒 2𝑥 | + 𝐶 2 Question #19 Evaluate the following integrals. Include absolute values only when needed. 𝑒2 𝑑𝑥 ∫𝑒 𝑥 ln3 𝑥 Answer: Given: 𝑒2 𝑑𝑥 ∫𝑒 𝑥 ln3 𝑥 To find- The value of the above integral. Concept Used: Substitution method can be used to find the value of the above integral. Explanation- Rewrite the given integral, 𝑒2 𝑑𝑥 ∫𝑒 𝑥 ln3 𝑥 𝑒2 1 Let, 𝐼 = ∫𝑒 𝑥 ⋅ 𝑑𝑥 ln3 𝑥 Now, substitute ln 𝑥 = 𝑡 and differentiate both sides with respect to x, we get, 1 𝑥 𝑑𝑥 = 𝑑𝑡 As we have substituted ln 𝑥 = 𝑡, so we have to change the limit which is as follows, at x=e, so 𝑡 = ln 𝑒 t=1 at 𝑥 = 𝑒 2 , so 𝑡 = ln 𝑒 2 t=2 So, the above integrals becomes, 1 𝐼 =∈12 3 𝑑𝑡 𝑡 2 = ∫1 𝑡 −3 𝑑𝑡 =[ =[ = = = = 𝑡 −3+1 2 ] −3+1 1 2 𝑡 −2 ] −2 1 −1 1 1 ⋅[ − 2] 2 22 −1 1 1 [ − 1] 2 4 −1 −3 2 3 ⋅( ) 4 8 𝑒2 Answer: Hence, the value of the integral ∫𝑒 𝑑𝑥 𝑥 ln3 𝑥 is 3 8 Question #20 Evaluate the following integrals. Include absolute values only when needed. 𝜋/2 sin 𝑥 ∫0 1+cos 𝑥 𝑑𝑥 Answer: Given, 𝜋/2 sin 𝑥 The integral ∫0 𝜋 2 Let I= ∫0 sin 𝑥 1+cos 𝑥 1+cos 𝑥 𝑑𝑥 𝑑𝑥 Use the substitute 𝑢 = 1 + cos 𝑥 𝜋 Then 𝑑𝑢 = − sin 𝑥 𝑑𝑥 , as 𝑥 = 0 ⇒ 𝑢 = 2 and 𝑥 = ⇒ 𝑢 = 1 1 1 𝐼 = ∫2 − 𝑑𝑢 11 = − ∫2 𝑢 𝑢 𝑑𝑢 = −[ln(𝑢)]12 = −[ln(1) − ln(2)] = −[0 − ln(2)] 2 = ln(2) =0.693 𝜋 sin 𝑥 Thus the value of the integral ∫02 𝑑𝑥 is 0.693 1+cos 𝑥