Questions and Answers #26 Applications of Integrals

Questions and Answers Sheet 26 Applications of Integrals Applications of Integrals Question #1 Evaluate the following integral. ∫ sin5 𝑥 cos 3 𝑥𝑑𝑥 Answer: Step 1 Consider the given integral. 𝐼 = ∫ sin5 𝑥 cos 3 𝑥𝑑𝑥 Put sin 𝑥 = 𝑡 in the integral. sin 𝑥 = 𝑡 cos 𝑥 ∗ 𝑑𝑥 = 𝑑𝑡 Put the value. 𝐼 = ∫ sin5 𝑥 cos 3 𝑥𝑑𝑥 = ∫ sin5 𝑥 cos 2 𝑥(cos 𝑥 𝑑𝑥) = ∫ 𝑡 5 (1 − sin2 𝑥)𝑑𝑡 = ∫ 𝑡 5 (1 − 𝑡 2 )𝑑𝑡 Step 2 Further, simplify. 𝐼 = ∫(𝑡 5 − 𝑡 7 )𝑑𝑡 =[ 𝑡6 𝑡8 − ]+𝑐 6 8 =[ sin6 𝑥 sin8 𝑥 − ]+𝑐 6 8 Question #2 Evaluate the given integral. 2 ∫ (3𝑥 3 − 2𝑥 + 5)𝑑𝑥 0 Answer: Step 1 Consider the given integral, 2 𝐼 = ∫ (3𝑥 3 − 2𝑥 + 5)𝑑𝑥 0 use the following formula of the integration, 𝑥 𝑛+1 ∫ 𝑥 𝑛 𝑑𝑥 = 𝑛+1 Step 2 using the above formula, 2 3𝑥 4 𝑥2 𝐼 = [( ) − 2 ( ) + 5(𝑥)] 4 2 0 4 (2)2 (0)2 3(2) 3(0)4 = [( ) − 2( ) + 5((2)) − {( ) − 2( ) + 5(0)}] 4 2 4 2 = [12 − 4 + 10] = 18 hence the value of the integral is 18. Question #3 Use the basic integration rules to find or evaluate the integral. 𝑒 ln 2 𝑥 ∫ 𝑑𝑥 𝑥 1 Answer: Step 1 𝑒 ln 2𝑥 Given: 𝐼 = ∫1 𝑥 𝑑𝑥 for evaluating given integral, we substitute ln(2𝑥) = 𝑡 … (1) now, differentiating equation(1) with respect to x 𝑑 𝑑𝑡 𝑑 1 (ln(2𝑥)) = (ln 𝑥) = ) (∵ 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑥 1 𝑑𝑡 (2) = 2𝑥 𝑑𝑥 𝑑𝑥 = 𝑑𝑡 𝑥 Step 2 now, replacing 𝑑𝑥 𝑥 with dt, ln(2𝑥) with t and change limits also in given integral so, 𝑒 𝐼=∫ ln(2𝑥) 𝑑𝑥 𝑥 1 1+ln 2 =∫ 𝑡𝑑𝑡 ln 2 2 1+ln 2 𝑡 =( ) 2 ln 2 1 = [(1 + ln 2)2 − (ln 2)2 ] (∵ (𝑎 + 𝑏)2 = 𝑎2 + 2𝑎𝑏 + 𝑏 2 ) 2 1 = [1 + 2 ln 2 + (ln 2)2 − (ln 2)2 ] 2 1 = [1 + 2 ln 2] 2 hence, given integral is equal to 1 2 (1 + 2 ln 2). Question #4 Find the indefinite integral ∫(2𝑡𝑖 + 𝑗 + 9𝑘)𝑑𝑡. Answer: Step 1 we have to find the value of the given indefinite integral ∫(2𝑡𝑖 + 𝑗 + 9𝑘)𝑑𝑡 . let the given integral be I. therefore, 𝐼 = ∫(2𝑡𝑖 + 𝑗 + 9𝑘)𝑑𝑡 Step 2 𝐼 = ∫(2𝑡𝑖 + 𝑗 + 9𝑘)𝑑𝑡 = (∫ 2𝑡 𝑑𝑡) 𝑖 + (∫ 𝑑𝑡) 𝑗 + (∫ 9 𝑑𝑡) 𝑘 = 𝑡 2 𝑖 + 𝑡𝑗 + 9𝑡𝑘 + 𝐶 where C is constant of integration. therefore the value of the given indefinite integral ∫(2𝑡𝑖 + 𝑗 + 9𝑘)𝑑𝑡 𝑖𝑠𝑡 2 𝑖 + 𝑡𝑗 + 9𝑡𝑘 + 𝐶 Question #5 Evaluate the integral. 𝑥+3 ∫ 𝑑𝑥 𝑥−1 Answer: Step 1 𝑥+3 Given: 𝐼 = ∫ 𝑥−1 𝑑𝑥 For evaluating given integral, first we simplify given expression then integrate it Step 2 So, 𝐼 = ∫ =∫ 𝑥+3 𝑥−1 𝑑𝑥 (𝑥 − 1 + 4) 𝑑𝑥 (𝑥 − 1) 𝑥−1 4 = ∫( + ) 𝑑𝑥 𝑥−1 𝑥−1 = ∫ (1 + 4 ) 𝑑𝑥 𝑥−1 = ∫ 𝑑𝑥 + 4 ∫ 𝑑𝑥 𝑥−1 (∵ ∫ 𝑑𝑥 = 𝑥 + 𝑐, ∫ 𝑑𝑥 = ln|𝑥 − 𝑎| + 𝑐) 𝑥−𝑎 = 𝑥 + 4 ln|𝑥 − 1| + 𝑐 Hence, given integral is equal to (𝑥 + 4 ln|𝑥 − 1| + 𝑐) . Question #6 Find the indefinite integral. (5 − 𝑒 𝑥 ) ∫ 𝑑𝑥 𝑒 2𝑥 Answer: Step 1 We have find indefinite integral: ∫ (5 − 𝑒 𝑥 ) 𝑑𝑥 𝑒 2𝑥 Separating each terms of the rational function of the integral, ∫ (5 − 𝑒 𝑥 ) 5 𝑒𝑥 𝑑𝑥 = ∫ ( − ) 𝑑𝑥 𝑒 2𝑥 𝑒 2𝑥 𝑒 2𝑥 = 5 ∫ 𝑒 −2𝑥 𝑑𝑥 − ∫ 𝑒 𝑥−2𝑥 𝑑𝑥 = 5 ∫ 𝑒 −2𝑥 𝑑𝑥 − ∫ 𝑒 −𝑥 𝑑𝑥 Step 2 We know the formula of integration, ∫ 𝑒 𝑥 𝑑𝑥 = 𝑒 𝑥 + 𝐶 ∫ 𝑒 −𝑎𝑥 𝑑𝑥 = 𝑒 −𝑎𝑥 +𝐶 −𝑎 So applying above formula, 5 ∫ 𝑒 −2𝑥 𝑑𝑥 − ∫ 𝑒 −𝑥 𝑑𝑥 = 5 ( 𝑒 −2𝑥 −2 )−( 𝑒 −𝑥 −1 )+𝐶 5 = − 𝑒 −2𝑥 + 𝑒 −𝑥 + 𝐶 2 Where, C is an arbitrary constant. 5 Hence, value of integration is − 𝑒 −2𝑥 + 𝑒 −𝑥 + 𝐶. 2 Question #7 Solve the integral. 𝑣 2 𝑑𝑣 ∫ 3 (49 − 𝑣 2 )2 Answer: Step 1 Given: ∫ 𝑣 2 𝑑𝑣 3 (49−𝑣 2 )2 To find- The value of the above integral. Concept Used- Above integral can be evaluated by substitution method. Step 2 Explanation- Rewrite the given expression as, 𝑣 2 𝑑𝑣 𝐼=∫ 3 (49 − 𝑣 2 )2 −1 substituting (49 − 𝑣 2 ) 2 = 𝑡 and differentiating both sides w.r.t. v, we get, −1 1 (−2𝑣)𝑑𝑣 = 𝑑𝑡 ∗ 3∗ 2 (49 − 𝑣 2 )2 1 (49 − 3 𝑣 2 )2 𝑑𝑣 = 𝑑𝑡 𝑣 From the above expression, we can write as, = ∫ 𝑣2 ∗ 𝑑𝑡 𝑣2 = ∫ 𝑑𝑡 =𝑡+𝐶 −1 = (49 − 𝑣 2 ) 2 + 𝐶 = 1 √(49 − 𝑣 2 ) +𝐶 Hence the value of the integral ∫ Question #8 Evaluate the following integral. sec 4 (ln 0) ∫ 𝑑0 0 Answer: Step 1 Given ∫ sec4 (ln 0) 0 𝑑0 Step 2 ∫ sec 4 (ln 0) 𝑑0 0 put 𝑢 = ln 0 𝑑𝑢 = 1 𝑑0 0 = ∫ sec 4 (𝑢) 𝑑𝑢 = ∫(1 + tan2 (𝑢)) sec 2 (𝑢) 𝑑𝑢 𝑣 = tan 𝑢 𝑑𝑣 = sec 2 𝑢𝑑𝑢 𝑣 2 𝑑𝑣 3 (49−𝑣 2 )2 is 1 √(49−𝑣 2 ) +𝐶 . = ∫ 1 + 𝑣 2 𝑑𝑣 = ∫ 1 𝑑𝑣 + ∫ 𝑣 2 𝑑𝑣 =𝑣+ 𝑣3 +𝐶 3 put 𝑣 = tan(ln 0) = tan(ln(0)) + ∫ tan3 (ln(0)) +𝐶 3 sec 4 (ln(0)) tan3 (ln(0)) 𝑑0 = tan(ln(0)) + +𝐶 0 3 Question #9 Evaluate the following integral. sec 2 𝑥 ∫ 𝑑𝑥 tan5 𝑥 Answer: Step 1 sec2 Evaluate, ∫ 5 tan 𝑥 𝑥 𝑑𝑥 Step 2 Simplification: We have sec2 𝑥 ∫ tan5 𝑥 𝑑𝑥 ...(1) Substitute tan 𝑥 = 𝑢 =⇒ sec 2 𝑥𝑑𝑥 = 𝑑𝑢 in (1), ∫ sec 2 𝑥 1 𝑑𝑥 = ∫ 5 𝑑𝑢 tan5 𝑥 𝑢 = ∫ 𝑢−5 𝑑𝑢 = 𝑢−5+1 +𝐶 −5 + 1 = 𝑢−4 +𝐶 −4 =− =− 1 +𝐶 4𝑢4 1 4 tan4 𝑥 + 𝐶 [Since 𝑢 = tan 𝑥 ] Step 3 Hence ∫ sec2 𝑥 tan5 𝑥 𝑑𝑥 = − 1 4 tan4 𝑥 +𝐶 Question #10 Evaluate the following integral. 2 2𝑥 ∫ ∫ 𝑥𝑦 𝑑𝑦 𝑑𝑥 𝑥2 0 Answer: Step 1 To determine Evaluate: 2 2𝑥 ∫ ∫ 𝑥𝑦 𝑑𝑦 𝑑𝑥 𝑥2 0 Step 2 Calculation 2 2𝑥 Consider ∫0 ∫𝑥 2 𝑥𝑦 𝑑𝑦 𝑑𝑥 2 =∫ 0 2 =∫ 0 2 =∫ 0 𝑥𝑦 2 2𝑥 | 2 𝑑𝑥 2 𝑥 𝑥 ((2𝑥)2 − (𝑥 2 )2 )𝑑𝑥 2 𝑥 (4𝑥 2 − 𝑥 4 )𝑑𝑥 2 1 2 = ∫ (4𝑥 3 − 𝑥 5 )𝑑𝑥 2 0 1 4𝑥 4 𝑥 6 2 = ( − ) |0 2 4 6 1 𝑥6 = (𝑥 4 − ) |20 2 6 1 26 = (24 − − 0 + 0) 2 6 1 64 = (16 − ) 2 6 1 96 − 64 = ( ) 2 6 1 32 = ( ) 2 6 = 16 6 = 8 3 2 2𝑥 8 Hence, ∫0 ∫𝑥 2 𝑥𝑦 𝑑𝑦 𝑑𝑥 = 3 Question #11 Evaluate the following integral. 1 𝑥 2 ∫ ∫ 2𝑒 𝑥 𝑑𝑦𝑑𝑥 0 0 Answer: Step 1 1 𝑥 2 Given, ∫0 ∫0 2𝑒 𝑥 𝑑𝑦𝑑𝑥 1 𝑥 2 1 2 On integrating, we get ∫0 ∫0 2𝑒 𝑥 𝑑𝑦𝑑𝑥 = ∫0 2𝑒 𝑥 ∗ [𝑦]0𝑥 𝑑𝑥 1 2 = ∫ 2𝑒 𝑥 ∗ [𝑥 − 0]𝑑𝑥 0 1 2 = ∫ 2𝑥𝑒 𝑥 𝑑𝑥 0 Step 2 Now put 𝑥2 = 𝑡 ⇒ 2𝑥 = 𝑑𝑡 𝑑𝑥 ⇒ 2𝑥𝑑𝑥 = 𝑑𝑡 When 𝑥 = 0, 𝑡 = 0 & when 𝑥 = 1, 𝑡 = 1 1 2 1 Therefore, ∫0 2𝑥𝑒 𝑥 𝑑𝑥 = ∫0 𝑒 𝑡 𝑑𝑡 = [𝑒 𝑡 ]10 = 𝑒1 − 𝑒 0 =𝑒−1 Question #12 2 Evaluate the integral of 𝑥𝑒 𝑥 𝑑𝑥 Answer: Step 1 To determine the value of the integral. Step 2 Given: 2 𝐼 = ∫ 𝑥𝑒 𝑥 𝑑𝑥 Step 3 Let 𝑢 = 𝑥 2 , 𝑑𝑢 = 2𝑥𝑑𝑥. 2 𝐼 = ∫ 𝑥𝑒 𝑥 𝑑𝑥 1 𝐼 = ∫ 𝑒 𝑢 𝑑𝑢 2 𝐼= 1 ∫ 𝑒 𝑢 𝑑𝑢 2 𝐼= 1 𝑢 (𝑒 ) + 𝐶 2 2 𝑒𝑥 𝐼= +𝐶 2 Step 4 Thus, the value of the integral is. 2 2 ∫ 𝑥𝑒 𝑥 𝑑𝑥 = 𝑒𝑥 +𝐶 2 Question #13 Evaluate the following integral. ∫ 𝑥 sec ℎ2 𝑥𝑑𝑥 Answer: Step 1 To evaluate the integral: ∫ 𝑥 sec ℎ2 𝑥𝑑𝑥 Solution: Integral by parts states that: ∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢 Evaluating the integral by parts. ∫ 𝑥 sec ℎ2 𝑥𝑑𝑥 = 𝑥 ∫ sec ℎ2 𝑥𝑑𝑥 − ∫ [ 𝑑 (𝑥) ∫ sec ℎ2 𝑥𝑑𝑥] 𝑑𝑥 𝑑𝑥 = 𝑥 tan ℎ (𝑥) − ∫ tan ℎ (𝑥)𝑑𝑥 = 𝑥 tan ℎ (𝑥) − ln(cos ℎ (𝑥)) + 𝐶 Step 2 Hence, required answer is 𝑥 tan ℎ (𝑥) − ln(cos ℎ (𝑥)) + 𝐶. Question #14 To find the equation: 2(2𝑥 3 − 𝑦 2 )𝑑𝑥 − 𝑥𝑦𝑑𝑦 = 0 Answer: Putting these values in (2), we get, 𝑣𝑑𝑣 4−3𝑣 2 = 𝑑𝑥 𝑥 ⇒− ⇒ 𝑑𝑝 𝑑𝑥 = 6𝑝 𝑥 𝑑𝑝 𝑑𝑥 = −6 ( ) 𝑝 𝑥 Now integrating both sides of the above equation, we get, ∫ ⇒∫ 𝑑𝑝 𝑝 =∫ −6𝑑𝑥 𝑥 𝑑𝑝 𝑑𝑥 = −6 ∫ 𝑝 𝑥 ⇒ log(𝑝) = −6 log(𝑥) + log(𝑐) [c is integrating constant] ⇒ log(𝑝) = log(𝑐𝑥 (−6) ) 𝑐 𝑥6 ⇒𝑝= 𝑐 ⇒ 4 − 3𝑣 2 = 𝑥6 [putting the value of p] 𝑐 𝑥6 ⇒ 3𝑣 2 = 4 − 1 4𝑥 6 − 𝑐 ⇒ 𝑣2 = ( ) 3 𝑥6 ⇒𝑣=± ⇒𝑣=± ⇒ 𝑦 𝑥 =± ⇒𝑦=± √4𝑥 6 − 𝑐 ) 𝑥3 √3 1 ( 1 √3𝑥 3 1 √3𝑥 3 (√4𝑥 6 − 𝑐) (√4𝑥 6 − 𝑐) [putting the value of v] 1 √3𝑥 2 (√4𝑥 6 − 𝑐) Hence the required solution is, 𝑦 = ± 1 √3𝑥 2 (√4𝑥 6 − 𝑐) Question #15 Find the general indefinite integrals integral of 𝑥 2 square root of (𝑥 3 + 1)𝑑𝑥 integral of 𝜋 sin 𝑥 cos 3 𝑥𝑑𝑥 Answer: Step 1 Given: 𝐼 = ∫ 𝑥 2√𝑥 3 +1 𝑑𝑥 for evaluating given integral, in given integral we substitute 𝑥 3 + 1 = 𝑡 … (1) now, differentiating equation (1) with respect to x 𝑑 3 𝑑𝑡 𝑑 𝑛 (𝑥 + 1) = (𝑥 ) = 𝑛𝑥 𝑛−1 ) (∵ 𝑑𝑥 𝑑𝑥 𝑑𝑥 3𝑥 2 + 0 = 3𝑥 2 = 𝑑𝑡 𝑑𝑥 𝑑𝑡 𝑑𝑥 𝑑𝑡 3 𝑥 2 𝑑𝑥 = Step 2 now, replace (𝑥 3 + 1) with t, 𝑥 2 𝑑𝑥 with so, ∫ 𝑥 2√𝑥 3 +1 𝑑𝑡 3 in given integral 1 𝑑𝑥 = ∫ √𝑡𝑑𝑡 3 1 1 𝑡 2+1 = ( )+𝑐 3 1+1 2 3 1 𝑡2 = ( )+𝑐 3 ( 3) 2 3 2𝑡 2 = +𝑐 9 3 2(𝑥 3 + 1)2 = +𝑐 9 3 Hence, given integral is equal to 2(𝑥 3 +1)2 9 + 𝑐. Question #16 Use the properties of integrals to evaluate integral 0 to 1 (4 + 3𝑥 2 )𝑑𝑥 Answer: Step 1 1 Given: 𝐼 = ∫0 (4 + 3𝑥 2 )𝑑𝑥 for evaluating given integral, first we separate terms then integrate it Step 2 1 1 1 so, ∫0 (4 + 3𝑥 2 )𝑑𝑥 = 4 ∫0 𝑑𝑥 + 3 ∫0 𝑥 2 𝑑𝑥 (∵ ∫ 𝑥 𝑛 𝑑𝑥 = 𝑥 𝑛+1 𝑛+1 + 𝑐) 1 = 4[𝑥]10 + 3 [ 𝑥3 ] 3 0 = 4(1 − 0) + (13 − 03 ) = 4(1) + 1 =4+1 =5 hence, value of given integral is equal to 5. Question #17 Write the integral as a sum of integrals without absolute values and evaluate. 𝜋 ∫ |cos 𝑥|𝑑𝑥 0 Answer: Step 1 𝜋 Given ∫0 |cos 𝑥|𝑑𝑥 To find Evaluating the above integral using proper identities. Concept used- As the given integral is in modulus, so break the limit as per the given function. Step 2 Explanation: As the above integral in modulus, so using the property of modulus, proceeding as follows, 𝜋 Let, 𝐼 = ∫0 |cos 𝑥|𝑑𝑥 Now, solving further, Splitting the function from 0 to for 𝜋 2 𝜋 2 and 𝜋 2 to 𝜋 and we know that cos 𝑥 is postive for 0 to to 𝜋 . 𝜋 2 𝜋 𝐼 = ∫ cos 𝑥 𝑑𝑥 − ∫ cos 𝑥 𝑑𝑥 𝜋 2 0 𝜋 = [sin 𝑥]02 − [sin 𝑥]𝜋𝜋 2 = [sin 𝜋 𝜋 − sin 0] − [sin 𝜋 − sin ] 2 2 =1−0−0+1 =2 𝜋 Answer- The value of the integral ∫0 |cos 𝑥|𝑑𝑥 is 2. Question #18 Evaluate each of the following integrals and simplify completely. ∫ 𝑥 ln(𝑥) 𝑑𝑥 Answer: Step 1 The detailed solution is as follows below: Step 2 ∫ 𝑥 ln(𝑥) 𝑑𝑥 𝑑 = ln(𝑥) ∫ 𝑥 𝑑𝑥 − ∫ ( (ln 𝑥) ∫ 𝑥𝑑𝑥) 𝑑𝑥 𝑑𝑥 = ln(𝑥) ∗ 𝑥2 1 𝑥2 − ∫ ∗ 𝑑𝑥 2 𝑥 2 = 𝑥2 1 ∗ ln(𝑥) − ∫ 𝑥 𝑑𝑥 2 2 = 𝑥2 1 1 ∗ ln(𝑥) − ∗ ∗ 𝑥 2 + 𝐶 2 2 2 𝜋 2 and negtaive 𝑥2 𝑥2 ∗ ln(𝑥) − + 𝐶 2 4 = where, C is the constant of integration. Question #19 Solve. 1 ∫ 𝑑𝑥 cos 4 𝑥 Answer: Step 1 To solve the integral. Step 2 To solve ∫ 1 cos4 𝑥 Using sec 𝑥 = 𝑑𝑥 1 cos 𝑥 , we have ∫ 1 cos4 𝑥 𝑑𝑥 = ∫ sec 4 𝑥𝑑𝑥 = ∫ sec 2 𝑥(tan2 𝑥 + 1)𝑑𝑥 [using, sec 2 𝑥 = tan2 𝑥 + 1] we substitute 𝑢 = tan 𝑥 ⇒ 𝑑𝑢 = sec 2 𝑥 𝑑𝑥 ⇒ 𝑑𝑥 = Thus, ∫ 1 𝑑𝑢 sec 2 𝑥 1 cos4 𝑥 𝑑𝑥 = ∫(𝑢 2 + 1)𝑑𝑢 = 𝑢3 3 +𝑢 Undo substitution 𝑢 = tan 𝑥 ; we have ∫ 1 cos4 𝑥 𝑑𝑥 = tan3 𝑥 3 + tan 𝑥 + 𝐶 where C is constant of integration. Question #20 Evaluate the double integral by first identifying it as the volume of a solid. 3 8 ∫−3 ∫3 2𝑑𝐴, 𝑅 = {𝑥, 𝑦| − 3 ≤ 𝑥 ≤ 3,3 ≤ 𝑥 ≤ 8} Answer: 3 8 ∫ ∫ 2𝑑𝑥𝑑𝑦 −3 3 3 2 ∫ [𝑦]83 𝑑𝑥 −3 3 3 2 ∫ 5𝑑𝑥 ⇒ 10 ∫ 𝑑𝑥 −3 ⇒ 60 units −3