Questions and Answers #24 Applications of Integrals

Questions and Answers Sheet 24 Applications of Integrals Applications of Integrals Question #1 Use properties of the indefinite integral to express the following integral in terms of the simplest integrals: ∫(−4𝑥 2 + 7𝑥)𝑑𝑥 Answer: Step 1 Let us evaluate the integral Given: ∫(−4𝑥 2 + 7𝑥)𝑑𝑥 Properties of the Indefinite Integral: ∫ 𝑐 𝑓(𝑥)𝑑𝑥 = 𝑐 ∫ 𝑓(𝑥)𝑑𝑥 ∫[𝑓(𝑥) + 𝑔(𝑥)]𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 + ∫ 𝑔(𝑥)𝑑𝑥 Step 2 Evaluate the integral = ∫(−4𝑥 2 + 7𝑥)𝑑𝑥 = ∫ − 4𝑥 2 𝑑𝑥 + ∫ 7𝑥 𝑑𝑥 = −4 ∫ 𝑥 2 𝑑𝑥 + 7 ∫ 𝑥 𝑑𝑥 = −4 × 𝑥3 𝑥2 +7× 3 2 ∫(−4𝑥 2 + 7𝑥)𝑑𝑥 = [∫ 𝑥 𝑛 𝑑𝑥 = 𝑥 𝑛+1 ] 𝑛+1 −4𝑥 3 7𝑥 2 + 3 2 Step 3 Answer ∫(−4𝑥 2 + 7𝑥)𝑑𝑥 = −4𝑥 3 7𝑥 2 + 3 2 Question #2 𝑥2 Find the indefinite integral of ∫ 𝑑𝑥 𝑥−9 Answer: Step 1 To evaluate ∫ Step 2 𝑥2 𝑥−9 𝑑𝑥 Now substitute x - 9 = u, x = u+9 And dx = du Step 3 Now substitute the value of x and dx in above integral, we get ∫ (𝑢+9)2 𝑢 Step 4 Now simplify the above integral, we get ∫ = ∫ (𝑢 + 18 + (𝑢+9)2 𝑢 𝑑𝑢 = ∫ 81 ) 𝑑𝑢 𝑢 𝑢2 + 18𝑢 + 81 ln|𝑢| + 𝑐 2 = Step 5 Now replace u = x-9, we get ∫ (𝑥 − 9)2 𝑥2 𝑑𝑥 = + 18(𝑥 − 9) + 81 ln|𝑥 − 9| + 𝑐 𝑥−9 2 Step 6 Answer: ∫ (𝑥 − 9)2 𝑥2 𝑑𝑥 = + 18(𝑥 − 9) + 81 ln|𝑥 − 9| + 𝑐 𝑥−9 2 Question #3 Find the indefinite integral. 2 𝑑𝑥 ∫ csc 𝑥 sec 𝑥 Answer: The given indefinite integral can be written as: Let us suppose 𝐼 = ∫ We know that 1 csc 𝑥 2 csc 𝑥 sec 𝑥 𝑑𝑥 = sin 𝑥 and 1 sec 𝑥 = cos 𝑥 So, the above integral can be written as : 𝐼 = ∫ 2 sin 𝑥 cos 𝑥 𝑑𝑥 𝐼 = ∫ sin 2 𝑥𝑑𝑥 𝐼= − cos 2𝑥 2 + 𝐶 where C is a constant of integration. Question #4 Find the indefinite integral. √𝑥 ∫ 3 𝑑𝑥 √𝑥 Answer: 𝑢2 +18𝑢+81 𝑢 𝑑𝑢 𝑑𝑢 Let us consider the integral as: √𝑥 𝐼 = ∫ 3 𝑑𝑥 √𝑥 1 𝐼=∫ 𝑥2 1 𝑑𝑥 𝑥3 1 1 𝐼 = ∫ 𝑥 2−3 𝑑𝑥 1 𝐼 = ∫ 𝑥 6 𝑑𝑥 1 𝑥 6+1 𝐼= +𝐶 1 +1 6 6 7 𝐼 = 𝑥6 + 𝐶 7 √𝑥 6 √𝑥 7 7 Hence, the value of indefinite integral 𝐼 = ∫ 3 𝑑𝑥 will be 𝐼 = 𝑥 6 + 𝐶 . Question #5 Evaluate the indefinate integral. ∫ 𝑥 2 sin(𝑥 3 ) 𝑑𝑥 Answer: Step 1 Evaluate indefinite integral. ∫ 𝑥 2 sin(𝑥 3 ) 𝑑𝑥 Consider, 𝑢 = 𝑥 3 𝑑𝑢 = 3𝑥 2 𝑑𝑥 𝑑𝑢 = 𝑥 2 𝑑𝑥 3 Step 2 Substitute the values in integral and evaluate. ∫ sin(𝑢) 𝑑𝑢 3 ⇒ 1 ∫ sin 𝑢 𝑑𝑢 3 ⇒ 1 (− cos 𝑢) 3 Substitute u value, ⇒ 1 (− cos(𝑥)3 ) + 𝐶 3 1 ⇒ − cos(𝑥)3 + 𝐶 3 Question #6 Evaluate the integral. 2𝑥 2 + 7𝑥 − 3 ∫ 𝑑𝑥 𝑥−2 Answer: Given The integral is ∫ 2𝑥 2 +7𝑥−3 𝑥−2 𝑑𝑥 Solution ∫ 2𝑥 2 + 7𝑥 − 3 2(𝑥 − 2)2 + 15(𝑥 − 2) + 19 𝑑𝑥 = ∫ 𝑑𝑥 𝑥−2 𝑥−2 let u=x-2; du=dx ∫ 2𝑥 2 + 7𝑥 − 3 2𝑢2 + 15𝑢 + 19 𝑑𝑥 = ∫ 𝑑𝑢 𝑥−2 𝑢 =∫ 2𝑢2 15𝑢 19 𝑑𝑢 + ∫ 𝑑𝑢 + ∫ 𝑑𝑢 𝑢 𝑢 𝑢 1 = 2 ∫ 𝑢 𝑑𝑢 + 15 ∫ 1 𝑑𝑢 + 19 ∫ 𝑑𝑢 𝑢 =2∗ 𝑢2 + 15𝑢 + 19 ln 𝑢 + 𝐶 2 put u=x-2 ∫ 2𝑥 2 + 7𝑥 − 3 𝑑𝑥 = (𝑥 − 2)2 + 15(𝑥 − 2) + 19 ln(𝑥 − 2) + 𝐶 𝑥−2 = 𝑥 2 − 4𝑥 + 4 + 15𝑥 − 30 + 19 ln(𝑥 − 2) + 𝐶 = 𝑥 2 + 11𝑥 − 26 + 19 ln(𝑥 − 2) + 𝐶 ∫ 2𝑥 2 + 7𝑥 − 3 𝑑𝑥 = 𝑥 2 + 11𝑥 + 19 ln(𝑥 − 2) + 𝐶 ′ (𝐶 ′ = 𝐶 − 26) 𝑥−2 Question #7 Use properties of the indefinite integral to express the following integral in terms of the simplest integrals: ∫(−4𝑥 2 + 7𝑥)𝑑𝑥 Answer: Step 1 Given ∫(−4𝑥 2 + 7𝑥)𝑑𝑥 = ∫ − 4𝑥 2 𝑑𝑥 + ∫ 7𝑥 𝑑𝑥 = −4 ∫ 𝑥 2 𝑑𝑥 + 7 ∫ 𝑥 𝑑𝑥 Step 2 ∵ ∫ 𝑥 𝑛 𝑑𝑥 = 𝑥 𝑛+1 +𝑐 𝑛+1 = −4 [ =− 𝑥 2+1 𝑥 1+1 ]+ 7[ ]+𝑐 2+1 1+1 4𝑥 3 7𝑥 2 + +𝑐 3 2 4 7 = − 𝑥3 + 𝑥2 + 𝑐 3 2 Question #8 Evaluate the indefinite integral. 𝑑𝑥 ∫ (𝑥 + 5)3 Answer: Step 1 Evaluating: 𝑑𝑥 (𝑥 + 5)3 ∫ Solution: Using Substitution rule: Let (x+5)=u dx=du Step 2 Substituting in the indefinite Integral: 𝑑𝑥 𝑑𝑢 = ∫ 3 = ∫ 𝑢−3 𝑑𝑢 (𝑥 + 5)3 𝑢 ∫ Applying the power rule of Integration: ∫ 𝑢−3 𝑑𝑢 = 𝑢−3+1 𝑢 −2 1 = =− 2 −3 + 1 −2 2𝑢 ∫ 𝑢−3 𝑑𝑢 = − 1 +𝐶 2𝑢2 Step 3 Substituting the value of u -back: We get: 𝑑𝑥 1 =− +𝐶 3 (𝑥 + 5) 2(𝑥 + 5)2 ∫ Hence, this is the final answer. Question #9 Evaluate the following iterated integrals. 2 1 ∫ ∫ 4𝑥𝑦 𝑑𝑥 𝑑𝑦 0 0 Answer: Step 1 Given integral 2 1 ∫ ∫ 4𝑥𝑦 𝑑𝑥 𝑑𝑦 0 0 2 = ∫ 4[𝑥𝑑𝑥]𝑦𝑑𝑦 0 2 =∫ 4 0 𝑥2 𝑦𝑑𝑦 2 2 = ∫ 2[𝑥 2 ]10 𝑦𝑑𝑦 0 2 = ∫ 2[12 − 02 ]𝑦𝑑𝑦 0 2 = ∫ 2𝑦𝑑𝑦 0 Step 2 2 𝑦2 = 2[ ] 2 0 = [𝑦 2 ]20 = 22 − 02 =4 2 1 Therefore, the result of ∫0 ∫0 4𝑥𝑦 𝑑 𝑥𝑑𝑦 is 4. Question #10 Find the indefinite integral of ∫(5 − 5𝑥 + 𝑥 5 )𝑑𝑥 Answer: Step 1 We have to find the indefinite integral of the function: ∫(5 − 5𝑥 + 𝑥 5 )𝑑𝑥 We know the formula of integration, ∫ 𝑥 𝑛 𝑑𝑥 = 𝑥 𝑛+1 +𝐶 𝑛+1 ∫ 𝑑𝑥 = 𝑥 + 𝐶 ∫ 𝑎 𝑥 𝑑𝑥 = 𝑎𝑥 +𝐶 ln(𝑎) Where, C is an arbitrary constant. Step 2 Applying above formula for the given integral, we get ∫(5 − 5𝑥 + 𝑥 5 )𝑑𝑥 = 5 ∫ 𝑑𝑥 − ∫ 5𝑥 𝑑𝑥 + ∫ 𝑥 5 𝑑𝑥 = 5𝑥 − 5𝑥 𝑥 5+1 + +𝐶 ln(5) 5 + 1 = 5𝑥 − 5𝑥 𝑥6 + +𝐶 ln(5) 6 Hence, value of the indefinite integral is 5𝑥 − 5𝑥 ln(5) + 𝑥6 6 + 𝐶. Question #11 Evaluate the integral. 1 ∫ (4𝑥 − 9𝑥 2 )𝑑𝑥 0 Answer: Step 1 Solve the integral ∫(4𝑥 − 9𝑥 2 )𝑑𝑥 by using the general power rule of the integration that is ∫ 𝑥 𝑛 = , where n is not equal to 1. ∫(4𝑥 − 9𝑥 2 )𝑑𝑥 = 𝑥 𝑛+1 𝑛+1 𝑑𝑥 4𝑥 1+1 9𝑥 2+1 − 1+1 2+1 4𝑥 2 9𝑥 3 − 2 3 = = 2𝑥 2 − 3𝑥 3 Step 2 Now, use the fundamental theorem of calculus and apply the upper and lower limit 1 and 0 in the integral 1 solution of the integral ∫0 (4𝑥 − 9𝑥 2 )𝑑𝑥 to calculate the final value. 1 1 ∫ (4𝑥 − 9𝑥 2 )𝑑𝑥 = 2𝑥 2 − 3𝑥 3|0 0 = 2(1)2 − 3(1)3 − [2(0)2 − 3(0)3 ] =2−3−0 = −1 Question #12 Evaluate the following definite integral. 1 3 𝑥 +1 ∫ 𝑑𝑥 0 𝑥+1 Answer: Step 1 Definite integrals can be computed using the fundamental theorem of calculus. It states that if 𝐹(𝑥) = 𝑏 ∫ 𝑓(𝑥)𝑑𝑥 then we have ∫𝑎 𝑓(𝑥)𝑑𝑥 = 𝐹(𝑏) − 𝐹(𝑎) . We will use the following identity for this problem: 𝑥 3 + 1 = (𝑥 + 1)(𝑥 2 − 𝑥 + 1) . Use this to simplify the integrand. Use the standard integral ∫ 𝑥 𝑛 𝑑𝑥 = Step 2 1 𝑥 3 +1 The integral to be computed is ∫0 1 ∫ 0 𝑥+1 𝑥 𝑛+1 𝑛+1 𝑓𝑜𝑟 𝑛 ≠ −1 . 𝑑𝑥 . Use information from step 1 to simplify and integrate. 1 (𝑥 𝑥3 + 1 + 1)(𝑥 2 − 𝑥 + 1) 𝑑𝑥 = ∫ 𝑑𝑥 𝑥+1 𝑥+1 0 1 = ∫ (𝑥 2 − 𝑥 + 1)𝑑𝑥 0 1 𝑥3 𝑥2 − + 𝑥) 3 2 0 1 1 = − +1 3 2 2−3+6 = 6 =( = 5 6 5 Hence, the definite integral is equal to . 6 Question #13 Find the indefinite integral. ∫(sec 𝑦)(tan 𝑦 − sec 𝑦)𝑑𝑦 Answer: Step 1 To evaluate the given integral ∫ sec 𝑦 (tan 𝑦 − sec 𝑦)𝑑𝑦 Step 2 To begin with, we remove the bracket using distributive law, ∫ sec 𝑦 (tan 𝑦 − sec 𝑦)𝑑𝑦 = ∫ sec 𝑦 tan 𝑦 𝑑𝑦 − ∫ sec 2 𝑦𝑑𝑦 Now, solve the integral individually, ∫ sec 𝑦 tan 𝑦 𝑑𝑦 = sec 𝑦 + 𝑐1 The second integral ∫ sec 2 𝑦𝑑𝑦 = tan 𝑦 + 𝑐2 Step 3 Substitute the values back to the original integral, ∫ sec 𝑦 tan 𝑦 𝑑𝑦 − ∫ sec 2 𝑦𝑑𝑦 = sec 𝑦 − tan 𝑦 + 𝐶 Step 4 Answer: ∫ sec 𝑦 (tan 𝑦 − sec 𝑦)𝑑𝑦 = sec 𝑦 − tan 𝑦 + 𝐶 Question #14 Find the indefinite integral ∫ √ 𝑡 1−𝑡 4 𝑑𝑡 Answer: Step 1 Given integral, ∫ √ 𝑡 1−𝑡 4 𝑑𝑡 Step 2 Now evaluating the given integral ∫ 𝑡 √1 − 𝑡 4 =∫ 𝑑𝑡 𝑡 √1 − (𝑡 2 )2 𝑑𝑡 Let 𝑡 2 = 𝑥 ⇒ 2𝑡 𝑑𝑡 = 𝑑𝑥 1 ⇒ 𝑡𝑑𝑡 = 𝑑𝑥 2 Now integral ∫ 1 1 𝑑𝑥 √1 − (𝑥)2 2 1 1 = ∫ 𝑑𝑥 2 √1 − (𝑥)2 1 = sin−1 𝑥 + 𝐶 2 1 = sin−1 (𝑡 2 ) + 𝐶 2 Step 3 Final answer: ∫ 𝑡 √1 − 𝑡 4 𝑑𝑡 = 1 −1 2 sin (𝑡 ) + 𝐶 2 Question #15 Evaluate the integrals in ∫ ln(𝑥)𝑑𝑥 𝑥√ln2 (𝑥)+1 Answer: Step 1 Consider the following integral: ∫ ln(𝑥) 𝑥√ln2 (𝑥) + 1 𝑑𝑥 = 𝐼 Substitute ln2 (𝑥) + 1 = 𝑢 ⇒ 𝐼=∫ 1 2√ 𝑢 2 ln(𝑥) 𝑥 𝑑𝑥 = 𝑑𝑢 in the above integral: 𝑑𝑢 1 1 = ∫(𝑢)−2 𝑑𝑢 2 1 1 𝑢−2+1 = × +𝐶 2 −1 + 1 2 1 1 𝑢2 = × +𝐶 1 2 2 = √𝑢 + 𝐶 Step 2 Substitute 𝑢 = ln2 (𝑥) + 1 in the above equation: 𝐼 = √ln2 (𝑥) + 1 + 𝐶 Hence, the solution is √ln2 (𝑥) + 1 + 𝐶 . Question #16 Evaluate the integral. 4 ∫ 3𝑥 2 + 4𝑥 + 3 + 𝑥𝑑𝑥 𝑥 Answer: Step 1 Assume the given integral be I, Rewrite integral and simplify it 𝐼 = ∫ 3𝑥 2 + 4𝑥 + 4 + 𝑥𝑑𝑥 𝑥3 = ∫ 3𝑥 2 𝑑𝑥 + ∫ 4𝑥 𝑑𝑥 + ∫ 4 𝑑𝑥 + ∫ 𝑥𝑑𝑥 𝑥3 = 3 ∫ 𝑥 2 𝑑𝑥 + 4 ∫ 𝑥𝑑𝑥 + 4 ∫ 𝑥 −3 𝑑𝑥 + ∫ 𝑥𝑑𝑥 Step 2 Integrate it by using: ∫ 𝑥 𝑛 𝑑𝑥 = 𝐼=3 𝑥 𝑛+1 +𝐶 𝑛+1 𝑥3 𝑥2 𝑥 −3+1 𝑥2 +4 +4 + +𝐶 3 2 −3 + 1 2 𝐼 = 𝑥 3 + 2𝑥 2 − 2 𝑥2 + +𝐶 𝑥2 2 Question #17 Use the form of the definition of the integral given in the theorem to evaluate the integral. 5 ∫ (12 − 6𝑥)𝑑𝑥 2 Answer: Step 1 According to the given information, it is required to evaluate the integral. 5 ∫ (12 − 6𝑥)𝑑𝑥 2 Step 2 Now solve the integral. 5 ∫ (12 − 6𝑥)𝑑𝑥 = 12𝑥 − 2 6𝑥 2 2 = [12𝑥 − 3𝑥 2 ]52 = (12(5) − 3(5)2 ) − (12(2) − 3(2)2 ) = (60 − 75) − (24 − 12) = −15 − 12 5 ∫ (12 − 6𝑥)𝑑𝑥 = −27 2 Question #18 Use integration by parts to find the integral: ∫ 7𝑥 ln 𝑥 𝑑𝑥 Answer: Step 1 Consider the integral ∫ 7𝑥 ln 𝑥 𝑑𝑥 Step 2 Integration by parts ∫ 𝑢𝑣 = 𝑢 ∫ 𝑣 𝑑𝑥 − ∫ 𝑢 ′ (∫ 𝑣 𝑑𝑥) 𝑑𝑥 ∫ 7𝑥 ln 𝑥 𝑑𝑥 Here 𝑢 = ln 𝑥 v=7x Now ∫ 7𝑥 ln 𝑥 𝑑𝑥 = ln 𝑥 ∫ 7𝑥𝑑𝑥 − ∫(ln 𝑥)′ (∫ 7𝑥𝑑𝑥)𝑑𝑥 = ln 𝑥 [ 7𝑥 2 1 7𝑥 2 1 ] − ∫( )[ ] 𝑑𝑥 (∵ (ln 𝑥)′ = ) 2 𝑥 2 𝑥 = 7𝑥 2 7 ln 𝑥 − ∫ 𝑥𝑑𝑥 2 2 = 7𝑥 2 7 𝑥2 ln 𝑥 − [ ] 2 2 2 = 7𝑥 2 7𝑥 2 ln 𝑥 − 2 4 ∫ 7𝑥 ln 𝑥 𝑑𝑥 = 7𝑥 2 7𝑥 2 ln 𝑥 − 2 4 Step 3 Answer: ∫ 7𝑥 ln 𝑥 𝑑𝑥 = 7𝑥 2 2 ln 𝑥 − 7𝑥 2 4 Question #19 Find the indefinite integral ∫(4𝑥 5 − 6√𝑥 + 2 cos 𝑥 − 4)𝑑𝑥 Answer: Step 1 Consider the given indefinite integral: ∫(4𝑥 5 − 6√𝑥 + 2 cos 𝑥 − 4)𝑑𝑥 Here, the objective is to evaluate the given indefinite integral. Step 2 Re-write the given indefinite integral as: 𝐼 = ∫(4𝑥 5 − 6√𝑥 + 2 cos 𝑥 − 4)𝑑𝑥 Use the sum rule of integration ∫(𝑓 (𝑥) ± 𝑔(𝑥))𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 ± ∫ 𝑔(𝑥)𝑑𝑥 1 𝐼 = 4 ∫ 𝑥 5 𝑑𝑥 − 6 ∫ 𝑥 2 𝑑𝑥 + 2 ∫ cos 𝑥 𝑑𝑥 − 4 ∫ 𝑑𝑥 Use ∫ 𝑥 𝑛 𝑑𝑥 = 𝑥 𝑛+1 𝑛+1 + 𝐶 and ∫ cos 𝑥 𝑑𝑥 = sin 𝑥 + 𝐶 3 4𝑥 6 6𝑥 2 𝐼= − + 2 sin 𝑥 − 4𝑥 + 𝐶 3 6 2 𝐼= 3 2𝑥 6 − 4𝑥 2 + 2 sin 𝑥 − 4𝑥 + 𝐶 3 Question #20 Solve the following indefinite integrals. cos 𝑥 ∫ 2 𝑑𝑥 sin 𝑥 Answer: To find: cos 𝑥 The indefinite integral of ∫ 2 𝑑𝑥 sin 𝑥 Calculation: Substitute 𝑡 = sin 𝑥 and 𝑑𝑡 = cos 𝑥 𝑑𝑥 in ∫ ∫ cos 𝑥 sin2 𝑥 cos 𝑥 𝑑𝑡 𝑑𝑥 = ∫ 2 2 sin 𝑥 𝑡 =( 𝑡 −2+1 )+𝐶 −2 + 1 1 =− +𝐶 𝑡 Substitute 𝑡 = sin 𝑥 in above integral: ∫ cos 𝑥 1 𝑑𝑥 = − +𝐶 2 sin 𝑥 sin 𝑥 cos 𝑥 1 Thus, the integral of ∫ 2 𝑑𝑥 is − +𝐶 sin 𝑥 sin 𝑥 𝑑𝑥