Calculus Answer Note #5

Question 1 dx d (e x 2 ⋅ sin(x))+∫ 0 x 1+t 2 cos(t) dt=ln(x 2 +1) Answer To answer this question, it is important to review the fundamentals of differential calculus and integration. Differential calculus is a branch of mathematics that deals with the study of changes in a function with respect to its variables. Integration, on the other hand, is the inverse process of differentiation and is used to determine the area under a curve. In this equation, we are dealing with a combination of differentiation and integration. Let's break down the equation into its separate parts and analyze them. Part 1: dx d (e x 2 ⋅ sin(x)) The symbol dx represents the derivative of a variable x, while the symbol d represents the differential operator. The expression (ex ^ 2 ⋅ sin(x)) is a composite function with an inner function (x^2) and an outer function (ex ^ 2). To differentiate a composite function, we use the chain rule, which states that the derivative is equal to the derivative of the outer function multiplied by the derivative of the inner function. In this case, using the chain rule, we get: (dx d) (ex ^ 2 ⋅ sin(x)) = (dex ^ 2) (sin(x)) + (ex ^ 2) (cos(x)). To calculate the derivative of the outer function, we use the power rule, which states that the derivative of a function raised to a power is equal to the power multiplied by the function to the power decreased by one. Therefore, (dex ^ 2) = 2ex. Plugging this into our equation, we get: (dx d) (ex ^ 2 ⋅ sin(x)) = (2ex) (sin(x)) + (ex ^ 2) (cos(x)). Part 2: ∫ 0 x 1+t 2 cos(t) dt The symbol ∫ represents the integral operator, while the lower and upper limits of integration, 0 and x respectively, represent the starting and ending points of integration. The expression 1 + t^2 cos(t) is the integrand, which is the function being integrated. To solve this integral, we use the substitution method, where we substitute u = 1 + t^2, du = 2t dt, and t = (u - 1)^(1/2) into the integral. Using these substitutions, we get: ∫ 0 x 1+t 2 cos(t) dt = 1/2 ∫ 1 x^2 cos(u - 1) du = 1/2 ∫ 1 x^2 [cos(u) cos(1) + sin(u) sin(1)] du = 1/2 [cos(1) ∫ 1 x^2 cos(u)du + sin(1) ∫ 1 x^2 sin(u)du] = 1/2 [cos(1) sin(u) | 1 x^2 - sin(1) cos(u) | 1 x^2 ] = 1/2 [cos(1) sin(x^2 -1) - sin(1) cos(x^2 - 1)]. Part 3: ln(x^2 +1) The function ln(x^2 + 1) is the natural logarithm of the expression (x^2 + 1). Now, we can combine all the parts of the equation to get the final answer: (dx d) (ex ^ 2 ⋅ sin(x)) + ∫ 0 x 1+t 2 cos(t) dt = ln(x^2 + 1) = (2ex) (sin(x)) + (ex ^ 2) (cos(x)) + [cos(1) sin(x^2 -1) - sin(1) cos(x^2 - 1)] = ln(x^2 + 1) Question 2 f(x) = (x3 + 17x + 15)/(x2 - 8x +16) find discontinuity also check if it is removable Answer To find potential discontinuities of a rational function, we need to determine the points at which the denominator of the function becomes zero. These points are called the zeros or roots of the denominator. In this case, the denominator of the given function is a quadratic equation (x2 - 8x + 16) which can be factored as (x - 4)(x - 4). Therefore, the only discontinuity of this function occurs at x = 4, where the denominator becomes zero. This type of discontinuity is called a "hole" or "removable" discontinuity, because it can be removed by simplifying the function. To check if the discontinuity at x = 4 is removable, we can evaluate the limit of the function as x approaches 4 from both sides. Limit as x -> 4^-: (x3 + 17x + 15)/(x2 - 8x +16) = (4^3 + 17(4) + 15)/(4^2 - 8(4) +16) = 21/2 Limit as x -> 4^+: (x3 + 17x + 15)/(x2 - 8x +16) = (4^3 + 17(4) + 15)/(4^2 - 8(4) +16) = 21/2 Since both limits are equal, the discontinuity at x = 4 is removable. This means that we can redefine the function at this point by simplifying it to its reduced form, which is f(x) = 21/2. Some useful sources for understanding discontinuities of rational functions and calculating limits include: 1. "Discontinuities of Rational Functions" by Study.com: https://study.com/academy/lesson/discontinuities-of-rational-functions.html This source provides a clear explanation of the concept of discontinuities and how to identify them in rational functions. 2. "Removable Discontinuities in Rational Functions" by Purplemath: https://www.purplemath.com/modules/remvble.htm This source includes examples and step-by-step solutions for finding and identifying removable discontinuities in rational functions. 3. "Defining and Finding Limits" by Paul's Online Math Notes: http://tutorial.math.lamar.edu/Classes/CalcI/LimitsIntro.aspx This source provides comprehensive explanations and examples of calculating limits, including limits of rational functions. By utilizing these sources and understanding the concepts and steps involved in finding and identifying discontinuities in rational functions, we can successfully answer the given question. Question 3 find the arclength of y =2x^2+4 on 0 less than or equal to x less than or equal to 3 Answer To find the arclength of a curve, we use the formula: Arclength = ∫√(1+(dy/dx)^2)dx In this case, the given curve is a parabola with equation y = 2x^2 + 4. We can rewrite this equation as: (dy/dx) = 4x Plugging this into our formula, we get: Arclength = ∫√(1+(4x)^2)dx = ∫√(1+16x^2)dx = ∫(√(1+16x^2))dx We can use the substitution method to solve this integral. Let u = 1+16x^2, then du = 32xdx. Substituting these values in our integral, we get: Arclength = (1/32)*∫√u du = (1/32)*(2/3)*u^(3/2) + C = (1/48)*(1+16x^2)^(3/2) + C Now, we can plug in the values of 0 and 3 for x to get the final answer: Arclength = (1/48)*[(1+16*3^2)^(3/2) - (1+16*0^2)^(3/2)] = (1/48)*[325]^(3/2) = (1/48)*113.02 = 2.35 units (rounded to two decimal places) Question 4 What is the surface area of revolution about the x-axis of y= 2sin(3x) over the interval 0 less than or equal to x less than or equal to pi/3? Answer According to the Fundamental Theorem of Calculus, the surface area of revolution about the x-axis can be calculated using the integral of the function 2πy√(1 + [f'(x)]^2) over the given interval 0 ≤ x ≤ π/3. Step-by-step calculation: 1. First, we need to find the derivative of the given function y = 2sin(3x). Using the chain rule, we get f'(x) = 6cos(3x). 2. Substitute this derivative into the formula: 2πy√(1 + [f'(x)]^2) to get: 2π(2sin(3x))√(1 + [6cos(3x)]^2) 3. Simplify the expression to get: 4πsin(3x)√(1 + 36cos^2(3x)) 4. Now, we can integrate this expression over the given interval of 0 ≤ x ≤ π/3. Using substitution method, let u = 3x, du = 3dx. And applying the limits of integration, we get: ∫0^(π/3) 4πsin(u)√(1 + 36cos^2(u)) * (1/3)du = (4/3)π ∫0^(π/3) sin(u)√(1 + 36cos^2(u)) du = (4/3)π ∫0^(π/3) √(1 + 36 - 36sin^2(u))du = (4/3)π ∫0^(π/3) √(37 - 36sin^2(u))du 5. Using the formula, ∫√(a^2 - x^2)dx = (1/2)(x√(a^2 - x^2) + a^2arcsin(x/a)) + C, we can integrate the above expression to get: (4/3)π ∫0^(π/3) √(37 - 36sin^2(u))du = (4/3)π * √(37)(1/2) * arcsin(6/ √37) + (4/3)π * (1/4) * (6/√37)√(37 - 36) + C = (2/3)π √37arcsin(6/√37) + (1/3)π √37 6. Therefore, the surface area of revolution about the x-axis of y = 2sin(3x) over the given interval is: (2/3)π √37arcsin(6/√37) + (1/3)π √37 = (2/3)π [(√37/2)arcsin(6/√37) + 1] Question 5 Find the surface area of revolution about the x-axis of y=5sin(4x) over the interval 0 less than or equal to x less than or equal to pi/4 Answer To find the surface area of revolution about the x-axis, we will use the formula: S = 2π∫𝑦√(1+(𝑦′)^2)𝑦𝑦 Where, 𝑦 is the curve, 𝑦′ is the derivative of the curve, 𝑦 is the variable of integration, and the limits of integration are the given interval. Rewriting the given equation in terms of 𝑦, we get: 𝑦 = 5sin(4𝑦) Taking the derivative of 𝑦 with respect to 𝑦, we get: 𝑦′ = 20cos(4𝑦) Therefore, substituting these values in the surface area formula, we get: S = 2π∫(5sin(4𝑦))√(1+(20cos(4𝑦))^2)𝑦𝑦 Integrating and simplifying, we get: S = 2π∫√(400sin^2(4𝑦)+1)𝑦𝑦 Using the trigonometric identity, sin^2(𝑦) = (1-cos(2𝑦))/2, we can rewrite the above equation as: S = 2π∫√(200(1-cos(8𝑦))+1)𝑦𝑦 Expanding the square root, we get: S = 2π∫√(200-200cos(8𝑦)+1)𝑦𝑦 Simplifying further, we get: S = 2π∫√(201-200cos(8𝑦))𝑦𝑦 Using the substitution, u = 8𝑦, 𝑦𝑦 = 8𝑦𝑦, the integral becomes: S = 2π/8∫√(201-200cos𝑦)𝑦𝑦 Using the trignometric identity, cos𝑦 = (e^𝑦𝑦+e^(-𝑦𝑦))/2, we can rewrite the above equation as: S = π/4∫√(201-200(e^𝑦𝑦+e^(-𝑦𝑦)))𝑦𝑦 Expanding the square root, we get: S = π/4∫(5√(9+4e^𝑦𝑦)+5√(9+4e^(-𝑦𝑦)))𝑦𝑦 Using the substitution, 𝑦 = 9+4e^𝑦, 𝑦𝑦 = 4e^𝑦𝑦𝑦, the integral becomes: S = π/4∫(5√𝑦+5√𝑦)𝑦𝑦 Simplifying further, we get: S = π/4∫10√𝑦 𝑦𝑦 Integrating, we get: S = [π/4(2/3(𝑦)^3/2)]∣ ∣ ∣ ∣ ∣ 9+4e^𝑦 Applying the limits of integration, we get: S = [π/2(3(9+4e^(pi/4))^3/2-3(9+4))^3/2] = 3π(√6+2)^3/2 - 27π/2 = 34.136 units squared (approx)