Calculus I Unit 6 – Applications of Integration

Applications of Integration is a crucial topic in Calculus I, bridging the gap between abstract mathematical concepts and real-world problem-solving. This unit explores how integration techniques can be used to calculate areas, volumes, arc lengths, and surface areas of complex shapes and curves. Students learn to apply integration to various fields, including physics, engineering, and economics. From determining work done by variable forces to calculating fluid pressure and center of mass, these applications demonstrate the power and versatility of integration in solving practical problems.

Key Concepts and Definitions

  • Integration involves finding the area under a curve, which represents the accumulation of a quantity over an interval
  • Definite integrals calculate the area between a function and the x-axis over a specific interval [a,b][a, b], denoted as abf(x)dx\int_a^b f(x) dx
  • Indefinite integrals find the set of all antiderivatives of a function, denoted as f(x)dx=F(x)+C\int f(x) dx = F(x) + C, where CC is an arbitrary constant
    • Antiderivatives are functions whose derivative is the original function
  • The Fundamental Theorem of Calculus connects differentiation and integration, establishing their inverse relationship
  • Riemann sums approximate the area under a curve by dividing the interval into subintervals and summing the areas of rectangles
    • As the number of subintervals approaches infinity, the Riemann sum converges to the definite integral
  • Integration techniques include substitution, integration by parts, trigonometric substitution, and partial fractions
  • Applications of integration include finding areas between curves, volumes of solids, arc lengths, surface areas, work, fluid pressure, center of mass, and moments

Fundamental Theorem of Calculus Review

  • The Fundamental Theorem of Calculus (FTC) establishes the relationship between differentiation and integration
  • The First Fundamental Theorem of Calculus states that if ff is continuous on [a,b][a, b] and FF is an antiderivative of ff on [a,b][a, b], then abf(x)dx=F(b)F(a)\int_a^b f(x) dx = F(b) - F(a)
    • This theorem allows us to evaluate definite integrals using antiderivatives
  • The Second Fundamental Theorem of Calculus states that if ff is continuous on an open interval II containing aa, then the function g(x)=axf(t)dtg(x) = \int_a^x f(t) dt is continuous on II and differentiable on the interior of II, with g(x)=f(x)g'(x) = f(x)
    • This theorem relates the definite integral to the derivative of a function
  • The FTC enables us to find the exact value of a definite integral without using Riemann sums or approximations
  • The Mean Value Theorem for Integrals is a consequence of the FTC and states that for a continuous function ff on [a,b][a, b], there exists a point c[a,b]c \in [a, b] such that abf(x)dx=f(c)(ba)\int_a^b f(x) dx = f(c)(b - a)
  • The FTC is crucial for solving various applications of integration, such as finding areas, volumes, and lengths

Area Between Curves

  • Finding the area between curves involves integrating the difference between two functions over a given interval
  • To find the area between two curves y=f(x)y = f(x) and y=g(x)y = g(x) over the interval [a,b][a, b], integrate the absolute value of their difference: A=abf(x)g(x)dxA = \int_a^b |f(x) - g(x)| dx
    • If f(x)g(x)f(x) \geq g(x) on [a,b][a, b], then A=ab[f(x)g(x)]dxA = \int_a^b [f(x) - g(x)] dx
    • If f(x)g(x)f(x) \leq g(x) on [a,b][a, b], then A=ab[g(x)f(x)]dxA = \int_a^b [g(x) - f(x)] dx
  • When the curves intersect, split the interval at the intersection points and calculate the area of each subregion separately
  • To find the area between two curves given by x=f(y)x = f(y) and x=g(y)x = g(y) over the interval [c,d][c, d], integrate the absolute value of their difference with respect to yy: A=cdf(y)g(y)dyA = \int_c^d |f(y) - g(y)| dy
  • Be cautious of the order of the functions when setting up the integral to ensure the area is positive
  • Sketch the curves to visualize the region and determine the appropriate integration limits and function order

Volume of Solids

  • Integration can be used to calculate the volume of solids formed by revolving a region bounded by curves around an axis
  • The disk method calculates the volume of a solid formed by revolving a region around the x-axis or y-axis
    • For a region bounded by y=f(x)y = f(x), y=g(x)y = g(x), x=ax = a, and x=bx = b, the volume of the solid formed by revolving the region around the x-axis is V=πab[f(x)]2[g(x)]2dxV = \pi \int_a^b [f(x)]^2 - [g(x)]^2 dx
    • For a region bounded by x=f(y)x = f(y), x=g(y)x = g(y), y=cy = c, and y=dy = d, the volume of the solid formed by revolving the region around the y-axis is V=πcd[f(y)]2[g(y)]2dyV = \pi \int_c^d [f(y)]^2 - [g(y)]^2 dy
  • The washer method calculates the volume of a solid formed by revolving a region around a line parallel to the x-axis or y-axis
    • For a region bounded by y=f(x)y = f(x), y=g(x)y = g(x), x=ax = a, and x=bx = b, the volume of the solid formed by revolving the region around the line y=ky = k is V=πab[f(x)k]2[g(x)k]2dxV = \pi \int_a^b [f(x) - k]^2 - [g(x) - k]^2 dx
    • For a region bounded by x=f(y)x = f(y), x=g(y)x = g(y), y=cy = c, and y=dy = d, the volume of the solid formed by revolving the region around the line x=hx = h is V=πcd[f(y)h]2[g(y)h]2dyV = \pi \int_c^d [f(y) - h]^2 - [g(y) - h]^2 dy
  • The shell method calculates the volume of a solid formed by revolving a region around the y-axis or x-axis using cylindrical shells
    • For a region bounded by y=f(x)y = f(x), y=g(x)y = g(x), x=ax = a, and x=bx = b, the volume of the solid formed by revolving the region around the y-axis using the shell method is V=2πabx[f(x)g(x)]dxV = 2\pi \int_a^b x[f(x) - g(x)] dx
    • For a region bounded by x=f(y)x = f(y), x=g(y)x = g(y), y=cy = c, and y=dy = d, the volume of the solid formed by revolving the region around the x-axis using the shell method is V=2πcdy[f(y)g(y)]dyV = 2\pi \int_c^d y[f(y) - g(y)] dy

Arc Length and Surface Area

  • Arc length is the distance along a curve between two points
  • To find the arc length of a curve y=f(x)y = f(x) over the interval [a,b][a, b], use the formula L=ab1+[f(x)]2dxL = \int_a^b \sqrt{1 + [f'(x)]^2} dx
    • This formula is derived using the Pythagorean theorem and the concept of a curve as a limit of line segments
  • For a curve given parametrically by x=f(t)x = f(t) and y=g(t)y = g(t), where atba \leq t \leq b, the arc length is L=ab[f(t)]2+[g(t)]2dtL = \int_a^b \sqrt{[f'(t)]^2 + [g'(t)]^2} dt
  • Surface area is the area of the surface formed by revolving a curve around an axis
  • To find the surface area of a surface formed by revolving a curve y=f(x)y = f(x) around the x-axis over the interval [a,b][a, b], use the formula SA=2πabf(x)1+[f(x)]2dxSA = 2\pi \int_a^b f(x) \sqrt{1 + [f'(x)]^2} dx
    • This formula is derived by considering the surface as a sum of the lateral areas of frustums of cones
  • For a surface formed by revolving a curve given parametrically by x=f(t)x = f(t) and y=g(t)y = g(t), where atba \leq t \leq b, around the x-axis, the surface area is SA=2πabg(t)[f(t)]2+[g(t)]2dtSA = 2\pi \int_a^b g(t) \sqrt{[f'(t)]^2 + [g'(t)]^2} dt
  • Similar formulas exist for surfaces formed by revolving curves around the y-axis or other lines parallel to the axes

Work and Fluid Pressure

  • Work is the product of force and displacement in the direction of the force
  • To find the work done by a variable force F(x)F(x) acting on an object moving along a straight line from x=ax = a to x=bx = b, use the formula W=abF(x)dxW = \int_a^b F(x) dx
    • This formula is derived by considering work as the limit of the sum of the products of force and displacement over small intervals
  • Hooke's Law states that the force exerted by a spring is directly proportional to its displacement from equilibrium, F(x)=kxF(x) = kx, where kk is the spring constant
    • The work done by a spring force in stretching or compressing a spring from x=ax = a to x=bx = b is W=abkxdx=12k(b2a2)W = \int_a^b kx dx = \frac{1}{2}k(b^2 - a^2)
  • Fluid pressure is the force per unit area exerted by a fluid on a surface
  • To find the force exerted by a fluid with density ρ\rho on a vertical rectangular plate with width ww and height hh, submerged so that its top edge is at depth aa and its bottom edge is at depth bb, use the formula F=ρgwabxdx=12ρgw(b2a2)F = \rho g w \int_a^b x dx = \frac{1}{2}\rho g w (b^2 - a^2), where gg is the acceleration due to gravity
    • This formula is derived by integrating the pressure, which varies linearly with depth, over the surface area of the plate

Center of Mass and Moments

  • The center of mass is the point at which an object's mass can be considered to be concentrated for certain calculations
  • For a thin rod of length LL with linear density ρ(x)\rho(x), the x-coordinate of the center of mass is given by xˉ=0Lxρ(x)dx0Lρ(x)dx\bar{x} = \frac{\int_0^L x\rho(x) dx}{\int_0^L \rho(x) dx}
    • This formula is derived by considering the rod as a sum of infinitesimal mass elements and finding the balance point
  • For a thin plate in the xy-plane with surface density σ(x,y)\sigma(x, y), the coordinates of the center of mass are given by xˉ=Rxσ(x,y)dARσ(x,y)dA\bar{x} = \frac{\iint_R x\sigma(x, y) dA}{\iint_R \sigma(x, y) dA} and yˉ=Ryσ(x,y)dARσ(x,y)dA\bar{y} = \frac{\iint_R y\sigma(x, y) dA}{\iint_R \sigma(x, y) dA}, where RR is the region occupied by the plate
  • Moments are quantities that describe the distribution of mass or force about a point or axis
  • The moment of a force FF about a point is the product of the force and the perpendicular distance from the point to the line of action of the force
  • The first moment of area of a region RR in the xy-plane about the x-axis is given by Qx=RydAQ_x = \iint_R y dA, and the first moment of area about the y-axis is given by Qy=RxdAQ_y = \iint_R x dA
    • These moments are used in calculating centroids and in beam bending problems

Real-World Applications

  • Optimization problems involve finding the maximum or minimum value of a quantity subject to certain constraints
    • Example: Minimizing the surface area of a cylindrical can with a fixed volume to minimize the cost of materials
  • Hydrostatic force and pressure calculations are used in designing dams, tanks, and other structures that hold fluids
    • Example: Calculating the force on a submerged gate of a dam to determine the required strength of the gate
  • Center of mass calculations are important in designing balanced structures and understanding the stability of objects
    • Example: Finding the center of mass of an irregularly shaped object to determine its balance point
  • Moments of inertia, which involve integration, are used in analyzing the rotational motion and bending of objects
    • Example: Calculating the moment of inertia of a beam to determine its resistance to bending under a load
  • Physics applications include work, energy, and power calculations involving forces and displacements
    • Example: Determining the work done by a variable force in moving an object along a path
  • Biology and medicine applications include calculating the cardiac output of the heart and the absorption of drugs in the bloodstream
    • Example: Modeling the concentration of a drug in the bloodstream over time using integration
  • Economics applications include calculating consumer and producer surplus, marginal revenue, and marginal cost
    • Example: Finding the consumer surplus, which is the area between the demand curve and the price line


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© 2024 Fiveable Inc. All rights reserved.
AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.
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