Calculus II Unit 2 – Applications of Integration

Applications of Integration is a crucial topic in Calculus II, bridging the gap between abstract mathematical concepts and real-world problem-solving. This unit explores how integrals can be used to calculate areas, volumes, arc lengths, and surface areas of complex shapes and curves. Students learn to apply integration techniques to physics problems, including work, fluid pressure, and center of mass calculations. These applications demonstrate the power of calculus in modeling and analyzing physical phenomena, providing a foundation for advanced studies in engineering and science.

Key Concepts and Definitions

  • Integration involves finding the area under a curve, the opposite of differentiation which finds the slope of a tangent line at a point
  • Definite integrals calculate the area under a curve between two specific points (lower and upper limits of integration)
  • Indefinite integrals find the general antiderivative of a function without specific limits
    • Represented as f(x)dx=F(x)+C\int f(x) dx = F(x) + C, where CC is the constant of integration
  • Riemann sums approximate the area under a curve by dividing it into rectangles and summing their areas
    • As the number of rectangles approaches infinity, the Riemann sum approaches the exact value of the definite integral
  • Integration by substitution is a technique for evaluating integrals by substituting a new variable to simplify the expression
  • Integration by parts is a method for integrating products of functions, using the formula udv=uvvdu\int u dv = uv - \int v du
  • Improper integrals are integrals with infinite limits or discontinuous integrands
    • Convergent improper integrals have finite values, while divergent improper integrals do not

Fundamental Theorem of Calculus Review

  • The Fundamental Theorem of Calculus (FTC) connects differentiation and integration, showing they are inverse operations
  • The First Fundamental Theorem of Calculus states that if ff is continuous on [a,b][a, b], then the function g(x)=axf(t)dtg(x) = \int_a^x f(t) dt is an antiderivative of ff on [a,b][a, b]
    • In other words, ddxaxf(t)dt=f(x)\frac{d}{dx} \int_a^x f(t) dt = f(x)
  • The Second Fundamental Theorem of Calculus provides a way to evaluate definite integrals using antiderivatives
    • If 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)
  • The FTC allows us to calculate definite integrals without using Riemann sums or other approximation methods
  • The Mean Value Theorem for Integrals is a consequence of the FTC, stating 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)

Area Between Curves

  • To find the area between two curves, integrate the difference of their functions over the interval where they intersect
    • If f(x)g(x)f(x) \geq g(x) on [a,b][a, b], then the area between the curves is ab[f(x)g(x)]dx\int_a^b [f(x) - g(x)] dx
  • When the curves intersect at more than two points, split the integral into subintervals where one function is consistently greater than the other
  • For curves defined parametrically, find the intersection points by setting their x and y components equal and solving for the parameter
  • When working with polar curves, use the formula A=12ab[r2(θ)2r1(θ)2]dθA = \frac{1}{2} \int_a^b [r_2(\theta)^2 - r_1(\theta)^2] d\theta to find the area between them
  • Be careful with the order of subtraction when setting up the integral, as subtracting in the wrong order can lead to a negative area

Volume of Solids

  • The disk method calculates the volume of a solid of revolution by integrating the area of circular cross-sections perpendicular to the axis of rotation
    • For a solid formed by rotating the region bounded by y=f(x)y = f(x), y=0y = 0, x=ax = a, and x=bx = b about the x-axis, the volume is V=abπ[f(x)]2dxV = \int_a^b \pi [f(x)]^2 dx
  • The washer method is similar to the disk method but subtracts the volume of an inner hole from each cross-section
    • For a solid formed by rotating the region bounded by y=f(x)y = f(x) and y=g(x)y = g(x) about the x-axis, the volume is V=abπ([f(x)]2[g(x)]2)dxV = \int_a^b \pi ([f(x)]^2 - [g(x)]^2) dx
  • The shell method calculates volume by integrating cylindrical shells parallel to the axis of rotation
    • For a solid formed by rotating the region bounded by x=f(y)x = f(y), x=g(y)x = g(y), y=cy = c, and y=dy = d about the y-axis, the volume is V=2πcdx[f(y)g(y)]dyV = 2\pi \int_c^d x[f(y) - g(y)] dy
  • When using the shell method, the radius of each shell is the distance from the shell to the axis of rotation
  • Choose the most appropriate method based on the geometry of the solid and the axis of rotation to simplify the integral

Arc Length and Surface Area

  • Arc length measures the distance along a curve between two points
    • For a function y=f(x)y = f(x) on [a,b][a, b], the arc length is L=ab1+[f(x)]2dxL = \int_a^b \sqrt{1 + [f'(x)]^2} dx
    • For a parametric curve (x(t),y(t))(x(t), y(t)) on [a,b][a, b], the arc length is L=ab[x(t)]2+[y(t)]2dtL = \int_a^b \sqrt{[x'(t)]^2 + [y'(t)]^2} dt
  • Surface area is the area of the surface formed by rotating a curve about an axis
    • For a curve y=f(x)y = f(x) rotated about the x-axis on [a,b][a, b], the surface area is SA=2πabf(x)1+[f(x)]2dxSA = 2\pi \int_a^b f(x) \sqrt{1 + [f'(x)]^2} dx
    • For a parametric curve (x(t),y(t))(x(t), y(t)) rotated about the x-axis on [a,b][a, b], the surface area is SA=2πaby(t)[x(t)]2+[y(t)]2dtSA = 2\pi \int_a^b y(t) \sqrt{[x'(t)]^2 + [y'(t)]^2} dt
  • When calculating surface area, the integrand represents the circumference of each circular strip multiplied by the arc length of the generating curve
  • Be careful to use the correct formula based on whether the curve is given as an explicit function or parametrically

Work and Fluid Pressure

  • Work is the product of force and displacement in the direction of the force
    • For a variable force F(x)F(x) acting on an object moving along a straight line from x=ax = a to x=bx = b, the work done is W=abF(x)dxW = \int_a^b F(x) dx
  • Hooke's Law states that the force exerted by a spring is proportional to its displacement from equilibrium, F(x)=kxF(x) = kx, where kk is the spring constant
    • The work done to compress or stretch a spring from aa to bb 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
    • The hydrostatic pressure at a depth hh in a fluid with density ρ\rho is P=ρghP = \rho gh, where gg is the acceleration due to gravity
  • The force exerted by fluid pressure on a vertical surface is F=abP(y)w(y)dyF = \int_a^b P(y) w(y) dy, where P(y)P(y) is the pressure at depth yy and w(y)w(y) is the width of the surface at that depth
  • Work and fluid pressure problems often involve setting up integrals based on physical principles and given information

Center of Mass and Moments

  • The center of mass is the point where an object's mass is evenly distributed, and the object would balance if supported at that point
    • For a thin rod with linear density ρ(x)\rho(x) along the x-axis from aa to bb, the x-coordinate of the center of mass is xˉ=abxρ(x)dxabρ(x)dx\bar{x} = \frac{\int_a^b x\rho(x) dx}{\int_a^b \rho(x) dx}
    • For a lamina (thin plate) with surface density σ(x,y)\sigma(x, y) over a region RR, the coordinates of the center of mass are 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}
  • Moments measure the tendency of a force to cause rotation about a point or axis
    • The moment of a point mass mm about the origin is r×mv\vec{r} \times m\vec{v}, where r\vec{r} is the position vector and v\vec{v} is the velocity vector
    • For a lamina with surface density σ(x,y)\sigma(x, y), the moments of inertia about the x and y axes are Ix=Ry2σ(x,y)dAI_x = \iint_R y^2 \sigma(x, y) dA and Iy=Rx2σ(x,y)dAI_y = \iint_R x^2 \sigma(x, y) dA
  • The parallel axis theorem relates the moment of inertia about an axis passing through the center of mass to the moment of inertia about a parallel axis
    • If IcmI_{cm} is the moment of inertia about an axis through the center of mass and dd is the perpendicular distance between the axes, then the moment of inertia about the parallel axis is I=Icm+md2I = I_{cm} + md^2

Real-World Applications and Examples

  • Calculating the volume of irregular objects (vases, bottles, or other containers) by rotating their cross-sections
  • Determining the work done by variable forces (spring compression, cable hoisting, or fluid displacement)
  • Finding the center of mass and moments of inertia for complex shapes (beams, plates, or mechanical components)
  • Modeling population growth or decay using exponential and logistic functions
  • Analyzing the flow of fluids through pipes or the pressure exerted on dams and tanks
  • Calculating the length of power cables or pipelines spanning uneven terrain
  • Determining the surface area of objects for coating, heating, or cooling applications
  • Estimating the mass of thin wires, rods, or sheets with non-uniform density
  • Optimizing the design of containers, beams, or other structures to minimize weight or maximize strength
  • Predicting the motion of objects under the influence of variable forces or moments


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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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