Applications of Integration and Polar Curves

A definite integral adds up many small pieces, and the choice of piece decides what it measures: thin rectangles give an area, thin discs give a volume, short straight segments give the length of a curve, and thin sectors give the area of a region described by an angle. A differential equation contains a derivative, and solving it means integrating back from the rate of change to the quantity itself.

What does integration measure in practice?

Integrating velocity over time gives displacement, and integrating acceleration gives velocity. A stretch of negative velocity subtracts, so the integral is the net change of position.

Distance counts every stretch as positive. Solve v = 0 to find where the velocity changes sign, integrate between those times, and add the absolute values.

See Velocity and displacement by integration and Total distance traveled.

How do you find the area between two curves?

At each x the gap is the upper height minus the lower height, so integrate that difference: area $=\int \left(\text{upper}-\text{lower}\right)dx$= ∫ (upper − lower) dx. The limits are where the curves cross, so first set the two expressions equal and solve.

If the curves swap over inside the region, split the integral at each crossing. Otherwise the parts cancel and the answer comes out too small.

See Area between two curves.

How do you find the volume of a solid of revolution?

Spin the region under a curve about the x-axis. Each thin slice is a disc of radius y and thickness dx, so its volume is $\pi y^{2}dx$πy² dx. About the y-axis, the radius is x and the thickness is dy.

If the region does not reach the axis, each slice is a ring with area $\pi R^2-\pi r^2$πR² − πr², so $V=\pi \int \left(R^2-r^2\right)dx$V = π ∫ (R² − r²) dx. Square each radius before subtracting; $\pi {\left(R-r\right)}^2$π(R − r)² is a different number.

A solid need not be made by spinning. If every slice has a known shape, integrate the slice area: $V=\int A\left(x\right)dx$V = ∫ A(x) dx. Squares standing on the region under $y=\sqrt{x}$y = √x have area x, so on [0, 4] the volume is $\int xdx=8$∫ x dx = 8, and no $\pi$π appears.

See Volumes of revolution, Volumes by washers and Volumes with known cross sections.

How do you find the length of a curve, or its average height?

A small piece of curve is the hypotenuse of a right triangle with legs dx and dy, so $ds=\sqrt{d{x}^2+d{y}^2}$ds = √(dx² + dy²). Factor out dx and integrate: $L=\int \sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx$L = ∫ √(1 + (dy/dx)²) dx.

The mean value of a function on an interval is its average height: the area under the curve divided by the width, $\left(\frac{1}{\left(b-a\right)}\right)\int f\left(x\right)dx$(1/(b − a)) ∫ f(x) dx.

See Arc length and The mean value of a function.

What if the region never ends?

An improper integral has infinity as a limit, or an integrand that becomes infinite at one end. Replace the troublesome limit with a variable t, integrate to t, and then take the limit.

The area under $\frac{1}{x^2}$1/x² from 1 to t is $1-\frac{1}{t}$1 − 1/t, which tends to 1: the integral converges, and a region of endless width has a finite area. The area under $\frac{1}{x}$1/x is ln t, which grows without bound, so that integral diverges. The same distinction runs through sequences and series.

See Improper integrals.

What if the integral cannot be done at all?

Some functions, such as $e^{-x^2}$e^(−x²), have no antiderivative that can be written down, yet the area under them is a definite number. Estimate it numerically.

The trapezium rule uses strips of width h with slanted tops: $\left(\frac{h}{2}\right)\left(y_0+2y_1+\cdots +2y_{n-1}+y_n\right)$(h/2)(y₀ + 2y₁ + … + 2yₙ₋₁ + yₙ), each end height once and each inner height twice.

Simpson's rule fits a parabola to each pair of strips, so it needs an even number of them. The weights are 1, 4, 2, 4, …, 4, 1, multiplied by $\frac{h}{3}$h/3. A parabola follows the bend, so the estimate is far closer, and it is exact for any cubic.

See Numerical integration and Simpson's rule.

What is a differential equation?

A differential equation contains a derivative, and its solution is a function. Forming one is a translation: "grows at a rate proportional to its size" makes the rate dP/dt and the size P, so dP/dt = kP.

A separable equation can be rearranged so that every y is on one side with dy and every x on the other with dx, and then both sides are integrated. For $\frac{dy}{dx}=ky$dy/dx = ky, divide by y and integrate: ln y = kx + c. An initial condition, a known value of y at a known x, fixes the constant.

A linear equation $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$dy/dx + P(x) y = Q(x) does not separate. Multiply through by the integrating factor $I=e^{\int Pdx}$I = e^(∫ P dx). The left side becomes $\frac{d}{dx}\left(Iy\right)$d/dx (Iy), a single derivative, so integrate both sides and divide by I.

See Forming differential equations, Separable differential equations and The integrating factor.

What if you cannot solve it exactly?

A slope field draws a short segment at each point with the gradient the equation gives there. A solution runs along the segments at every point it passes through, so the field shows every solution at once.

Euler's method turns the field into numbers. Start at the known point, follow the tangent for a width h, use the equation to find the new gradient, and repeat. On a curve that bends upward each tangent sits below it, so the estimate lands low, and a smaller step reduces the error.

See Slope fields and Euler's method.

How do you solve a second-order equation?

The equation a y'' + b y' + c y = 0 asks for a function whose derivatives are multiples of itself, so try $y=e^{mx}$y = e^(mx). Dividing out the common factor $e^{mx}$e^(mx) gives the auxiliary equation $a{m}^2+bm+c=0$a m² + b m + c = 0.

Two real roots give $y=A{e}^{m_{1}x}+B{e}^{m_{2}x}$y = A e^(m₁x) + B e^(m₂x). A repeated root gives $y=\left(A+Bx\right)e^{mx}$y = (A + Bx) e^(mx). Complex roots $p\pm qi$p ± qi give $y=e^{px}\left(A\cos qx+B\sin qx\right)$y = e^(px)(A cos qx + B sin qx).

With a nonzero right side f(x), the complementary function solves the equation with 0 on the right, and the particular integral produces f(x): try a function of the same form with unknown coefficients, substitute, and match coefficients. Add the two.

See The auxiliary equation and Complementary function and particular integral.

Why does this describe oscillation?

In simple harmonic motion the acceleration is proportional to the displacement and points back toward the center: ẍ $=-{\omega }^{2}x$= −ω²x. The auxiliary equation $m^2+{\omega }^2=0$m² + ω² = 0 has roots $\pm \omega i$±ωi, so $x=A\cos \omega t+B\sin \omega t=R\cos \left(\omega t-\alpha \right)$x = A cos ωt + B sin ωt = R cos(ωt − α) with amplitude $R=\sqrt{A^2+B^2}$R = √(A² + B²). The period is $\frac{2\pi }{\omega }$2π/ω, so a larger swing takes exactly as long.

Resistance that grows with speed adds a term kẋ, and the discriminant $k^2-4{\omega }^2$k² − 4ω² decides the motion: light damping oscillates inside a shrinking envelope, heavy damping returns slowly with no oscillation, and critical damping is the fastest return that never crosses the center.

See Simple harmonic motion and Damped oscillations.

What if two quantities depend on each other?

Two quantities are coupled when the rate of each depends on both, so neither equation can be solved alone.

Try $X=e^{\lambda t}v$X = e^(λt) v. Cancel $e^{\lambda t}$e^(λt), and the guess works exactly when v is an eigenvector of M with eigenvalue $\lambda$λ. With two real eigenvalues the general solution is $A{e}^{{\lambda }_{1}t}{v}_1+B{e}^{{\lambda }_{2}t}{v}_2$A e^(λ₁t) v₁ + B e^(λ₂t) v₂, and the initial values fix A and B. The eigenvalue work is in matrices and Markov chains.

A phase portrait plots y against x, so each solution is one path, and the signs of the eigenvalues predict its shape: both negative, every path runs into the origin (a stable node); both positive, every path runs away (an unstable node); complex, the paths spiral.

A second-order equation is a coupled system in disguise: name the derivative $v=\frac{dx}{dt}$v = dx/dt, and one equation becomes the pair $\frac{dx}{dt}=v$dx/dt = v and $\frac{dv}{dt}$dv/dt in terms of x and v. The characteristic polynomial of that matrix is the auxiliary equation.

See Coupled differential equations, Solving a coupled system with eigenvalues, Phase portraits and equilibrium types and A second-order equation as a coupled system.

What are polar coordinates?

Polar coordinates name a point by its distance r from the origin and the angle $\theta$θ turned from the positive x-axis. The right triangle converts them: $x=r\cos \theta$x = r cos θ and $y=r\sin \theta$y = r sin θ. Going back, Pythagoras gives r and $\tan \theta =\frac{y}{x}$tan θ = y/x gives $\theta$θ, checked against the quadrant.

A polar curve lets r depend on $\theta$θ. r = 2 is a circle, and $r=2\cos \theta$r = 2 cos θ is a circle through the origin. To sketch one, read r at the quarter turns, mark those points, and find where r reaches 0. Differentiating r with respect to $\theta$θ shows where the curve moves in toward the pole.

A sector of angle $d\theta$dθ is the fraction $\frac{d\theta }{2\pi }$dθ/2π of a disc of radius r, so its area is ½$r^{2}d\theta$r² dθ, and adding the sectors gives area = ½ $\int r^{2}d\theta$∫ r² dθ between the two angles. Check it on a quarter disc of radius 2: ½ $\times 4\times \frac{\pi }{2}=\pi$× 4 × π/2 = π, a quarter of $4\pi$4π. Between two curves, subtract the squares: ½ $\int \left(R^2-r^2\right)d\theta$∫ (R² − r²) dθ.

See Polar coordinates, Sketching polar curves, The rate of change of a polar function and The area of a polar region.

The mistakes worth naming

• Giving displacement when distance was asked. Find where v = 0, integrate between those times, and add the absolute values.
• Subtracting the radii before squaring. The area of a ring is $\pi \left(R^2-r^2\right)$π(R² − r²), not $\pi {\left(R-r\right)}^2$π(R − r)².
• Wrong limits for an enclosed area. Solve for the crossing points first, and split where the curves swap over.
• An odd number of strips in Simpson's rule. One parabola covers each pair of strips, so the strips must pair up.
• Using $\int ydx$∫ y dx for a polar area. A polar region is swept in sectors, so the area is ½ $\int r^{2}d\theta$∫ r² dθ.

Learn this properly in the app

Every lesson linked above is in Math Challenge, with a diagram, a worked example and practice questions. The integration methods are in techniques of integration, convergence in sequences and series, and eigenvalues in matrices and Markov chains.