Questions and Answers #33 Applications of Integrals

Questions and Answers Sheet Applications of Integrals Applications of Integrals Question #1 Use the Change of Variables Formula to evaluate the definite integral. 4 ∫0 𝑥√𝑥 2 + 9 𝑑𝑥 Answer: Step 1 Using Change of variables we have to evaluate the definite integral. 4 ∫ 𝑥 √𝑥 2 + 9 𝑑𝑥 0 Step 2 4 ∫0 𝑥√𝑥 2 + 9 𝑑𝑥 Let us substitute 𝑥 2 + 9 by t So, when 𝑥 = 0, 𝑡 = 9 and when 𝑥 = 4, 𝑡 = 25 Differentiating 𝑥 2 + 9 = 𝑡 ⇒ 2𝑥𝑑𝑥 = 𝑑𝑡 ⇒ 𝑥𝑑𝑥 = 4 So we can write the integral ∫0 𝑥√𝑥 2 + 9 𝑑𝑥 as 3 25 \𝑓𝑟𝑎𝑐12 ∫9 Now 27 3 = √𝑡𝑑𝑡 = 𝑡 2 25 | 3 9 98 3 4 So, ∫0 𝑥√𝑥 2 + 9 𝑑𝑥 = 98 3 Question #2 Evaluate the indefinite integral. ∫ 𝑎 + 𝑏𝑥 2 √3𝑎𝑥 + 𝑏𝑥 3 𝑑𝑥 Answer: Step 1 Given integral: ∫ 𝑎 + 𝑏𝑥 2 √3𝑎𝑥 + 𝑏𝑥 3 𝑑𝑥 Step 2 Now, Apply u-substitution: = 125 3 𝑑𝑡 2 25 ∫ 2 9 1 √𝑡𝑑𝑡 𝑢 = 3𝑎𝑥 + 𝑏𝑥 3 Differentiate both sides: 𝑑𝑢 = 3𝑎𝑑𝑥 + 3𝑏𝑥 2 𝑑𝑥 𝑑𝑢 = (𝑎 + 𝑏𝑥 2 )𝑑𝑥 3 Therefore, ∫ 𝑎 + 𝑏𝑥 2 √3𝑎𝑥 + 1 𝑏𝑥 3 𝑑𝑥 = ∫ 𝑑𝑢 3√ 𝑢 −1 = ∫ 𝑢 2 𝑑𝑢 3 Apply the Power Rule: ∫ 𝑥 𝑎 𝑑𝑥 = 𝑥 𝑎+1 𝑎+1 1 1 − +1 𝑢 2 3 − +1 = ∗ 1 2 2 = √𝑢 3 Substitute back 𝑢 = 3𝑎𝑥 + 𝑏𝑥 3 : 2 = √3𝑎𝑥 + 𝑏𝑥 3 3 Add a constant to the solution: 2 = √3𝑎𝑥 + 𝑏𝑥 3 + 𝐶 3 Question #3 Evaluate the indefinite integrals. 3 ∫ 3𝑥 2 𝑒 𝑥 𝑑𝑥 Answer: Step 1 We have to find the given integral. Step 2 Solution: we have to find the given integral. 3 Let 𝐼 = ∫ 3𝑥 2 𝑒 𝑥 𝑑𝑥 let 𝑡 = 𝑥 3 ⇒ 𝑑𝑡 = 3𝑥 2 𝑑𝑥 ⇒ 𝐼 = ∫ 𝑒 𝑡 𝑑𝑡 ⇒ 𝐼 = 𝑒𝑡 + 𝐶 3 ⇒ 𝐼 = 𝑒 𝑥 + 𝐶 where 'c' is constant. , 𝑎 ≠ −1 Question #4 Determine the following integrals. 𝑦 (𝑦 2 +1)5/4 𝑑𝑦 Answer: Step 1 Our Aim is to evaluate the integral given below:𝑦 ∫ (𝑦2+1)5/4 𝑑𝑦 ...(i) Step 2 Consider the integral given by equation-(i), we have:∫ 𝑦 5 (𝑦 2 +1)4 𝑑𝑦-(i) 𝑦 For the integrand 5 , we will substitute 𝑡 = 𝑦 2 + 1 and 𝑑𝑡 = 2𝑦𝑑𝑦 (𝑦 2 +1)4 1 = ∫ 2 1 5 𝑑𝑡 𝑡4 Since, the integral of = −2 4 √𝑡 1 5 𝑡4 −4 𝑖𝑠 4 √𝑡 + constant Answer = −2 4 √𝑦 2 +1 + constant Question #5 In the given equation as follows , use a table of integrals with forms involving ln u to find the indefinite integral:∫(cos 𝑥)𝑒 sin 𝑥 𝑑𝑥 Answer: Step 1 The given integral can be evaluated by the method of substitution. We know that, ∫ 𝑒 𝑥 𝑑𝑥 = 𝑒 𝑥 + 𝑐 𝑑 sin 𝑥 = cos 𝑥 𝑑𝑥 Step 2 The given integral is, 𝐼 = ∫ cos 𝑥 𝑒 sin 𝑥 𝑑𝑥 . Put, sin 𝑥 = 𝑢 𝑑𝑢 = cos 𝑥 𝑑𝑥 Then, 𝐼 = ∫ 𝑒 𝑢 𝑑𝑢 = 𝑒𝑢 + 𝑐 = 𝑒 sin 𝑥 + 𝑐 Hence, the required expression is, 𝑒 sin 𝑥 + 𝑐 . Question #6 Evaluate the following integrals. 1 ∫ 𝑥3𝑥 2 +1 𝑑𝑥 0 Answer: Step 1 1 Given: ∫0 𝑥3𝑥 2 +1 𝑑𝑥 for evaluating given integral we substitute 𝑥 2 + 1 = 𝑡...(1) so, at, 𝑥 = 0, 𝑡 = 1 𝑥 = 1, 𝑡 = 2 now differentiating equation (1) with respect to t 2𝑥 2𝑥 𝑑𝑥 𝑑𝑡 +0=1 𝑑𝑥 =1 𝑑𝑡 𝑥𝑑𝑥 = 𝑑𝑡 2 Step 2 now in given integral replace 𝑥 2 + 1 with t, xdx with so, 1 ∫ 𝑥 ∗ 3𝑥 2 +1 𝑑𝑥 = 0 (∵ ∫ 𝑎 𝑥 𝑑𝑥 = 𝑎𝑥 + 𝑐) ln 𝑎 2 1 3𝑡 = ( ) 2 ln 3 1 = 1 2 𝑡 ∫ 3 𝑑𝑡 2 1 1 (32 − 31 ) 2 ln 3 𝑑𝑡 2 and change limits also = 1 (9 − 3) 2 ln 3 = 6 2 ln 3 = 3 ln 3 hence, given integral is equal to 3 ln 3 . Question #7 Evaluate the following indefinite integral. ∫ cos 4 (𝑥) sin(𝑥) 𝑑𝑥 Answer: Step 1 See attached file for a step by step explanation to find the given integral. Step 2 Solution ∫ cos 4 (𝑥) sin(𝑥) 𝑑𝑥 ⇒ ∫(cos(𝑥))4 sin(𝑥) 𝑑𝑥 Let cos(𝑥) = 𝑡 differentiate − sin(𝑥) 𝑑𝑥 = 𝑑𝑡 sin(𝑥) 𝑑𝑥 = −𝑑𝑡 ⇒ ∫ 𝑡 4 (−𝑑𝑡) ⇒ − ∫ 𝑡 4 𝑑𝑡 ⇒− 𝑡5 +𝑐 5 put 𝑡 = cos(𝑥) ⇒− (cos 𝑥)5 cos 5 𝑥 + 𝑐𝑜𝑟 − +𝑐 5 5 Question #8 Evaluate the integrals. 5 ∫1 𝑑𝑥 Answer: Step 1 5 We have to evaluate ∫1 𝑑𝑥 Solution: We know that ∫ 𝑑𝑥 = 𝑥 Evaluating the given integral, 5 ∫ 𝑑𝑥 = [𝑥]15 1 = [5 − 1] =4 Step 2 Hence, the required answer is 4. Question #9 Evaluate the indefinite integral. ∫ 5𝑡 sin(5𝑡 ) 𝑑𝑡 Answer: Step 1 The given integral is ∫ 5𝑡 sin 5𝑡 𝑑𝑡 Step 2 Obtain the value of the given integral as shown below: ∫ 5𝑡 sin 5𝑡 𝑑𝑡 = ∫ = = 1 ln 5 sin 𝑢 ln 5 𝑑𝑢 (Substituting 𝑢 = 5𝑡 ) (− cos 𝑢) + 𝐶 − cos 5𝑡 +𝐶 ln 5 Question #10 Evaluate the integrals: ∫ 𝑥2 𝑥 𝑑𝑥 − 25 Answer: Step 1 Given integral, ∫ 𝑥2 𝑥 𝑑𝑥 − 25 Step 2 Let 𝐼 = ∫ 𝑥 𝑥 2 −25 𝑑𝑥 putting 𝑢 = 𝑥 2 − 25 𝑑𝑢 = 2𝑥 𝑑𝑥 ⇒ 1 𝑑𝑢 = 𝑥𝑑𝑥 2 1 Now 𝐼 = ∫ 2 𝑑𝑢 𝑢 1 = ln|𝑢| + 𝐶 2 1 = ln|𝑥 2 − 25| + 𝐶 2 Question #11 Evaluate the following integral: 1−𝑥 ∫√ 1−𝑥 2 𝑑𝑥 Answer: Step 1: Given that Evaluate the following integral: ∫ 1−𝑥 √1 − 𝑥 2 𝑑𝑥 Step 2: Solve We have, ∫ 1−𝑥 √1 − 𝑥2 𝑑𝑥 = ∫ = sin−1 (𝑥) − ∫ 1 √1 − 𝑥 × √𝑡 𝑥2 𝑑𝑥 − ∫ 𝑥 √1 − 𝑥 2 𝑑𝑥 𝑑𝑡 −2𝑥 1 = sin −1 1 𝑡 −2+1 𝑥+ [ ]+𝐶 2 −1 +1 2 = sin−1 𝑥 + √𝑡 + 𝐶 = sin−1 𝑥 + √1 − 𝑥 2 + 𝐶 Question #12 Use a table of integrals to find the indefinite integral ∫ 𝑒 3𝑥 /(1 + 𝑒 𝑥 )3 𝑑𝑥 Answer: Step 1: Given that Use a table of integrals to find the indefinite integral ∫ 𝑒 3𝑥 /(1 + 𝑒 𝑥 )3 𝑑𝑥 Step 2: Solve We have, 𝑒 3𝑥 𝐼 = ∫ (1+𝑒 𝑥)3 𝑑𝑥 Substitute, 1 + 𝑒𝑥 = 𝑡 𝑒 𝑥 ∗ 𝑑𝑥 = 𝑑𝑡 Plugging all the values into the integral we obtain, 𝐼=∫ 𝑒 3𝑥 𝑑𝑡 × 𝑡3 𝑒𝑥 =∫ 𝑒 2𝑥 𝑑𝑡 𝑡3 =∫ (𝑡 − 1)2 𝑑𝑡 𝑡3 =∫ 𝑡 2 − 2𝑡 + 1 𝑑𝑡 𝑡3 =∫ 𝑑𝑡 𝑑𝑡 1 − 2 ∫ 2 + ∫ 3 𝑑𝑡 𝑡 𝑡 𝑡 = ln|𝑡| − 2 ( 𝑡 −2+1 𝑡 −3+1 )+( )+𝐶 −2 + 1 −3 + 1 2 1 = ln|𝑡| + − 2 + 𝐶 𝑡 2𝑡 = ln|1 + 𝑒 𝑥 | + 2 1+𝑒 𝑥 − 1 2(1+𝑒 𝑥 )2 +𝐶 Question #13 Evaluate the given integral. 𝑥−1 ∫ 𝑥+2 𝑑𝑥 Answer: Step 1 𝑥−1 Use Substitution method to determine the value of integral ∫ 𝑑𝑥 . 𝑥+2 Step 2 substitute 𝑥 − 1 = 𝑥 + 2 − 3 in the denominator and simplify the rational fraction as below, ∫ (𝑥 + 2) − 3 (𝑥 + 2) 3 𝑑𝑥 = ∫ − 𝑑𝑥 𝑥+2 𝑥+2 𝑥+2 = ∫1− 3 𝑑𝑥 𝑥+2 = ∫ 1𝑑𝑥 − ∫ 3 𝑥+2 𝑑𝑥 Step 3 Substitute 𝑥 + 2 = 𝑡 and 𝑑𝑥 = 𝑑𝑡 ∫ 1𝑑𝑥 − ∫ 3 3 𝑑𝑥 = 𝑥 − ∫ 𝑑𝑡 𝑥+2 𝑡 = 𝑥 − 3 ln 𝑡 + 𝑐 = 𝑥 − 3 ln(𝑥 + 2) + 𝑐 hence, the value of integral is 𝑥 − 3 ln(𝑥 + 2) + 𝑐. Question #14 1 2 Find the iterated integral ∫0 ∫𝑦 (𝑥 + 2𝑦)𝑑𝑥𝑑𝑦 Answer: Step 1 We have given a double integral; 1 2 ∫0 ∫𝑦 (𝑥 + 2𝑦)𝑑𝑥𝑑𝑦 Step 2 We shall integrate this according to the order of dx and dy. 1 2 ∫ ∫ (𝑥 + 2𝑦)𝑑𝑥𝑑𝑦 0 𝑦 1 =∫ [ 0 1 =∫ [ 0 1 =∫ [ 0 2 𝑥2 + 2𝑦𝑥] 𝑑𝑦 2 𝑦 22 − 𝑦 2 + 2𝑦(2 − 𝑦)] 𝑑𝑦 2 4 − 𝑦2 + 4𝑦 − 2𝑦 2 ] 𝑑𝑦 2 1 = ∫ [2 − 0 5𝑦 2 + 4𝑦] 𝑑𝑦 2 1 = [2𝑦 − 5𝑦 3 𝑦2 +4 ] 6 2 0 5 = [2 − + 2] 6 =[ = 24 − 5 ] 6 19 6 Question #15 Find the integral, full solution ∫(2𝑥 + 3)2 𝑑𝑥 Answer: Step 1 Given integral, ∫(2𝑥 + 3)2 𝑑𝑥 Step 2 ∫(2𝑥 + 3)2 𝑑𝑥 = ∫(4𝑥 2 + 12𝑥 + 9)𝑑𝑥 = ∫ 4𝑥 2 𝑑𝑥 + ∫ 12𝑥𝑑𝑥 + ∫ 9𝑑𝑥 = 4 ∫ 𝑥 2 𝑑𝑥 + 12 ∫ 𝑥𝑑𝑥 + 9 ∫ 𝑑𝑥 =4 = 𝑥3 3 + 12 𝑥2 2 + 9𝑥 + 𝐶 Since ∫ 𝑥 𝑛 𝑑𝑥 = 4𝑥 3 + 6𝑥 2 + 9𝑥 + 𝐶 3 Question #16 Find integral. ∫ sin10 𝑥 cos 𝑥 𝑑𝑥 Answer: Step 1 To find the integral 𝐼 = ∫ sin10 𝑥 cos 𝑥 𝑑𝑥 Step 2 𝐼 = ∫ sin10 𝑥 cos 𝑥 𝑑𝑥 By using substitution method, 𝑡 = sin 0 𝑑𝑡 = cos 0 𝑑0 𝐼 = ∫ sin10𝑥 cos 𝑥 𝑑𝑥 𝐼 = ∫ 𝑡 10 𝑑0 𝑈= 𝐼= 𝑡 11 +𝑐 11 sin11 0 11 +𝑐 Question #17 Evaluate the integral. 1 ∫ 𝑥(𝑥−𝑎) 𝑑𝑥 Answer: Step 1 Given data: 1 The integral expression is: ∫ 𝑑𝑥 𝑥(𝑥−𝑎) 𝑥 𝑛+1 𝑛+1 + 𝐶, 𝑛 ≠ 1 The given integral can be expressed as, 1 ∫ 𝑥(𝑥−𝑎) 𝑑𝑥 = ∫ = 1 𝑥 𝑎 𝑎𝑥( −1) 𝑑𝑥 1 1 ∫ 𝑥 𝑑𝑥 𝑎 𝑥 ( − 1) 𝑎 Assume 𝑥/𝑎 = 𝑢, and differentiate it with respect to x. 1 𝑑𝑥 = 𝑑𝑢 𝑎 𝑑𝑥 = 𝑎𝑑𝑢 Step 2 Substitute the above-calculated values in the expression. 1 1 1 ∫ 𝑥(𝑥−𝑎) 𝑑𝑥 = 𝑎 ∫ (𝑎𝑢)(𝑢−1) (𝑎𝑑𝑢) = 1 𝑑𝑢 ∫ 𝑎 𝑢(𝑢 − 1) = 1 −1 1 ∫( + ) 𝑑𝑢 𝑎 𝑢 𝑢−1 1 = (− ln|𝑢| + ln|𝑢 − 1|) + 𝐶 𝑎 1 𝑢−1 𝑎 𝑢 = ln ∣ ∣ +C Substitute 𝑥/𝑎 for u in the expression. 1 1  x( x − a)dx = a ln | ( x / a) − 1 1 x−a | +C = ln | | +C ( x / a) a x Thus, the integration of the given expression is (ln|(𝑥 − 𝑎)𝑥|)𝑎 + 𝐶 . Question #18 Evaluate the integral. 𝑑𝑥 ∫ cos 𝑥−1 Answer: Step 1 𝑑𝑥 to evaluate: ∫ cos 𝑥−1 Solution: We have formula: 1 − cos 2 𝑥 = 2 sin2 𝑥 Therefore, 𝑥 1 − cos 𝑥 = 2 sin2 ( ) 2 And also we have ∫ csc 2 𝑥𝑑𝑥 = − cot 𝑥 Step 2 Evaluate the above integral: ∫ 𝑑𝑥 1 =∫ 𝑑𝑥 cos 𝑥 − 1 −(1 − cos 𝑥) =∫ 1 𝑥 𝑑𝑥 − sin2 ( ) 2 𝑥 = ∫ − csc 2 ( ) 𝑑𝑥 2 𝑥 cot ( ) 2 + 𝑐 (𝑈𝑠𝑖𝑛𝑔 ∫ − csc 2 𝑥 = cot 𝑥) = 1 ( ) 2 𝑥 = 2 cot ( ) + 𝑐 2 𝑑𝑥 𝑥 Hence, required answer is ∫ = 2 cot ( ) + 𝑐 . cos 𝑥−1 2 Question #19 Find indefinite integral. ∫(𝑥 1/2 + 3𝑥 −2/3 )𝑑𝑥 Answer: Step 1 Let the given integral be y – 𝑦 = ∫(𝑥 1/2 + 3𝑥 −2/3 )𝑑𝑥 Identity to be used – ∫ 𝑥 𝑛 𝑑𝑥 = 𝑥 (𝑛+1) 𝑛+1 +𝐶 Step 2 Using identity in the given function – 1 𝑦=[ 𝑦=[ ( +1) 𝑥 2 1 ( +1) 2 ]+3[ 2 (−3+1) 2 (− +1) 3 𝑥 ]+𝐶 𝑥 3/2 𝑥 1/3 ] +3[ ]+𝐶 3/2 1/3 𝑦= 2 3/2 𝑥 + 3(3)𝑥 1/3 + 𝐶 3 𝑦= 2 3/2 𝑥 + 9𝑥 1/3 + 𝐶 3 Hence, the integration of given integral is – 2 ∫(𝑥 1/2 + 3𝑥 −2/3 )𝑑𝑥 = 3 𝑥 3/2 + 9𝑥 1/3 + 𝐶 Question #20 Use partial fractions to evaluate the definite integral. 1 ∫ 0 𝑥2 − 𝑥 𝑑𝑥 𝑥2 + 𝑥 + 1 Answer: Step 1: Consider the provided definite integral, 1 𝑥 2 −𝑥 ∫0 𝑥 2 +𝑥+1 𝑑𝑥 Now, use the partial fraction to reduce the integral and after that integrate it. 1 ∫ 0 1 𝑥2 − 𝑥 −2𝑥 − 1 𝑑𝑥 = ∫ ( 2 + 1) 𝑑𝑥 2 𝑥 +𝑥+1 0 𝑥 +𝑥+1 1 =∫ 0 1 −2𝑥 − 1 𝑑𝑥 + ∫ 1𝑑𝑥 𝑥2 + 𝑥 + 1 0 = [− ln(𝑥 2 + 𝑥 + 1)]10 + [𝑥]10 = [− ln(1 + 1 + 1) − ln 0] + 1 = − ln 3 + 1 Step 2: 1 𝑥 2 −𝑥 ∫0 𝑥 2 +𝑥+1 𝑑𝑥 ≈ − ln 3 + 1 Hence, the result is verified. The result of given definite integral is true.