Questions and Answers #11 Integrals

Questions and Answers Sheet 11 Integrals Question #1 Use the Comparison Theorem to determine whether the integral is convergent or divergent. integral 0 to 𝑥 infinity 3 + 1𝑑𝑥 𝑥 Answer: ∞ 1 𝑥 ∫0   𝑥 𝑑𝑥 ≤ ∫0 𝑥 3 +1 ∞ 𝑥 ∫0 𝑥 3+1 𝑑𝑥 ≤ ∞ 𝑥 𝑑𝑥 + ∫1 𝑥 3 +1 1 𝑥 ∫0 𝑥 3 +1 𝑑𝑥 + 𝑑𝑥 𝑥 3 +0 ∞ 1 ∫1 𝑥 2 𝑑𝑥 x x 1 dx   3 dx + − ]1 0 x +1 x +1 x 1 3 0 ∞ 1 𝑥 ∫0 𝑥 𝑑𝑥 ≤ ∫0 𝑥 3 +1 ∞ 𝑥 ∫0 𝑥 3+1 𝑑𝑥 ≤ 𝑑𝑥 + [− 𝑥 3 +1 1 𝑥 ∫0 𝑥 3 +1 𝑑𝑥 1 ∞ 1 + ] 1 +1 Note that the RHS is finite, because 𝑥 𝑥 3 +1 is finite on the interval [0,1] Hence LHS is finite because it is less than RHS Result: Converges Question #2 4 √2/2 Evaluate the following integrals in cylindrical coordinates. ∫0 ∫0 √1 ∫𝑥 −𝑥𝑒 −𝑥 2 −𝑦 2 𝑑𝑦𝑑𝑥𝑑𝑧 Answer: We rewrite the integral in cylindrical coordinates. We know that 𝑟 2 = 𝑥 2 + 𝑦 2, 0 ≤ 𝑟 ≤ 1,0 ≤ 𝑧 ≤ 4,0 ≤ 𝜃 ≤ 𝜋 . Therefore, 4 𝜋 4 1 4 𝜋 4 2 1 2 ∫ ∫ ∫ (𝑒 −𝑟 )𝑟𝑑𝑧𝑑𝑟 𝑑𝜃 = ∫ ∫ 𝑟𝑒 −𝑟 [𝑧]40 𝑑𝑟𝑑𝜃 0 0 𝜋 4 0 1 0 = ∫ ∫ 4𝑟𝑒 0 𝜋 −𝑟 2 0 𝑑𝑟𝑑𝜃 0 −1 = ∫04 (4 ∫0 − 𝜋 4 𝑒𝑢 2 𝑑𝑢) 𝑑 𝜃 0 = ∫ (2 ∫ 𝑒 𝑢 𝑑𝑢) 𝑑𝜃 0 𝜋 4 −1 = ∫ (2[𝑒 𝑢 ]0−1 )𝑑𝜃 0 𝜋 1 = ∫04 2 (1 − ) 𝑑 𝜃 𝑒 1 𝜋 = 2 (1 − ) [𝜃]04 𝜋 𝑒 1 = (1 − ) 2 Result: 𝑒 𝜋 2 1 (1 − ) 𝑒 Question #3 Evaluate the surface integral ∫ ∫𝑆(𝑥 2 𝑧 + 𝑦 2 𝑧)𝑑𝑠 where S is the hemisphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 9, 𝑧 ≥ 0 Answer: Answer: The surface integral ∫ ∫𝑆(𝑥 2 𝑧 + 𝑦 2 𝑧)𝑑𝑠 (for the surface represented by the hemisphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 9, 𝑧 ≥ 0 ) is ∫ ∫𝑆(𝑥 2 𝑧 + 𝑦 2 𝑧)𝑑𝑠 = 18 𝜋 Explanation: The surface S has parametric equations 𝑥 = 𝑟 sin(𝜃) cos(𝜙) 𝑦 = 𝑟 sin(𝜃) sin(𝜙) 𝑧 = 𝑟 cos(𝜃) 𝜋 where 0 ≤ 𝜃 ≤ and 0 ≤ 𝜙 ≤ 2𝜋 and S 2 is the hemisphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 9 in the form 𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑟 2 where r is the radius of the sphere, thus r=3 And the parametric equations becomes 𝑥 = 3 sin(𝜃) cos(𝜙) 𝑦 = 3 sin(𝜃) sin(𝜙) 𝑧 = 3 cos(𝜃) Thus, we get 𝑥 2 𝑧 + 𝑦 2 𝑧 = (3 sin 𝜃 cos 𝜙)2 (3 cos 𝜃) + (3 sin 𝜃 sin 𝜙)2 (3 cos 𝜃) = (9 sin2 𝜃 cos 2 𝜙)(3 cos 𝜃) + (9 sin2 𝜃 sin2 𝜙)(3 cos 𝜃) = 27 sin2 𝜃 cos 𝜃 (cos 2 𝜙 + sin2 𝜙) = 27 sin2 𝜃 cos 𝜃 (1) = 27 sin2 𝜃 cos 𝜃 Thus, the surface integral ∫ ∫𝑆(𝑥 2 𝑧 + 𝑦 2 𝑧)𝑑𝑠 becomes 𝜋/2 ∫ ∫(𝑥 2 𝑧 + 𝑦 2 𝑧)𝑑𝑠 = ∫ 𝑆 𝜋/2 2 𝜋(27 sin2 𝜃 cos 𝜃)𝑑𝜙] 𝑑𝜃 [∫ 𝜃=0 𝜙=0 (27 sin2 𝜃 cos 𝜃)[𝜙]2𝜋 𝜙=0 𝑑𝜃 ∫ 𝜃=0 2𝜋 ∫ 𝜋/2 = 27[𝜙]𝜙=0𝜃=0 sin2 𝜃 cos 𝜃 𝑑𝜃 𝜋/2 Finally, we avaluate the second integral given by ∫𝜃=0 sin2 𝜃 cos 𝜃 𝑑 𝜃 as follows 𝜋/2 ∫ 𝜋/2 sin2 𝜃 cos 𝜃 𝑑𝜃 = ∫ 𝜃=0 [sin 𝜃]2 cos 𝜃 𝑑𝜃 𝜃=0 𝜋/2 [sin 𝜃]3 ] 3 𝜃=0 1 𝜋/2 = [(sin 𝜃)3 ]𝜃=0 3 Thus, the surface integral ∫ ∫𝑆(𝑥 2 𝑧 + 𝑦 2 𝑧)𝑑𝑠 becomes =[ 2𝜋 ∫ 𝜋/2 ∫ ∫(𝑥 2 𝑧 + 𝑦 2 𝑧)𝑑𝑠 = 27[𝜙]𝜙=0𝜃=0 sin2 𝜋 1 3 2 = 27[𝜙]2𝜋 𝜙=0 [(sin 𝜃) ]𝜃=0 3 𝜋/2 [(sin = 9[𝜙]2𝜋 𝜃)3 ]𝜃=0 𝜙=0 = 9[2𝜋 − 0][(sin 𝜋 /2)3 − (sin 0)3 ] = 9[2𝜋][(sin 𝜋 /2)3 ] = 9[2𝜋][(1)3 ] = 18𝜋 Question #4 Calculate the double integral. 1 𝑑𝐴 , 𝑅 = [1,3] × [1,2] ∫ ∫𝑅 1+𝑥+𝑦 𝜃 cos 𝜃 𝑑𝜃 Answer: Once again use the fact that: ∫ 𝑓 ′(𝑥) 𝑓(𝑥) 𝑑𝑥 = ln|𝑓(𝑥)| + 𝐶 It doesn't matter do you opt to solve the first integral with respect to x or with respect to y, Here, we opt to solve the integral with respect to y first: 3 2 ∫1 ∫1 1 1+𝑥+𝑦 3 𝑑𝑦𝑑𝑥 = ∫1 ln(𝑥 + 3) − ln(𝑥 + 2) Now using the integration by parts we obtain that the integral is equal to: ((𝑥 + 3) ln(𝑥 + 3) − (𝑥 + 3)) − ((𝑥 + 2) ln(𝑥 + 2) − (𝑥 + 2))|13 = 6 ln 6 − 5 ln 5 − 4 ln 4 + 3 ln 3 Result: 6 ln 6 − 5 ln 5 − 4 ln 4 + 3 ln 3 Question #5 Apply Green’s Theorem to evaluate the integral. ∮(6𝑦 + 𝑥)𝑑𝑥 + (𝑦 + 2𝑥)𝑑𝑦 𝐶 C: The circle (𝑥 − 2)2 + (𝑦 − 3)2 = 4 Answer: Green's theorem - Tangential form 𝑑𝑁 ∮𝐶 𝑀𝑑𝑥 + 𝑁𝑑𝑦 = ∫ ∫𝑅 ( 𝑑𝑥 − 𝑑𝑀 𝑑𝑦 ) 𝑑𝑥𝑑𝑦 Therefore (6𝑦 + 𝑥)𝑑𝑥 + (𝑦 + 2𝑥)𝑑𝑦 = ∫ ∫𝑅 ( 𝑑(𝑦+2𝑥) 𝑑𝑥 − 𝑑(6𝑦+𝑥) 𝑑𝑥 ) 𝑑𝑥𝑑𝑦 = ∫ ∫𝑅 2 − 6𝑑𝑥𝑑𝑦 = −4 ∫ ∫𝑅 𝑑𝑥𝑑𝑦 = −4 × Area of CIrcle with radius 2 = −4 × 𝜋 ⋅ 22 = −16 𝜋 Question #6 Evaluate the improper integral. ∞ ∫0 6𝑒 (−6𝑥) 𝑑𝑥 Answer: int_0∞ 6𝑒 (−6𝑥) 𝑑𝑥 = 6𝑖𝑛𝑡0∞ 𝑒 (−6𝑥) 𝑑𝑥 = ∞ = 6𝑥[(𝑒 (−6𝑥) ]/(−6)]0 = ∞ = 6/(−6)𝑥[1/𝑒 (6𝑥) ]0 = = (−1)𝑥[1/𝑒 (6(∞)) – 1/𝑒 (6×0) ] = = (−1)𝑥[1/𝑒 ∞ – 1/𝑒 0 ] = = (−1)𝑥[1/∞– 1/1] = = (−1)𝑥[0 − 1] = = (−1)𝑥 × (−1) = 1 Question #7 Determine whether the expression is true or false: ∫ 𝑡𝑎𝑛(2𝑥) × 𝑑𝑥 = −1/2𝑙𝑛 × | cos (2𝑥)| + 𝐶 Answer: ∫ 𝑡𝑎𝑛(2𝑥) × 𝑑𝑥 = −1/2𝑙𝑛 × | cos (2𝑥)| + 𝐶 𝐼 = ∫ 𝑡𝑎𝑛(2𝑥) × 𝑑𝑥, substitute 2x = t, dx = dt 𝐼 = 1/2 ∫ 𝑡𝑎𝑛(𝑡)𝑑𝑡 𝐼 = 1/2 ln|sec(𝑥)| +𝐶, sec(𝑥) = 1/ cos(𝑥) = (cos(𝑥))−1 , 𝐼 = −1/2 𝑙𝑛| cos (𝑥)| + 𝐶. Thus, the solution is true. Question #8 Evaluate the iterated integral: 3 𝜋/3 4 ∫0 𝑟 × 𝑐𝑜𝑠 θdrdθdz a) ∫0 ∫0 𝑝𝑖/2 b) ∫0 c) d) 3 4–𝑧 ∫0 ∫0 zdrdzdθ 𝜋/2 𝜋/2 2 ∫0 ∫0 ∫0 𝜌2 𝑑 𝑝𝑖/4 𝑝𝑖/4 cos(𝜙) ∫0 ∫0 ∫0 𝜌𝑑𝜃𝑑 𝜙 𝑐𝑜𝑠 θ 𝑑 𝜌𝑑 𝜙𝑑𝜃 Answer: 3 𝜋/3 4 a) ∫0 ∫0 ∫0 𝑟 × 𝑐𝑜𝑠𝜃 drdθdz = 3 𝑝𝑖/3 cos 𝜃 [𝛾 2/2 ]0 dθdz = 3 𝜋/3 4 = ∫0 ∫0 = ∫0 ∫0 8 cos 𝜃 × 𝑑𝜃 × 𝑑𝑧 = 𝜋/3 = 𝑖𝑛𝑡03 8 × [sin 𝜃]0 𝑑𝑧 = 3 = ∫0 8 × √(3)/2𝑑𝑧 = 3 = 4√3 ∫0 𝑑𝑧 = = 4√(3)[𝑧]30 = = 12√3 𝜋/2 b) ∫0 3 4–𝑧 ∫0 ∫0 zdrdzdθ = 𝑝𝑖/2 3 = ∫0 ∫0 𝑧 × [𝛾]4–𝑧 0 𝑑𝑧 × 𝑑𝜃 = 𝜋/2 3 = ∫0 ∫0 𝑧 × [(4– 2)– 0]𝑑𝑧𝑑𝜃 3 (𝜋/2) ∫0 4𝑧 = 𝑖𝑛𝑡0 – 𝑧 2 × 𝑑𝑧𝑑𝜃 = 3 𝜋/2 = ∫0 [(4𝑧 2/2 )– 𝑧 3/3 ]0 𝑑𝜃 = 𝜋/2 = ∫0 (18 − 9) × 𝑑𝜃 = 𝜋/2 = 9∫ = 𝑑𝜃 = 0 (𝜋/2) = 9[𝜃]0 = = (9𝜋)/2 𝜋/2 c) ∫0 𝜋/2 = ∫0 𝜋/2 ∫0 2 𝜋/2 [𝜌3/3 ]0 𝑑𝜃𝑑 𝜙 = ∫0 𝑝𝑖/2 =∫ 2 ∫0 𝜌2 𝑑 𝜌𝑑𝜃𝑑 𝜙 = 𝜋/2 ∫ 8/3 𝑑𝜃𝑑 𝜙 = 0 = = 0 𝜋/2 𝜋/2 8/3 ∫0 [𝜃]0 𝑑 𝜙 = 𝜋/2 8/3 × 𝜋/2 ∫0 𝑑 𝜙 = 𝜋/2 8/3 × 𝜋/2[𝜙]0 = = = 8/3 × 𝜋/2 × 𝜋/2 = = (2𝜋 2 )/3 𝜋/4 d) ∫0 = = 𝜋/4 ∫0 𝜋/4 𝜋/4 ∫0 ∫0 𝜋/4 𝜋/4 ∫0 ∫0 cos(𝜙) ∫0 cos 𝜃 𝑑𝜌𝑑𝜙𝑑𝜃 = cos 𝜙 cos 𝜃 [𝜌]0 𝑑 𝜙𝑑𝜃 = cos 𝜃 𝑐𝑜𝑠 𝜙 × 𝑑𝜙𝑑𝜃 = 𝜋/4 𝜋/4 cos 𝜃 [sin 𝜙]0 𝑑𝜃 = = ∫0 𝜋/2 = ∫0 1/√2 cos 𝜃 × 𝑑𝜃 = (𝜋/4) = 1/√2[sin 𝜃]0 = = 1/√2 × 1/√2 = = 1/2 Question #9 1 1 Please, find the integral ∫0 ∫0 2𝑥 𝑑𝑥 𝑑𝑦 Answer: 1 1 We are given ∫0 ∫0 2𝑥 𝑑𝑥 𝑑𝑦 ∫(𝑦 = 0)1 ∫(𝑥 = 0)1 2𝑥 𝑑𝑥 𝑑𝑦 = ∫(𝑦 = 0)1 (∫(𝑥 = 0)1 2𝑥𝑑𝑥)𝑑𝑦 = 𝑥 1+1 = ∫(𝑦 = 0)1 2 [ 1 ] 𝑑𝑦 = 1+1 0 1 1 𝑥2 = ∫𝑦=0 2 [ ] 𝑑𝑦 = 2 0 1 1 = ∫𝑦=0 [2 ∗ ] 𝑑𝑦 = 2 1 = ∫0 1𝑑𝑦 = 1 − 0 = 1 1 1 The answer is: ∫0 ∫0 2𝑥 𝑑𝑥 𝑑𝑦 =1 Question #10 10 Please, find the integral ∫ (𝑥−1)(𝑥 2 𝑑𝑥 +9) Answer: 10 We are given: ∫ (𝑥−1)(𝑥 2 𝑑𝑥 +9) 10 ∫ (𝑥−1)(𝑥 2+9) 𝑑𝑥 = 1 = 10 ∗ ∫ (𝑥−1)(𝑥 2 = 10 ∗ ∫ +9) −𝑥−1 10(𝑥 2 +9) −𝑥−1 = 10 ∗ (∫ 𝑑𝑥 = +∫ 10(𝑥 2 +9) 1 10(𝑥−1) 𝑑𝑥 + ∫ 𝑑𝑥 = 1 10(𝑥−1) 2 𝑑𝑥) = = 10*(1/10(−1/ 2ln | x + 9 | −1/ 3arctan( x / 3)) + 1/10ln | x − 1|= = −1/ 2ln | x 2 + 9 | −1/ 3arctan( x / 3)) + ln | x − 1|= = −1/ 2ln | x 2 + 9 | −1/ 3arctan( x / 3)) + ln | x − 1| + The answer is: 10  ( x − 1)( x 2 + 9) dx = −1/ 2 ln | x 2 + 9 | −1/ 3arctan( x / 3)) + ln | x − 1| Question #11 1 2 Please, find the integral ∫ ( ) 𝑑𝑥 √(2𝑥−1) Answer: ( 1 1 ) 2 dx =  )dx 2x −1 (2 x − 1) By using substitution u=2x-1 we have du=2dx and so dx=du/2. So, the given integral becomes: ∫ 1/(2𝑢)𝑑𝑢 = 1/2 ∗ ∫ 1/𝑢𝑑𝑢 = 1/2 ln|𝑢| Having back the value of u we get: 2 1 ∫ (√(2𝑥−1)) 𝑑𝑥 = 1/2 ln|2𝑥 − 1| + 𝐶 Question #12 Compute, please, definite integral which have a absolute value function like these below: 3 ∫−2|𝑥|𝑑𝑥 3 ∫−2|𝑥 − 1|𝑑𝑥 2𝜋 ∫−2𝜋|sin 𝑥|𝑑𝑥 Answer: You should consider the fact that |x| is one linear function for 𝑥 ∈ (−∞; 0] and a linear function for 𝑥 ∈ [0; ∞) to break the interval of integration into subintervals that are easier. 3 0 3 ∫ |𝑥|𝑑𝑥 = ∫ −𝑥𝑑𝑥 + ∫ 𝑥𝑑𝑥 −2 −2 0 Question #13 Solve, please: 9 ∫7 (9 + 3𝑥)𝑑𝑥 = ? Answer: 9(9) + 1.5(92 ) − 9(7) − 1.5(72 ) = 66 Here you go. Question #14 𝑥 𝑓(𝑥, 𝑦) = ∫𝑦 𝑔(𝑡)𝑑𝑡 g(t) continuous for all t. Evaluate the derivatives of f. Answer: Let a be a real. Thus, 𝑥 𝑦 𝑓(𝑥, 𝑦) = ∫𝑎 𝑔(𝑡)𝑑𝑡 − ∫𝑎 𝑔(𝑡)𝑑𝑡 = 𝐺(𝑥) − 𝐺(𝑦) 𝑋 𝐺(𝑋) = ∫𝑎 𝑔(𝑡)𝑑𝑡 G'(X)=g(X) Since g is continuous at R, 𝑓𝑥 (𝑥, 𝑦) = 𝐺 ′ (𝑥) = 𝑔(𝑥) 𝑓𝑦 (𝑥, 𝑦) = 𝐺 ′ (𝑦) = −𝑔(𝑦) Question #15 Find the area enclosed by the curve 𝑥 = 𝑡 2 − 2𝑡, 𝑦 = √𝑡 and the y-axis Answer: Given data 𝑥 = 𝑡 2 − 2𝑡, 𝑦 = √𝑡, y -axis The curve will intersect y-axis when x=0 𝑡 2 − 2𝑡 = 0 t(t-2)=0 t=0,2 𝑡=0 ⇒𝑥=0 y=0 𝑡 =2 ⇒𝑥 =4−4=0 𝑦 = √2 Area A = ∫ −𝑥 𝑑𝑦 1 where 𝑦 = √𝑡 = 𝑡 2 1 1 𝑑𝑦 = 𝑦 2−1 𝑑𝑡 2 1 𝑑𝑦 = 𝐴= 𝑑𝑡 2√𝑡 2 (𝑡 2 −2𝑡)𝑑𝑡 − ∫0 2√𝑡 1 3 1 2 1 1 = − ∫0 (𝑡 2 𝑡 −2 − 2𝑡𝑡 −2 ) 𝑑𝑡 2 1 = − ∫ −02 (𝑡 2 − 2𝑡 2 ) 𝑑𝑡 2 3 1 𝑡2 = − [3 2 2 1 2 − +1 5 𝑡2 5 2 =− [ 2 1 +1 +1 2𝑡 2 0 3 2 𝑡2 3 2 0 −2 ] 5 1 2 ] 1 +1 2 4 3 = − [ (2)2 − (2)2 − 0] 2 5 3 5 3 1 6(2)2 −20(2)2 =− [ 2 =− =− 1 30 1 30 15 ] 3 [(2)2 [6(2) − 20]] [2√2(−8)] Area A = 8√2 15 Question #16 Use the Integral Test to determine whether the series is convergent or divergent. ∞ ∑ 𝑛=1 𝑛 𝑛2 + 1 Answer: ∑∞ 𝑛=1 𝑛 𝑛2 +1 𝑥 𝑓(𝑥) = 𝑥 2 +1 is decrasing since the degree of the denominator is higher than the degree of the numerator. It is also positive and continuous for x\geq0 so we can use the Integral Test. ∞ 𝑥 𝑥 ∫1 𝑥 2+1 𝑑𝑥 = lim 𝑥 2+1 𝑑𝑥 1 𝑏→∞ = lim ln(𝑥 2 + 1) |1𝑏 𝑏→∞ 2 1 = lim (ln(𝑏 2 + 1) − ln(2)) 𝑏→∞ 2 =∞ The integral diverges, so by the Integral Test, the series also diverges. Result: divergent Question #17 Convert the line integral to an ordinary integral with respect to the parameter and evaluate it. ∫𝐶 (𝑦 − 𝑧)𝑑𝑠 C is the helix 𝑟(𝑡) =< 3 cos 𝑡 , 3 sin 𝑡 , 𝑡 >, for 0 ≤ 𝑡 ≤ 2𝜋 Answer: ∫𝐶 (𝑦 − 𝑧)𝑑𝑠 𝑟(𝑡) =< 3 cos 𝑡 , 3 sin 𝑡 , 𝑡 > 0 ≤ 𝑡 ≤ 2𝜋 𝑟 ′ (𝑡) =< −3 sin 𝑡 , 3 cos 𝑡 + 1 > |𝑟 ′ (𝑡)| = √9 sin2 𝑡 + 9 cos 2 𝑡 + 1 |𝑟 ′ (𝑡)| = √9(cos 2 𝑡 + sin2 𝑡) + 1 = √10 2𝜋 = ∫0 (3 sin 𝑡 − 𝑡)√10 𝑑𝑡 2𝜋 √10 ∫0 (3 sin 𝑡 − 𝑡)𝑑𝑡 Integrate 𝑡2 2𝜋 = √10 [−3 cos 𝑡 − ] 2 0 (2𝜋)2 = √10 [−3 cos 2 𝜋 − 2 + 3 cos(0)] = √10[−3 − 2𝜋 2 + 3] = −2𝜋 2√10 Question #18 Complete the table by identifying u and du for the integral. ∫ 𝑓(𝑔(𝑥))𝑔′ (𝑥)𝑑𝑥 , 𝑢 = 𝑔(𝑥), 𝑑𝑢 = 𝑔′ (𝑥)𝑑𝑥 ∫(5𝑥 2 + 1)2 (10𝑥)𝑑𝑥 Answer: Let u=g(x) and let g' be continuous over an interval [a,b], and let f be continuous over the range of u=g(x). Then: 𝑏 𝑔(𝑏) ∫𝑎 𝑓(𝑔(𝑥))𝑔′ (𝑥)𝑑𝑥 = ∫𝑔(𝑎) 𝑓(𝑢)𝑑𝑢 We have: ∫ 𝑓(𝑔(𝑥))𝑔′ (𝑥)𝑑𝑥 = ∫(5𝑥 2 + 1)2 10𝑥𝑑𝑥 Use the formula. Let it be: 𝑢 = 𝑔(𝑥) = 5𝑥 2 + 1 Now, we differentiate both sides and we have: du=10xdx Result: 𝑢 = 5𝑥 2 + 1 du=10xdx Question #19 For a certain incompressible, two-dimensional flow field the velocity component in the y direction is given by the equation 𝑣 = 3𝑥𝑦 + 𝑥 2 𝑦 Determine the velocity component in the x direction so that the volumetric dilatation rate is zero. Answer: Given: 𝑣 = 3𝑥𝑦 + 𝑥 2 𝑦 Solution: In order to satisfy the continuity equation, the following equation must be writtem: Equation one: 𝑑𝑢 𝑑𝑥 + 𝑑𝑢 + 𝑑𝑥 𝑑𝑣 =0 𝑑𝑦 We can see that we only have given the v component of velocity, so now we can derive it: Equation two: 𝑑𝑣 𝑑𝑦 = 3𝑥 + 𝑥 2 We can now rewrite equation one: 𝑑𝑢 𝑑𝑥 𝑑𝑢 𝑑𝑥 𝑑𝑢 𝑑𝑥 =− 𝑑𝑣 𝑑𝑦 = −(3𝑥 + 𝑥 2 ) = −3𝑥 − 𝑥 2 Final step is to integrate the upper equation: ∫ 𝑑𝑢 = ∫ −3𝑥 − 𝑥 2 3 𝑥3 2 3 𝑢 = − ⋅ 𝑥2 − + 𝑓(𝑦) Question #20 Find the indefinite integral and check the result by differentiation. ∫(𝑥 + 7)𝑑𝑥 Answer: First, we use the Sum Rule: ∫(𝑥 + 7)𝑑𝑥 = ∫ 𝑥𝑑𝑥 + ∫ 7𝑑𝑥 These integrals are fairly easy to solve. The first is the integral of variable, while the second is the integral of a constant: ∫ 𝑥𝑑𝑥 = 𝑥2 2 + 𝐶1 ∫ 7𝑑𝑥 = 7𝑥 + 𝐶2 Let 𝐶 = 𝐶1 + 𝐶2 . Now we can find the indefinite integral: ∫(𝑥 + 7)𝑑𝑥 = ∫ 𝑥𝑑𝑥 + ∫ 7𝑑𝑥 = 𝑥2 2 + 𝐶1 + 7𝑥 + 𝐶2 = To check the result, let's differentiate 2 2 𝑥2 2 𝑥2 2 + 7𝑥 + 𝐶 + 7𝑥 + 𝐶 . First, we use the Sum Rule: 𝑑 𝑥 𝑑 𝑥 𝑑 𝑑 (7𝑥) + (𝐶). ( + 7𝑥 + 𝐶) = ( )+ 𝑑𝑥 2 𝑑𝑥 2 𝑑𝑥 𝑑𝑥 We can solve the first two derivatives by using the Constant Multiple Rule and the Power Rule: 𝑑 𝑑𝑥 𝑑 𝑑𝑥 𝑥2 1 𝑑 2 2 𝑑𝑥 𝑑 ( )= ⋅ (7𝑥) = 7 ⋅ 1 (𝑥 2 ) = ⋅ 2𝑥 = 𝑥 𝑑𝑥 2 (𝑥) = 7 Meanwhile, the third derivative is 𝑑 𝑑𝑥 ( 𝑥2 2 + 7𝑥 + 𝐶) = 𝑥 + 7 ⇒ 𝑥2 2 𝑑 𝑑𝑥 (𝐶) = 0. Therefore, now we have: + 7𝑥 + 𝐶 = ∫(𝑥 + 7)𝑑𝑥, which confirms our solution.