Questions and Answers #1 Integrals

Questions and Answers Sheet 1 Integrals Question #1 A small plane is flying a banner in the shape of a rectangle. The area of the banner is 144 square feet. 1 The width of the banner is the length of the banner. What are the dimensions of the banner? 4 Answer: Step 1 First, we will choose two numbers that have a product of 144 because it's the area of the banner, where 1 the width of the banner is the length of the banner. 4 Guess: 6 feet width and 24 feet length. Then, we will check those numbers. Is the width of the banner 1 1 24 4 1 × 24 = × 4 = 1×24 4×1 = 24 4 1 4 the length of the banner? =6 So, our guess is correct. The dimensions of the banner are 6 feet and 24 feet. Question #2 Evaluate the indefinite integral as an infinite series. ∫ 𝑒 𝑥 − 1/𝑥𝑑𝑥 Answer: We know that the Maclaurin series for 𝑒 𝑥 is: 𝑒 𝑥 = ∑∞ 𝑛=0 𝑥𝑛 𝑛! = 1 + ∑∞ 𝑛=1 𝑥𝑛 𝑛! ⇒ 𝑒 𝑥 − 1 = ∑∞ 𝑛=1 𝑥𝑛 𝑛! Furthermore, x can be considered as a constant in regards to n.That's why we heve (when 𝑥 ≠ 0 ): 𝑒 𝑥 −1 𝑥 1 1 𝑥𝑛 𝑥 𝑥 𝑛! = ⋅ (𝑒 𝑧 − 1) = ⋅ ∑∞ 𝑛=1 = ∑∞ 𝑛=1 𝑥 𝑛−1 𝑛! Integrating is now easy: ∫ = 𝑒 𝑥 −1 𝑑𝑥 = ∫ ∑∞ 𝑛=1 𝑥 1 𝑥𝑛 ∑∞ 𝑛=1 𝑛! 𝑛 +𝐶 = 𝑥 𝑛+1 𝑥 𝑛−1 1 𝑛−1 = ∑∞ 𝑛=1 𝑛! ∫ 𝑥 𝑛! 𝑛 𝑥 ∑∞ 𝑛=1 𝑛⋅𝑛! + 𝐶 𝑑𝑥 = ∑∞ 𝑛=0 (𝑛+1)⋅(𝑛+1)! + 𝐶 ∫ 𝑒 𝑥 −1 𝑥 𝑥 𝑛+1 𝑑𝑥 ∑∞ 𝑛=0 (𝑛+1)⋅(𝑛+1)! + 𝐶 Question #3 The cdf a certain college library checkout duration X is as follows 0  2  4x F ( x) =   49 1 x0 0  x  3.5 3.5  x Use this to compute th efollowing. a) 𝑃(𝑥 ≤ 1) b) 𝑃(0.5 ≤ 𝑥 ≤ 1) c) 𝑃(𝑋 > 0.5) d) The median checkout duration 𝜇̃ [solve, 𝑆 = 𝐹(𝜇̃)] e) 𝐹 ′ (𝑥) to obtain the density function f(x) f) 𝐸(𝑋) g) 𝑉(𝑋) 𝜎𝑥 h) If the borrower is charged an amount ℎ(𝑋) = 𝑥 2 when checkout duration is X, compute the expected charge 𝐸[ℎ(𝑋)] Answer: Step 1 Given: x0 0  2  4x F ( x) =   49 1 0  x  3.5 3.5  x 𝐹(𝑥) = 𝑃(𝑥 ≤ 𝑥) a) 𝑃(𝑥 ≤ 1) = 𝐹(1) = 4(1)2 49 4 = 49 Step 2 b) 𝑃(0.5 ≤ 𝑥 ≤ 1) = 𝑃(𝑥 ≤ 1) − 𝑃(𝑥 ≤ 0.5) = 𝐹(1) − 𝐹(0.5) = 4 49 − 4×(0.5)2 49 = 4 49 1 3 4 49 (1 − ) = Step 3 c) 𝑃(𝑥 > 0.5) = 1 − 𝑃(𝑥 ≤ 0.5) = 1 − 𝐹(0.5) = 1− 4𝑥(0.5)2 49 = 48 49 Question #4 Evaluate the integral by interpreting it in terms of areas. integral: 0 ∫−3(1 + √9 − 𝑥 2 )𝑑𝑥 Answer: Step 1 First we break the integral into two parts:  0 −3 0 0 −3 −3 (1 + 9 − x 2 )dx =  1dx +  9 − x 2 dx 0 The first term ∫−3 1𝑑𝑥 represents area of rectangle with base 𝑏 = 3 and height ℎ = 1 0 ∫−3 1𝑑𝑥 = 𝑏 × ℎ = 3 0 ∫−3 √9 − 𝑥 2 𝑑𝑥 represent area of quarter circle with radius 𝑟 = 3 0 1 0 (1 + 9 − x )dx =  1dx +  ∫−3 √9 − 𝑥 2 𝑑𝑥 = 4 𝜋 × 32 =  −3 0 2 4 0 0 −3 −3 ∫ (1 + √9 − 𝑥 2 ) 𝑑𝑥 = 3 + −3 9𝜋 9𝜋 4 9 − x 2 dx Question #5 What's the integral of ∫ sec 𝑥 tan 𝑥 𝑑𝑥 ? Answer: ∫ sec(𝑥) tan(𝑥) 𝑑𝑥 = sec(𝑥) + 𝐶 Here is the explanation: ∫ sec(𝑥) tan(𝑥) 𝑑𝑥 =∫ 1 sin(𝑥) × 𝑑𝑥 cos(𝑥) cos(𝑥) =∫ sin(𝑥) 1 𝑑𝑥 = +𝐶 (cos(𝑥))2 cos(𝑥) = sec(𝑥) + 𝐶 Question #6 Evaluate the definite integral. 2 1 𝑒 𝑥4 ∫ 5 𝑑𝑥 1 𝑥 Answer: Step 1 To evaluate the definite integral. Step2 Given information: 1 2 𝑒 𝑥4 ∫1 ( 𝑥 5 ) 𝑑𝑥 Step 3 Calculation: Integrate the function with respect to x. 1 2 𝑒 𝑥4 ∫1 ( 𝑥 5 ) 𝑑𝑥 Let 1 𝑥4 =𝑢 so, differentiate with respect to x. 𝑑 1 𝑑 (𝑢) ( )= 𝑑𝑥 𝑥 4 𝑑𝑥 −4 𝑑𝑢 = 𝑥 5 𝑑𝑥 1 1 ( 5 ) 𝑑𝑥 = 𝑑𝑢 𝑥 −4 [put in the given integral] Step 4 So, the integral become, 1 2 𝑒 𝑥4 ∫𝑥=1 ( 𝑥 5 ) 𝑑𝑥 2 𝑒𝑢 = ∫𝑥=1 ( ) 𝑑𝑢 −4 1 1 1 [𝑝𝑢𝑡 ( 5 ) 𝑑𝑥 = ( −4 ) 𝑑𝑢, 𝑎𝑛𝑑 𝑢 = 4 ] 𝑥 𝑥 𝑥 1 𝑢 2 = − [𝑒 ]𝑥=1 4 [∵ ∫ 𝑒 𝑥 𝑑𝑥 = 𝑒 𝑥 ] 1 1 2 = − [𝑒 𝑥 4 ] 4 𝑥=1 [𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑢 = 1 ] 𝑥4 1 1 1 = − [𝑒 2(4) − 𝑒 1(4) ] 4 1 2 𝑒 𝑥4 1 1 ∫ ( 5 ) 𝑑𝑥 = − [𝑒 16 − 𝑒] 𝑥 4 𝑥=1 Step 5 Thus, the integration is shown below. 1 2 𝑒 𝑥4 1 1 ∫ ( 5 ) 𝑑𝑥 = − [𝑒 16 − 𝑒] 𝑥 4 𝑥=1 Question #7 Integrate the following using Partial Fractions. 𝑥𝑑𝑥 ∫ 𝑥 2−3𝑥−4 Answer: 𝑥𝑑𝑥 Given: ∫ 2 𝑥 −3𝑥−4 To Find- The value of the above integration using partial fractions. Explanation- Rewrite the given expression as, 𝐼=∫ 𝑥𝑑𝑥 𝑥 2 −3𝑥−4 Now, simplifying the above expression as follows,we get, 𝑥𝑑𝑥 =∫ 4 𝑥 − 4𝑥 + 𝑥 − 4 𝑥 = ∫ (𝑥+1)(𝑥−4) 𝑑𝑥 The above expression can be simplified using partial fractions, we get, 𝑥 𝐴 𝐵 = + … (1) (𝑥 + 1)(𝑥 − 4) (𝑥 + 1) (𝑥 − 4) 𝑥 𝐴(𝑥 − 4) + 𝐵(𝑥 + 1) = (𝑥 + 1)(𝑥 − 4) (𝑥 + 1)(𝑥 − 4) 𝑥 = 𝐴(𝑥 − 4) + 𝐵(𝑥 + 1) To Find the value of A and B, we have to substitute the value of x as 4 in the above expression, we get, 4 =0+5⋅𝐵 Further, we can write as follows, 4 𝐵= 5 Similarily, substituting the value of x as -1 in the above expression, we get, 1 𝐴= 5 So, the equation (1) can be written as follows, 𝑥 1 1 4 1 ∫ = ∫ 𝑑𝑥 + ∫ 𝑑𝑥 (𝑥 + 1)(𝑥 − 4) 5 (𝑥 + 1) 5 (𝑥 − 4) 1 4 = ln|𝑥 + 1| + ln|𝑥 − 4| + 𝐶 5 5 𝑥𝑑𝑥 1 4 Answer- Hence, the solution of the integral ∫ 2 is ln|𝑥 + 1| + ln|𝑥 − 4| + 𝐶 𝑥 −3𝑥−4 5 5 Question #8 I want to reduce the two following fractions: 2𝑥 + 2𝑦 𝑥+𝑦 3𝑎𝑏 2 12𝑎𝑏 I fully understand the concept of reduce fractions of this type: 15 20 but i do not know what steps to take for reducing fractions like the two above. Anyone that can explain the steps needed, or point me to a website explaining it? Answer: For the first fraction: 2𝑥 − 2𝑦 2(𝑥 + 𝑦) = 𝑥+𝑦 𝑥+𝑦 = 2 assuming (𝑥 + 𝑦) ≠ 0 and dividing both numerator and denominator by (𝑥 + 𝑦) For the second fraction: 𝑎𝑏 2 3𝑎𝑏 × 𝑏 = 12𝑎𝑏 3𝑎𝑏 × 4 𝑏 = assuming 3𝑎𝑏 ≠ 0 and dividing both numerator and denominator by (3𝑎𝑏) 4 Question #9 2 2 2 4 4 4 + + ⋯+ = + + ⋯+ ⏟ 3 3 3 ⏟ 5 5 5 A copies of 2 3 B copies of 4 5 20 ≤ 𝐴 + 𝐵 ≤ 24 𝐴 + 𝐵 =? I have A copies of 2 3 on the left-hand side of the equality, and B copies of 4 5 on the right. In total, I've written down between 20 and 24 fractions. Exactly how many fractions are there? Answer: Well you have 2 3 4 5 5 6 𝐴= 𝐵⇒𝐵= 𝐴 so 11 𝐴+𝐵 = 𝐴 6 Since A+B is an integer and is between 20 and 24, we must have that 11 𝐴 + 𝐵 = (12) = 22 6 Question #10 𝑉𝑠 + 𝑅𝑖 𝑉0 𝑅𝑓 =0 𝑉𝑠 , 𝑅𝑖, 𝑉0 and Rf are just variables for voltage and resistance. After some rearanging, it becomes: 𝑉0 𝑉𝑠 =− 𝑅𝑓 𝑅𝑖 How did the fractions ''swap''? Answer: 𝑉𝑠 𝑉𝑜 + =0 𝑅𝑖 𝑅𝑓 𝑉𝑜 𝑉𝑠 =− 𝑅𝑓 𝑅𝑖 𝑉𝑜 × 𝑅𝑖 = −𝑉𝑠 𝑅𝑓 𝑉𝑜 × 𝑅𝑖 = −𝑉𝑠 × 𝑅𝑓 𝑉𝑜 × 𝑅𝑖 = −𝑅𝑓 𝑉𝑠 𝑉𝑜 𝑅𝑓 =− 𝑉𝑠 𝑅𝑖 Question #11 How to add two compound fractions with fractions in numerator like this one: 1 𝑥 2 + 2 3𝑥 𝑥 or fractions with fractions in denominator like this one: 𝑥 2 𝑥 + 1 𝑥 𝑥 Answer: One easy way to figure this out is that dividing by x is the same as multiplying by 1/𝑥 (but all bets are off when 𝑥 = 0, as division by 0 is undefined). So 𝑎 𝑏 = 1𝑎 = 𝑎 𝑐 𝑐 𝑏 𝑏𝑐 𝑎 1 𝑐 𝑎𝑐 =𝑎 =𝑎 = 𝑏 𝑏 𝑏 𝑏 𝑐 𝑐 Question #12 What is the integral of 1 1+𝑥 2 ? Answer: 1 ∫ 1+𝑥 2 𝑑𝑥 = tan−1 𝑥 + 𝐶 Explanation: 𝑑𝑢 ∫ 1+𝑢2 = tan−1 𝑢 + 𝐶 ⇒ Where u is a function of x Proof: 𝑑𝑢 ∫ 1+𝑢2 Integration by Trigonometric Substitution 𝑢 = tan 𝜃 ⇒ 𝑑𝑢 = sec 2 𝜃𝑑(𝜃) 𝑑𝑢 sec 2 𝜃𝑑(𝜃) ∫ = ∫ 1 + 𝑢2 1 + tan2 𝜃 2 sec 𝜃 = 1 + tan2 𝜃 (1 + tan2 𝜃)𝑑(𝜃) sec 2 𝜃𝑑(𝜃) = ∫ 1 + tan2 𝜃 1 + tan2 𝜃 = ∫ 𝑑(𝜃) = Reverse the Substitution 𝑢 = tan 𝜃 ⇒ 𝜃 = tan ^{−1}𝑢 ∫ ∴∫ 𝑑𝑢 1+𝑢2 = tan−1 𝑢 + 𝐶 Simply by Substituting in this relation 𝑑𝑥 ∫ = tan−1 𝑥 + 𝐶 1 + 𝑥2 arctan 𝑥 + 𝐶 Question #13 How do you integrate 𝑦 = 𝑥𝑒 −𝑥 ? Answer: Explanation: IBP: ∫ 𝑢𝑣 ′ = 𝑢𝑣 − ∫ 𝑢′ 𝑣 here 𝑢 = 𝑥, 𝑢′ = 1 𝑣 ′ = 𝑒 −𝑥 , 𝑣 = −𝑒 −𝑥 so −𝑥𝑒 −𝑥 − ∫ 𝑑𝑥(−𝑒 −𝑥 ) = −𝑥𝑒 −𝑥 − (𝑒 −𝑥 ) + 𝐶 = −𝑒 −𝑥 (𝑥 + 1) + 𝐶 Question #14 What is the integral of (cos 𝑥)2 ? Answer: Explanation: We will use the cosine double-angle identity in order to rewrite 𝑐𝑜𝑠 ^2𝑥 . (Note that cos 2 𝑥 = (cos 𝑥)2 , they are different ways of writing the same thing.) cos(2𝑥) = 2 cos 2 𝑥 − 1 2 This can be solved for cos x : cos(2𝑥) + 1 cos 2 𝑥 = 2 Thus, cos(2𝑥)+1 𝑑𝑥 ∫ cos 2 𝑥𝑑𝑥 = ∫ 2 Split up the integral: 1 1 = ∫ cos(2𝑥) 𝑑𝑥 + ∫ 𝑑𝑥 2 2 The second integral is the ''perfect integral:'' ∫ 𝑑𝑥 = 𝑥 + 𝐶 1 1 2 2 = ∈ cos(2𝑥) 𝑑𝑥 + 𝑥 The constant of integration will be added upon evaluating the remaining integral. For the cosine integral, use substitution. Let 𝑢 = 2𝑥 , implying that 𝑑𝑢 = 2𝑑𝑥 Multiply the integrand 2 and the exterioe of the integral by 1 1 2 . 1 = ∫ 2 cos(2𝑥) 𝑑𝑥 + 𝑥 4 2 Substiute in u and du: 1 1 = ∫ cos(𝑢) 𝑑𝑢 + 𝑥 4 2 Note that ∫ cos(𝑢) 𝑑𝑢 − sin(𝑢) + 𝐶 1 1 4 2 = sin(𝑢) + 𝑥 + 𝐶 Since 𝑢 = 2𝑥 : 1 1 4 2 = sin(2𝑥) + 𝑥 + 𝐶 Note that this can be many different ways, since sin(2𝑥) = 2 sin 𝑥 cos 𝑥 . Question #15 How do you integrate ∫ sin 2 𝑥𝑑𝑥 Answer: Using integration by substitution together with the known integral 1 sin(𝑥) 𝑑𝑥 = ∫ sin(2𝑥) 2𝑑𝑥 2 1 = ∫ sin(𝑢) 𝑑𝑢 2 1 = (− cos(𝑢)) + 𝐶 2 1 = − cos(2𝑥) + 𝐶 2 Question #16 4 ∫ (√𝑥 + 9 2 1 √𝑥 ) 𝑑𝑥 Answer: Compute the definite integral: For the integrand ( x + 1 2 ) , do long division: x 9 1 = x + + 2.dx. x 4 Integrate the sum term by term and factor out constants: 9 9 9 1 = x.dx. +  .dx. + 21.dx. x 4 4 4 Apply the fundamental theorem of calculus. x2 : 2 The antiderivative of x is = x2 right bracketing bar 2 9 4 1 +  egral49 .dx. + 2  egral491.dx. x Evaluate the antiderivative at the limits and subtract. x2 right bracketing bar 2 9 9 4 = 92 42 65 − = : 2 2 2 9 65 1 = +  .dx. + 21.dx. 2 4x 4 Apply the fundamental theorem of calculus. The antiderivative of = 1 is log( x ) : x 65 + log( x) right bracketing bar 2 Evaluate the antiderivative at the limits and subtract. log( x) right bracketing bar 9 4 9 = log(9) − log(4) = log( ) : 4 9 = 65 9 + log( ) + 21.dx. 2 4 4 Apply the fundamental theorem of calculus. The antiderivative of 1 is x: = 65 9 + log( ) + 2 x right bracketing bar 2 4 9 4 Evaluate the antiderivative at the limits and subtract. 2x right bracketing bar 9 4 = 29 − 24 = 10 : Answer: = 85 9 + log( ) 2 4 Question #17 Change from rectangular to cylindrical coordinates. (Let 𝑟 ≥ 0 and 0 ≤ 𝜃 ≤ 2𝜋.) (A) (−2,2,2) (B) (−9,9√3,6) (C) Use cylindrical coordinates. Evaluate ∫ ∫𝐸 ∫ 𝑥 𝑑𝑉 where E is enclosed by the planes 𝑧 = 0 and 𝑧 = 𝑥 + 𝑦 + 10 and by the cylinders 𝑥 2 + 𝑦 2 = 16 and 𝑥 2 + 𝑦 2 = 36. D) Use cylindrical coordinates. Find the volume of the solid that is enclosed by the cone 𝑧 = √𝑥 2 + 𝑦 2 and the sphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 8. Answer: Question #18 Find the exact length of the curve. y = x − x 2 + sin −1 x Answer: Question #19 Evaluate the integral ∫ arctan(𝑥) 𝑑𝑥 A) − arctan(𝑥) + ln(𝑥 2 + 1) + 𝐶 B) 1 𝑥 2 +1 +𝐶 1 C) 𝑥 arctan(𝑥) − ln(𝑥 2 + 1) + 𝐶 2 D) ln | arccos (𝑥)| + 𝐶 E) 𝑥 arctan(𝑥) + 𝐶 Answer: Question #20 To evaluate ∫ tan3 (𝜋𝑥) 𝑑𝑥 Answer: