Questions and Answers #12 Integrals

Questions and Answers Sheet 12 Question #1 Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. g(s)=integral 5 to 𝑥(𝑡 − 𝑡 2 )8 𝑑𝑡 Answer: Recall part 1 of fundament theorem of calculus: 𝑑 𝑥 ∫ 𝑓(𝑡)𝑑𝑡 𝑑𝑥 𝑎 = 𝑓(𝑥) Therefore 𝑑 𝑠 𝑔′ (𝑠) = ∫5 (𝑡 − 𝑡 2 )8 𝑑𝑡 = (𝑠 − 𝑠 2 )8 𝑑𝑠 Result: 𝑔′ (𝑠) = (𝑠 − 𝑠 2 )8 Question #2 Evaluate the line integral along the path C given by x = 2t, y = 10t, where 0 ≤ 𝑡 ≤ 1 ∫ 𝑐 𝑥𝑦𝑑𝑥 + 𝑦𝑑𝑦 Answer: ∫𝐶 𝑥𝑦𝑑𝑥 + 𝑦𝑑𝑦 Given that x=2t and y=4t Therefore, dx=2dt and dy=4dt Substitute the value to get ∫𝐶 (2𝑡 × 4𝑡)(2𝑑𝑡) + (4𝑡)(4𝑑𝑡) Given that 0 ≤ 𝑡 ≤ 1 , therefore we will integrate from 0 to 1 16𝑡 3 1 ∫0 16𝑡 2 + 16𝑡𝑑𝑡 = [ =[ = = = 16𝑡 3 16 2 3 + 16𝑡 2 2 1 ] 1 + 8𝑡 ] 0 0 +8 3 16+24 40 3 3 Question #3 Solve the integral: ∫ 𝑒 8𝑥 (3𝑥)𝑑𝑥 = Answer: 𝑑 ∫ sin 3 𝑥𝑒 8𝑥 𝑑𝑥 = sin 3 𝑥 ∫ 𝑒 8𝑥 𝑑𝑥 − ∫ (𝑑𝑥 (sin 3 𝑥) ∫ 𝑒 8𝑥 𝑑𝑥) 𝑑𝑥 Let I = sin 3 x( 1 e8 x e8 x −  3cos 3 x dx 8 8 3 𝐼 = 𝑒 8𝑥 sin 3 𝑥 − ∫ 𝑒 8𝑥 cos 3 𝑥𝑑𝑥 8 8 Similarly 𝑑 ∫ 𝑒 8𝑥 cos 3 𝑥 = cos 3 𝑥 ∫ 𝑒 8𝑥 𝑑𝑥 − ∫ (𝑑𝑥 (cos 3 𝑥) ∫ 𝑒 8𝑥 𝑑𝑥) 𝑑𝑥 = 𝑒 8𝑥 8 3 cos 3 𝑥 + ∫ 𝑒 8𝑥 sin 3 𝑥𝑑𝑥 8 𝑒 8𝑥 = 8 3 cos 3 𝑥 + 𝐼 8 Substitute this in above integration 𝑒 8𝑥 𝐼= 8 𝑒 8𝑥 𝐼= 𝐼+ 73 64 8 9 64 𝐼= sin 3 𝑥 − 𝑒 8𝑥 𝐼= 𝑒 8𝑥 𝑒 8𝑥 𝐼= 3 𝑒 8𝑥 sin 3 𝑥 − ( 73 8 8 8 3 64 8 𝑒 8𝑥 sin 3 𝑥 − sin 3 𝑥 − − 3 cos 3 𝑥 + 𝐼) cos 3 𝑥 − 3 64 3 64 8 9 64 𝐼 𝑒 8𝑥 cos 3 𝑥 𝑒 8𝑥 cos 3 𝑥 (8 sin 3 𝑥 − 3 cos 3 𝑥) + 𝐶 Question #4 Evaluate the definite integral. 1 3 ∫ √1 + 7𝑥 𝑑𝑥 0 Answer: Compute the definite integral 1 3 ∫0 √7𝑥 + 1𝑑𝑥 3 For the integrand √7𝑥 + 1 , substitute u=7x+1 and du=7dx This gives a new lower bound 𝑢 = 7×0+1= 1 and upper bound 𝑢 = 7 × 1 + 1 = 8 1 8 = ∫ 3√𝑢 𝑑𝑢 7 1 Apply the fundamental theorem of calculus 3 The antiderivative of √𝑢 is 𝑢𝑛+1 {∫ 𝑢𝑛 𝑑𝑢 = } 𝑛+1 =[ 3𝑢4/3 28 3𝑢4/3 4 8 ] 1 Evaluate the antiderivative at the limits and subtract 3𝑢4/3 [ 28 8 ] = 1 3×84/3 28 − 3×14/3 28 = 45 28 Therefore, 1 3 45 ∫0 √1 + 7𝑥 𝑑𝑥 = 28 Question #5 Evaluate the integral. (Use C for the constant of integration.) ∫ 𝑒 4𝜃 sin(5𝜃) 𝑑𝜃. Answer: 4𝜃 We have to find the value of integral ∫ sin(5𝜃) 𝑑𝜃 . To find the value we use the formula ILATE which states 𝑑𝑢 ∫ 𝑢𝑣 = 𝑢 ∫ 𝑣𝑑𝑥 − ∫ [𝑑𝑥 , ∫ 𝑣] 𝑑𝑥 Using ILATE we get 𝑢 = sin 5 𝜃 𝑎𝑛𝑑 𝑣 = 𝑒 4𝜃 substituting the values in the above formula we get sin 5 𝜃𝑒 4𝜃 𝑑𝜃 = sin 5 𝜃 ∫ 𝑒 4𝜃 𝑑𝑥 𝑒 4𝜃 5𝑒 4𝜃 cos(5𝜃) 25 ∫ sin 5 𝜃𝑒 4𝜃 𝑑𝜃 = sin 5 𝜃 −( + ∫ sin 5 𝜃𝑒 4𝜃 𝑑𝜃) 4 16 16 ∫ sin 5 𝜃𝑒 4𝜃 𝑑𝜃 + ∫ sin 5 𝜃𝑒 4𝜃 𝑑𝜃 = 25 ∫ sin 5 𝜃𝑒 4𝜃 𝑑𝜃 16 4 sin 5𝜃𝑒 4𝜃 cos(5𝜃) 41 Question #6 Let 𝐹(𝑥, 𝑦, 𝑧) = 𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘. Evaluate the integral of F alomg each of the following paths. 𝑐(𝑡) = (𝑡, 𝑡, 𝑡), 0 ≤ 𝑡 ≤ 3 Answer: The given path is 𝑐(𝑡) = (𝑡, 𝑡, 𝑡), 0 ≤ 𝑡 ≤ 3 Obtain the value of line integral of F as follows: 3 𝑑𝑐 ∫𝐶 ⋅ 𝑑𝑠 = ∫0 𝐹(𝑡, 𝑡, 𝑡) ⋅ 𝑑𝑡 𝑑𝑡 3 = ∫0 (𝑡, 𝑡, 𝑡) ⋅ (1,1,1)𝑑𝑡 3 = ∫0 3𝑡𝑑𝑡 𝑡2 3 3( ) 2 0 9 = 3 ( − 0) 2 27 = 2 Thus, the value of line integral of F is 27 2 Question #7 Two components of a minicomputer have the following joint pdf for their lifetimes X and Y: 𝑓(𝑥, 𝑦) = 𝑥𝑒 −𝑥(1+𝑦) , where 𝑥 ≥ 0, 𝑦 ≥ 0 f(x,y)=0 otherwise What is the probability that the lifetime X of the first component exceeds 3? What are the marginal pdf's of X and Y? Are the two lifetimes independent? Answer: The marginal pdf of X, by definition, should be (for 𝑥 ≥ 0 ) ∞ ∞ ∞ 𝑓(𝑥) = ∫ 𝑓(𝑥, 𝑦)𝑑𝑦 = ∫ 𝑥𝑒 −𝑥(1+𝑦) 𝑑𝑦 = −𝑒 −𝑥 ∫0 −∞ 𝑑(𝑒 −𝑥𝑦 ) = 𝑒 −𝑥 0 Thus, the probability that the life time X of the first component exceeds 3 is equal to ∞ ∞ ∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑒 −𝑥 𝑑𝑥 = 𝑒 −3 3 3 Similarly, the marginal pdf for Y should be (for 𝑦 ≥ 0 and using integration by parts)   −   0 0 f ( x, y )dx =  xe − x (1+ y ) dx =  −x 1 d (e − x (1+ y ) ) = 1+ y 1 + y)2 It is easy to check that 𝑓(𝑥, 𝑦) ≠ 𝑓(𝑥)𝑓(𝑦) in general, and hence the two life times are not independent. Question #8 Calculate this triple integral over D: ∫ ∫ ∫ √10 + 𝑦 2 − 𝑥 2 𝑑𝑥𝑑𝑦𝑑𝑧 over the region bounded D: 𝐷 = {(𝑥, 𝑦, 𝑧) ∣ 𝑥 2 + 𝑧 2 ≤ 𝑦 2 + 10, 𝑥 2 + 𝑦 2 ≤ 9} Answer: First of all we see from 𝑥 2 + 𝑦 2 = 9 that y lives in J_y=[-3,3] Now we can compute using Fubini: ∫ ∫ ∫𝐷 √10 + 𝑦 2 − 𝑥 2 𝑑𝑥𝑑𝑦𝑑𝑧 =∫ 𝑑𝑦 ∫ =∫ √10 + 𝑦 2 − 𝑥 2 𝑑𝑥 ∫ 𝑥∈𝐽𝑥 (𝑦) 𝑦∈[−3,3] 𝑧∈𝐽𝑧 (𝑥,𝑦) 𝑑𝑥 ⋅ 2√10 + 𝑦 2 − 𝑥 2 ⋅ √10 + 𝑦 2 − 𝑥 2 𝑑𝑦 ∫ 𝑥∈𝐽𝑥 (𝑦) 𝑦∈[−3,3] 𝑑𝑥 ⋅ (10 + 𝑦 2 − 𝑥 2 ) = 2 ∫𝑦∈[−3,3] 𝑑𝑦 ∫𝑥∈𝐽 𝑥 (𝑦) 1 𝑥=+√9+𝑦 2 3 𝑥=−√9−𝑦 2 = 2 ∫𝑦∈[−3,3] 𝑑𝑦 [10𝑥 + 𝑥𝑦 2 − 𝑥 3 ] 1 = 4 ∫𝑦∈[−3,3] 𝑑𝑦√9 − 𝑦 2 (10 + 𝑦 2 − (9 − 𝑦 2 )) 3 = 3 4 ∫−3 4 𝑦2 √9 − 2 (7 + 𝑦 ) 𝑑𝑦 3 Substitution 𝑦 = 3 sin 𝜋/2 4 = 4 ∫−𝜋/2 3 cos 𝑡 (7 + 9 sin2 𝑡) 3 cos 𝑡 𝑑𝑡 3 𝜋/2 = 3 ⋅ 32 ⋅ 2 ∫ cos 2 𝑡(7 + 12 sin2 𝑡)𝑑𝑡 0 = 180𝜋 Question #9 Evaluate the line integral, where C is the given curve. integral C 𝑥𝑒 𝑦𝑧𝑑𝑠 C is the line segment from (0, 0, 0) to (1, 2, 3) Answer: Step 1 Let C the line segment from (0, 0, 0) to (1, 2, 3) . The parametric equation for this line in vector form is r(t)=r_{0}+td where we chose 𝑟0 =≪ 0,0,0 ≫ 𝑟 and the direction vector d is 𝑑 =≪ 1,2,3 ≫≺< 0,0,0 ≫=≪ 1,2,3 ≫ Hence we have 𝑟(𝑡) =≪ 𝑡, 2𝑡, 3𝑡 ≫ and we have in scalar form the parametric equations for C : 𝑥 = 𝑡, 𝑦 = 2 , 𝑧 = 3 , 0 ≤ 𝑡 𝑙𝑒𝑠𝑙𝑎𝑛𝑡 1 To determine the line integral: 𝐼 = ∫𝐶 𝑥𝑒 𝑦𝑧 ds we first determine ds: 𝑑𝑥 2 𝑑𝑥 2 𝑑𝑧 2 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑠 = √( ) + ( ) + ( ) 𝑑𝑡 √(1)2 (2)2 = + = = √1 + 4 + 9𝑑𝑡 = √14𝑑𝑡 (3)2 𝑑𝑡 Step 2 1 ∫𝑥𝑒 𝑦𝑧 𝑑𝑠 = ∫ 𝑡𝑒 6𝑡 𝐶 2 √14 1 2 𝑑𝑡 = √14 ∫ 𝑒 6𝑡 𝑡𝑑𝑡 0 0 Now changing to the variable 𝑢 = 6𝑡 2 , we have du=12t dt=12t dt and: 6 1 𝐼 = √14 ∫0 12 14 u = e 12 = √14 12 (𝑒 6 𝑒 𝑢 𝑑𝑢 = 6 0 − 1) = √14 12 6 ∫0 𝑒 𝑢 𝑑𝑢 14 6 0 (e − e ) 12 Finally: ∫𝑐𝑥 𝑒 𝑦𝑧 𝑑𝑠 = √14 12 (𝑒 6 − 1) Question #10 Evaluate the integral by making an appropriate change of variables. double integral 𝑥 − 2𝑦/3𝑥 − 𝑦 𝑑𝐴 , where R is the parallelogram enclosed by the lines x-2y=0 x-2y=4, 3x-y=1 and 3x-y=8 Answer: Step 1 1) Let's start by substituting u=x-2y and v=3x-y . From there we get 𝑣−3𝑢 𝑦= 𝑥= 5 2𝑣−𝑢 5 2) R is a rectangle defined by 0 ≤ 𝑢 ≤ 4 and 1 ≤ 𝑣 ≤ 8 . We will need this later as bounds for integration. 3) Now we are ready to find the Jacobian. ( x, y) −1/ 5 2 / 5 =| | (u, v) −3 / 5 1/ 5 = 1 5 4) Now we can evaluate the integral. 4 𝑥−2𝑦 8𝑢 1 ∫ ∫𝑅 3𝑥−𝑦 𝑑𝐴 = ∫0 ∫1 | | 𝑑𝑣 𝑑𝑢 𝑣 5 5) Notice that we can separate the integrals. Thus getting the following. 81 1 4 = ∫0 𝑢 𝑑𝑢 ∫1 𝑑𝑣 5 1 1 4 2 0 𝑣 = ⋅ [ 𝑢2 ] ⋅ [ln 𝑣]18 5 8 = ln 8 5 Question #11 Find the length of the curve over the given interval. 𝑟 = 8 + 8 cos 𝜃 on the interval 0 ≤ 𝜃 ≤ 𝜋 Answer: Step 1 The given function is 𝑟 = 8 + 8 𝑐𝑜𝑠 𝜃 on the interval 0 ≤ 𝜃 ≤ 𝜋 The length of the polar curve is given by L= b a (r ( )) 2 + (r ( )) 2 d First, find the derivative: 𝑟 ′ (𝜃)(8 cos(𝜃) + 8)′ = −8 sin(𝜃) Step 2 Finally, calculate the integral L=  (8cos( ) + 8) 2 + (−8sin( )) 2 d 0  =  8 2 cos( ) + 1d 0   =  8 2 2 cos 2 ( ) d 0 2   =  16 cos( )d 0 2   = 32(sin( )) |( = ) −32(sin( )) |( =0) 2 2 = 32 Question #12 Find the charge flowing through a device if the current is: a) 𝑖(𝑡) = 3𝐴, 𝑞(0) = 1𝐶 b) 𝑖(𝑡) = (2𝑡 + 5)𝑚𝐴, 𝑞(0) = 0(𝑐)𝑖(𝑡) = 20 cos(10𝑡 + 𝜋/6) 𝜇𝐴, 𝑞(0) = 2𝜇𝐶(𝑑)𝐴, 𝑖(𝑡) = 10𝑒 −30𝑡sin 4 0𝑡𝐴, 𝑞(0) = 0 Answer: Let's first define that: 0 𝑞(𝑡) = 𝑞(0) + ∫𝑡 𝑖(𝑡)𝑑𝑡 Solving for a 0 𝑞(𝑡) = ∫𝑡 3𝑑𝑡 + 1 = 3𝑡|0𝑡 + 1 = (3𝑡 + 𝑞)𝑐 Solving for b 𝑖(𝑡) = (2𝑡 + 5)𝑚𝐴, 𝑞(0) = 0 0 𝑞(𝑡) = ∫𝑡 (2𝑡 + 5)𝑑𝑡 + 0 = ( 2𝑡 2 2 + 5𝑡) \𝐵𝑖𝑔 0𝑡 = (𝑡 2 + 5𝑡)𝑚𝐶 Solving for c 𝜋 𝑖(𝑡) = 20𝑐𝑜𝑠 (10𝑡 + ) 𝜇𝐴, 𝑞(0) = 2𝜇𝑐 𝑞(𝑡) = 20 10 6 0 𝜋 ∫𝑡 20𝑐𝑜𝑠 (10𝑡 + 6 ) 𝑑𝑡 𝜋 0 +2= 𝑠𝑖𝑛 (10𝑡 + ) + 2 = 6 𝑡 𝜋 0 2𝑠𝑖𝑛 (10𝑡 + ) + 2 = 6 𝑡 𝜋 𝜋 2𝑠𝑖𝑛 (10𝑡 + ) − 2𝑠𝑖𝑛 (10(0) + ) + 2 = 6 𝜋 6 1 2𝑠𝑖𝑛 (10𝑡 + ) − + 2 = 6 2 𝜋 (2𝑠𝑖𝑛 (10𝑡 + ) + 1) 𝜇𝐶 6 Solving for d 𝑖(𝑡) = (10𝑒 −30𝑡 sin 4 0𝑡)𝐴, 𝑞(0) = 0 𝑡 𝑞(𝑡) = 10 ∫0 𝑒 −30𝑡 sin 4 0𝑡𝑑𝑡 + 0 Solving this integration by parts we have ∫ 𝑢𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣𝑑𝑢 so taking 𝑢 = sin 4 0𝑡 𝑑𝑣 = 𝑒 −30𝑡 𝑑𝑢 = 40 cos 4 0𝑡 𝑣= 𝑒 −30𝑡 −30 𝑞(𝑡) = 10 [( sin 40𝑡𝑒 −30𝑡 −30 𝑡 𝑒 −30𝑡 ) |𝑡0 − ∫0 −30 4 − cos 4 0𝑡𝑑𝑡] Simplifying: sin 40𝑡𝑒 −30𝑡 𝑡 |0 3 𝑡 40 ∫0 𝑒 −30𝑡 cos 4 0𝑡𝑑𝑡 Again solving this integration by parts: so taking 𝑢 = cos 4 0𝑡 𝑑𝑣 = 𝑒 −30𝑡 𝑑𝑢 = −40 sin 4 0𝑡 v=\frac{e^{-30t}}{-30} − 𝑞(𝑡) = − + 3 sin 40𝑡𝑒 −30𝑡 𝑡 |0 3 + 40 cos 40𝑡𝑒 −30𝑡 𝑡 |0 [ 3 −30 𝑡 𝑒 −30𝑡 − ∫0 −30 (−40 sin 4 0𝑡)𝑑𝑡] Note that the last integration we got is the original one which is equal to q(t) so we will substitute with q(t) instead 𝑞(𝑡) = − 𝑞(𝑡) + 16 9 𝑞(𝑡) [1 + 𝑞(𝑡) = 9 sin 40𝑡𝑒 −30𝑡 3 𝑞(𝑡) = − 16 9 4 ]=− [ − 4 9 sin 40𝑡𝑒 −30𝑡 3 sin 40𝑡𝑒 −30𝑡 3 sin 40𝑡𝑒 −30𝑡 25 9 1 𝑞(𝑡) = ( [4 − 𝑒 25 4 16 9 9 − cos 4 0𝑡𝑒 −30𝑡 + − 3 −30𝑡 (3 𝑞(𝑡) 4 − cos 4 0𝑡𝑒 −30𝑡 + 9 4 − cos 4 0𝑡𝑒 −30𝑡 9 4 − cos 4 0𝑡𝑒 9 −30𝑡 + 4 9 4 9 ] sin 4 0𝑡 + 4 cos 4 0𝑡)]) 𝐶 Question #13 Which of the following integrals are improper? (Select all that apply.) 5 1 ∫4 𝑑𝑥 5𝑥−1 1 1 ∫0 2х−1 𝑑𝑥 −∞ 𝑠𝑖𝑛(𝑥) ∫∞ 1+3𝑥 2 𝑑𝑥 3 ∫1 𝑙𝑛(𝑥 − 1)𝑑𝑥 Answer: An improper integral is a definite integral that has either or both limits infinite or an integrand that approaches infinity at one or more points in the range of integration. We will each integral one by one. 5 1) ∫4 1 5𝑥−1 since any of the limits is not equal to infinity, so we check for the integrand. 1 using ∫ = 𝑙𝑛|𝑥| 𝑥 5 ∫4 1 1 5𝑥−1 1 = 𝑙𝑛|5𝑥 − 1||54 5 = 𝑙𝑛|24| − 𝑙𝑛|19| 5 the integrand is not approaching to infinity, hence given integral is not improper. 1 1 2) ∫0 2𝑥−1 1 1 2 2 plugging 𝑥 = , the integral apptoaches to 0, and 𝑥 = is within the range of integration [0,1]. So this is an improper integration. ∞ 3) ∫−∞ 𝑠𝑖𝑛𝑥 1+3𝑥 2 𝑑𝑥 limits of integration are infinity, hence by definition it is an improper integral. 3 4) ∫1 𝑙𝑛(𝑥 − 1)𝑑𝑥 at x=1, the integral becomes undefined and x=1 is within the range of integration. so it is an improper integration. Therefore, option (2),(3) and (4) are examples of improper integration. Question #14 Find the exact length of the curve. 𝑌 = 𝑦 4 /8 + 1/4𝑦 2 , 1 less than or equal to Y less than or equal 2. Answer: Given: 𝑌= 𝑦4 + 8 1 ,1 ≤ 𝑦 ≤ 2 4𝑦 2 As, the length of the curve, L is given by: 2 𝑏 𝑑𝑦 𝐿 = ∫𝑎 √1 + ( ) 𝑑𝑥 𝑑𝑥 According to the given question, we have: a=1 , b=2 Since, 𝑌 = 𝑦4 + 8 1 4𝑦 2 Therefore, 𝑑𝑌 𝑑𝑦 𝑑𝑌 𝑑𝑦 𝑑𝑌 𝑑𝑦 = = = 𝑑 𝑑𝑦 43 8 𝑦3 2 ( 𝑦4 8 − = + 1 4𝑦 2 ) 2 4𝑦 3 1 2𝑦 3 Therefore, the length of the given curve is: 3 2 𝑦 𝐿 = ∫1 √1 + ( − 2 1 2𝑦 3 2 ) 𝑑𝑦 1 𝑦 6 −1 2 𝐿= 2 ∫1 √1 𝐿= 2 4𝑌6+(𝑦 6 −1) ∫1 ( 4𝑦 6 𝐿= 2 4𝑦 6 +𝑦 1 2−2𝑦 6 +1 2 ) 𝑑𝑦 ∫1 ( 4𝑦 6 + ( 4 ) 𝑑𝑦 𝑦3 1 2 2 ) 𝑑𝑦 1 1 2 𝑦 1 2+2𝑦 6 +1 2 𝐿 = ∫1 ( 𝐿= 4𝑦 6 1 2 2 2 (𝑦 6 +1) ∫1 ( (2𝑦3 )2 2 𝑦 6 +1 𝐿 = ∫1 ( 𝐿= ) 𝑑𝑦 2𝑦 3 2 ∫ (𝑦 3 2 1 ) 𝑑𝑦 1 1 𝑦4 𝐿= [ 2 4 1 16 𝐿= [ 2 1 4 − ) 𝑑𝑦 + 1 1 𝑦3 2 ) 𝑑𝑦 ] 2𝑦 2 1 1 1 1 − − + ] 8 1 4 2 𝐿 = [4 + ] 2 1 𝐿= × 𝐿= 2 33 8 33 8 16 𝐿 = 2.0625 Hence, the exact length of the given curve is 2.0625. Question #15 Find the points on the curve 𝑦 = 2𝑥 3 , 3𝑥 2 − 12𝑥4 where the tangent line is horizontal Answer: Find the points on the curve where the tangent line is horizontal as follows. The given curve is 𝑦 = 2𝑥 3 + 3𝑥 2 – 12𝑥 + 4 Compute the derivative of the function as follows. 𝑑 𝑦` = (2𝑥 3 + 3𝑥 2 − 12𝑥 + 4) 𝑑𝑥 = 6𝑥 2 + 6𝑥 − 12 Equate the derivative equal to zero as follows. 6𝑥 2 + 6𝑥 − 12 = 0 𝑥2 + 𝑥 − 2 = 0 (x + 2) (x - 1) = 0 x = -1 or x = 1 For the smaller value of x that is , the value of y is as follows. 𝑦 = 2(−2)3 + 3(−2)2 − 12(−2) + 4 = - 16 + 12 + 24 + 4 = 24 For the larger value of x that is x = 1, the value of y is as follows. 𝑦 = 2(1)3 + 3(1)2 − 12(1) + 4 = -3 Thus, for the smaller value of x, (x, y) = (-2, 24) and for the larger values of x, (x, y) = (1, -3) Question #16 Calculate the double integral 4𝑥 = ∫ ∫𝑅 𝑑𝐴 , 𝑅 = [0,4] × [0,1] 1+𝑥𝑦 Answer: Step 1 The given integrals is = ∫ ∫𝑅 4𝑥 1+𝑥𝑦 𝑑𝐴 where = 𝑅 = [0,4] × [0,1] Evaluate as follows Step 2 = ∫ ∫𝑅 4𝑥 1+𝑥𝑦 1 4 1 4 4𝑥 𝑑𝐴 = ∫0 ∫0 1+𝑥𝑦 𝑑𝑥 𝑑𝑦 1 1 = 4∫ ∫ ( − ) 𝑑𝑥 𝑑𝑦 𝑦(1 + 𝑥𝑦) 0 0 𝑦 4 1 𝑥 ln(1 + 𝑥𝑦) = 4∫ [ − ] 𝑑𝑦 𝑦2 0 𝑦 0 1 4 = 4 ∫0 ( − = ln(1+𝑥𝑦) ) 𝑑𝑦 𝑦 𝑥2 1 4 ln(1+4𝑦) 4 ∫0 2 (4𝑦 − ) 𝑦 4 = 4[5 𝐼𝑛(5) − 4] (by uv meth) = 4(5 𝐼𝑛(5) − 4) 4𝑥 (5 𝐼𝑛 (5) − 4). Thus, = ∫ ∫𝑅 1+𝑥𝑦 Question #17 2 Is there really no way to integrate 𝑒 −𝑥 ? 2 We encountered the function 𝑒 −𝑥 , and I was told that it was not integrable. 2 2 I was very surprised. Is there really no way to find the integral of 𝑒 −𝑥 ? Graphing 𝑒 −𝑥 , it appears as though it should be 2 ∞ ∫−∞ 𝑒 −𝑥 𝑑𝑥 = √𝜋 This is from -infinity to infinity. If the function can be integrated within these bounds, I'm unsure why it can't be integrated with respect to (a,b). 2 Is there really no way to find the integral of 𝑒 −𝑥 , or are the methods to finding it found in branches higher than second semester calculus? Answer: That function is integrable. As a matter of fact, any continuous function (on a compact interval) is Riemann integrable (it doesn't even actually have to be continuous, but continuity is enough to guarantee integrability on a compact interval). 2 The antiderivative of 𝑒 −𝑥 (up to a constant factor) is called the error function, and can't be written in terms of the simple functions you know from calculus, but that is all. Question #18 Find the area of the surface. The part of the plane 3𝑥 + 2𝑦 + 𝑧 = 6 that lies in the first octant. Answer: Step 1 Remember that: Area of the surface which is part of z=f(x, y) is 2 𝐴(𝑆) = ∫ ∫𝐷 √1 + [𝑓𝑥 ]2 + [𝑓𝑦 ] 𝑑𝑥𝑑𝑦 Where D is the projection of the surface on the xy plane. Step 2 In the given problem f(x, y)=6-3x-2y Area of the surface is 𝐴(𝑆) = ∫ ∫𝐷 √1 + (−3)2 + (−2)2 𝑑𝐴 𝐴(𝑆) = ∫ ∫𝐷 √1 + 9 + 4 𝑑𝐴 𝐴(𝑆) = √14 ∫ ∫𝐷 𝑑𝐴 Note that D is the projection of the surface on the xy-plane Step 3 Intercept form of plane The x, y, z intercepts of the following plane are a, b, c respectively: 𝑥 𝑦 𝑧 + + =1 𝑎 𝑏 𝑐 Write the given equation of the plane in intercept form: 𝑥 𝑦 𝑧 + + =1 2 3 6 The x-intercept is 2 and y-intercept is 3 Projection of part of plane in 1st octant on the xy-plane is A triangle with vertices (0, 0), (2, 0) and (0, 3) Step 4 Note that: ∫ ∫𝐷 𝑑𝐴 is the area of the region inside D 1 1 2 2 Area of the triangle = × Base × Height = × 2 × 3 = 3 Therefore 𝐴(𝑆) = √14 ∫ ∫𝐷 𝑑𝐴 = 3√14 Question #19 Determine a region whose area is equal to the given limit. Do not evaluate the limit. lim x tends to infinity summation 𝜋 4𝑛 × tan 𝑖𝜋 4𝑛 Answer: Step 1 𝑏 Remember that ∫𝑎 𝑓(𝑥)𝑑𝑥 = lim Δ𝑥 ∑𝑛𝑖=1 𝑓(𝑎 + 𝑖Δ𝑥) 𝑛⇒∞ Where Δ𝑥 = Step 2 𝑏−𝑎 𝑛 lim 𝑛 𝜋 𝑛⇒∞ ∑𝑖=1 𝑖𝜋 ( ) tan 4𝑛 4𝑛 On comparing we see that, 𝜋 Δ𝑥 = 4𝑛 a=0 Δ𝑥 = 𝑏−𝑎 𝑛 = 𝜋 4𝑛 ⇒𝑏= 𝜋 4 and 𝑓(𝑥) = tan 𝑥 𝜋 Therefore, lim 𝑛⇒∞ ∑𝑛 𝑖=1 tan( 𝑖𝜋 ) 4𝑛 = ∫04 tan 𝑥 𝑑𝑥 We conclude that, This sum represent the area under the curve 𝑦 = tan 𝑥 and above x-axis in the interval 𝑥 ∈ [0. 𝜋 4 ] The sum represent the area of the blue region in the graph below: Question #20 Calculate the double integral 𝑅𝑦𝑒 −𝑥𝑦 𝑑𝐴, 𝑅 = [0,2] × [0,3] Answer: You want to integrate with respect to x here first to ease the calculation. In essence it doesn't matter but recognize that the function under the integral is derivative of −𝑒 −𝑥𝑦 with respect to x, hence we solve one integral without too much computation: 3 2 3 5 ∫0 ∫0 𝑦𝑒 −𝑥𝑦 𝑑𝑥𝑑𝑦 = ∫0 −𝑒 −2𝑦 + 1𝑑𝑦 = 2 + Result: 5 2 + 𝑒 −6 2 𝑒 −6 2