Applications of Integration

Velocity and Displacement by Integration #

The motion chain, run backwards.

Integrating acceleration recovers velocity, and integrating velocity recovers displacement

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Differentiating took s to v to a. Integrating reverses each step: a to v, then v to s.

v = 2t integrates to $s=t^2+C$s = t² + C, and starting at s = 0 when t = 0 pins C at 0.

Where v is below the axis the motion is backward: the integral gives the displacement, and distance adds the absolute values.

For $v=3t^2$v = 3t² from t = 0 to 2, integrate to get $t^3$t³ and evaluate at both ends: 8 − 0 = 8.

Total Distance Traveled #

Integrate the speed, not the velocity.

Integrating the speed rather than the velocity counts every stretch as a positive amount

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Velocity here is positive, then negative, then positive: the particle turns twice.

The signed integral gives displacement — the trip back subtracts from the trip out.

For total distance integrate the speed |v|, so no stretch can cancel another.

In practice: find where v = 0, integrate between those times, and add the absolute values.

Here the pieces are $\frac{4}{3}$4/3, $-\frac{4}{3}$−4/3 and $\frac{4}{3}$4/3: the distance is 4, but the displacement is only $\frac{4}{3}$4/3.

Area Between Two Curves #

Integrate the gap: upper minus lower.

Subtracting the lower curve from the upper one before integrating gives the gap

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Two curves cross, and the region between them is what is wanted.

At every x the gap is one height minus the other, so integrate that difference.

The points where they meet are the limits, so solve for those before integrating.

Three crossings here: which curve is on top changes, so split at each crossing.

Volumes of Revolution #

Spinning a curve about an axis sweeps a solid.

Spinning a curve about an axis sweeps out a solid made of circular discs

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Take the region under a curve and spin it about the x-axis.

A thin slice becomes a disc of radius y, so its area is $\pi y^2$πy².

Add the discs by integrating: the volume is $\pi \int y^{2}dx$π ∫ y² dx.

Spin about the y-axis instead and x and y change places: the radius is x, and the thickness is dy.

Volumes by Washers #

Each slice is a disc with a hole punched out.

Spinning a region that misses the axis leaves a hole, so every slice is a ring

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Spin the region between y = x and $y=x^2$y = x² about the x-axis, rather than the region under a single curve.

Each slice is a ring: a disc of radius R with a disc of radius r punched out.

Its face has area $\pi R^2-\pi r^2$πR² − πr²: the outer disc minus the hole.

Square each radius before subtracting — $\pi {\left(R-r\right)}^2$π(R − r)² is a different number entirely.

So $V=\pi \int \left(R^2-r^2\right)dx$V = π ∫ (R² − r²) dx, both radii measured out from the axis of the spin.

For y = x and $y=x^2$y = x² on [0, 1]: $\pi \left(\frac{1}{3}-\frac{1}{5}\right)=\frac{2\pi }{15}$π(1/3 − 1/5) = 2π/15.

About another line, each radius is the distance from that line down to the curve.

Volumes with Known Cross Sections #

Stack the slices; nothing has to spin.

Multiplying each slice area by its thickness and integrating gives a volume without spinning

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Start with a flat base region — here everything under $y=\sqrt{x}$y = √x from 0 to 4.

On each strip stands a square, and its side s is the strip’s height $\sqrt{x}$√x.

The square’s face is $s^2=x$s² = x, so a slab of thickness dx holds A(x) dx.

Add the slabs: $V=\int A\left(x\right)dx$V = ∫ A(x) dx, which is 8 here, and no $\pi$π appears at all.

Change the slice and only A changes: a rectangle uses sh, and a semicircle uses $\left(\frac{\pi }{8}\right)s^2$(π/8)s².

If the base sits between two curves, the side s is the gap: upper minus lower.

Arc Length #

Pythagoras on a tiny piece of curve.

Pythagoras on a tiny piece of curve gives a formula for its whole length

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Zoomed in far enough, a small piece of curve is almost straight, and Pythagoras gives ds from dx and dy.

Factor out dx, and integrating adds every small piece into the total length.

If x and y are both given in terms of a parameter t, take dt out of the same root instead.

The Mean Value of a Function #

The integral divided by the width.

Dividing the integral by the width gives the average height of a curve

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The integral is the area. Now ask what constant height would give the same area.

A horizontal line at that height encloses the same area: the mean value is the integral divided by the width.

Work it out: the area is $\frac{64}{3}$64/3, and spread over a width of 4 the height is $\frac{16}{3}$16/3.

Improper Integrals #

Run to a finite t, then let t grow forever.

An integral running to infinity is defined as the limit of ordinary ones

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This region extends forever to the right, so there is no upper limit to substitute.

Integrate to a finite t, then take the limit as t grows without bound.

Work it out: the area out to t is $1-\frac{1}{t}$1 − 1/t, and $\frac{1}{t}$1/t shrinks to 0, leaving exactly 1.

An endless region can still have a finite area, if it thins out fast enough.

$\frac{1}{x}$1/x falls too slowly: the area under it grows without bound, while the area under $\frac{1}{x^2}$1/x² converges.

The other kind is finite-width but infinitely tall, blowing up at an end point.

Start at t, integrate, then take the limit as t approaches the singularity.

Numerical Integration #

No antiderivative? Trapezia still get close.

When no antiderivative exists the area can still be estimated by trapezia

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This bell-shaped curve has no antiderivative that can be written down, yet the area under it is a definite number.

Cut the region into strips and let each top slant with the curve, not sit flat.

Adding those trapezia estimates the area, and thinner strips estimate it better.

Add the strips and the formula follows: h over 2, times the two end heights plus twice the inner heights.

Simpson's Rule #

Parabolas over the strips beat straight lines.

Fitting parabolas across the strip tops estimates an area far better than straight lines

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On a curve that bulges upward, a straight top cuts across it and undershoots the area.

Simpson’s rule takes the strips in pairs: three points fix one parabola for the top.

Add the areas of those parabolas and the weights come out as 1, 4, 2, 4, …, 4, 1, all multiplied by $h/3$h / 3.

With the same strips the answer is far closer, because a parabola follows the bend.

Separable Differential Equations #

Each variable on its own side, then integrate.

Gathering each variable on its own side lets both halves be integrated

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An equation linking a rate to the quantity itself describes growth or decay.

Separate the variables and integrate each side, which gives ln y = kx + c.

Undo the log: $e^{kx+c}$e^(kx + c) splits into $e^c$e^c times $e^{kx}$e^(kx), and that $e^c$e^c is the A.

Solving a Differential Equation by Substitution #

A given substitution, and the variables separate.

A given substitution turns the equation into one in which the variables separate

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$\frac{dy}{dx}={\left(x+y\right)}^2$dy/dx = (x + y)² does not separate. Start from the substitution given: differentiate it to link $\frac{du}{dx}$du/dx with $\frac{dy}{dx}$dy/dx.

Replace $\frac{dy}{dx}$dy/dx and every x + y. The new equation has only u and x, and it separates.

Integrate each side, then put x + y back for u and make y the subject.

For y = vx, the product rule gives $\frac{dy}{dx}=v+x\frac{dv}{dx}$dy/dx = v + x dv/dx, because v changes with x too.

Separate and integrate, then multiply by x to return to y: y = x ln|x| + Cx.

Forming Differential Equations #

A sentence about rates becomes an equation.

A sentence about a rate translates into a differential equation before any solving starts

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The sentence names both sides: "grows at a rate" is $\frac{dP}{dt}$dP/dt, and "its size" is P.

A hot drink cools toward room temperature R: $\frac{dT}{dt}=-k\left(T-R\right)$dT/dt = −k(T − R). The minus sign shows that T falls.

Name the rate, name what drives it, and join the two with a constant k.

Writing the equation is half the work, and separating the variables then solves it.

Slope Fields #

Every solution of an equation, drawn at once.

Drawing the gradient an equation prescribes at each point shows every solution at once

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$\frac{dy}{dx}=x$dy/dx = x names a gradient at every point. Draw a short piece of it at each one.

A solution is a curve that runs along the segment at every point it passes through.

Shift that curve up and it still fits, because the field holds the whole family $y=\frac{x^2}{2}+c$y = x²/2 + c.

When the gradient depends on y, the segments change as you go up a column, not across.

Flat segments mark stationary points, and a column of identical segments means the gradient does not depend on y.

Euler’s Method #

A differential equation stepped into numbers.

Stepping along the tangent a short way at a time turns a differential equation into numbers

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Some equations will not separate. The gradient at the starting point is still known.

Follow the tangent for a width h, then use the equation to find a new gradient.

Three steps, each one straight, track the true curve that runs above them.

For $\frac{dy}{dx}=y$dy/dx = y starting at 1, the steps give 1.5, 2.25 and 3.375. The true value at x = 1.5 is about 4.48.

On a curve bending upward each tangent sits below it, so the estimate always lands low.

The Integrating Factor #

One factor turns the left side into a product rule.

Multiplying a linear equation by a chosen factor turns its left side into one product rule

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This form does not separate: y is multiplied by a function of x, and $\frac{dy}{dx}$dy/dx stands on its own.

The left side is almost a product rule. Find the factor I that turns it into exactly one derivative, $\frac{d}{dx\left(Iy\right)}$d/dx(Iy).

That condition is itself separable, and solving it gives I as e to the integral of P.

Multiply through by I and the left side collapses to one derivative, then integrate.

Here P is 2 over x, so the integrating factor is $x^2$x², and the rest is a single integral.

The Auxiliary Equation #

A second-order equation becomes a quadratic.

Trying an exponential in a second-order equation turns it into a quadratic

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A second-order equation asks for a function whose derivatives are multiples of itself.

Substituting leaves a common exponential factor, and that factor is never zero.

A positive discriminant gives two real roots, so the solution is two exponentials.

A repeated root gives only one exponential, so an x is attached to build a second one.

A negative discriminant gives complex roots, and the solution becomes an exponential times sines and cosines.

With complex roots the solution oscillates, and the exponential sets the envelope.

Complementary Function and Particular Integral #

One part kills the left side, one builds the right.

A second-order equation with a right-hand side needs one solution of its own added on

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With a nonzero right side, the exponentials from the auxiliary equation no longer satisfy the equation on their own.

Split the answer in two: one part makes the left side 0, the other gives the right side.

The first part solves the equation with 0 on the right: the complementary function.

For the second part, try a function shaped like the right side and match the coefficients.

Solve for p and q, and the answer is the two parts added: complementary plus particular.

The trial function copies the form of the right side, with unknown coefficients in place of the given numbers.

Simple Harmonic Motion #

Pulled back in proportion, so it swings forever.

When acceleration is proportional to displacement and points back, the auxiliary roots are $\pm \omega i$±ωi and the motion is a wave

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Acceleration proportional to displacement and pointing back is $\ddot{x}=-{\omega }^{2}x$ẍ = −ω²x.

The roots are $\pm \omega i$±ωi. The real part is 0, so there is no decay: the motion is a steady oscillation.

x = 2 cos 2t oscillates between 2 and −2 forever, and its amplitude never changes.

Combine the two terms into $R\cos \left(\omega t-\alpha \right)$R cos(ωt − α), where $R=\sqrt{A^2+B^2}$R = √(A² + B²) is the amplitude.

The period is $\frac{2\pi }{\omega }$2π/ω. Only $\omega$ω sets it, so a larger amplitude takes exactly as long.

Damped Oscillations #

Light, critical or heavy — the discriminant says.

A resistance term adds a middle term km to the auxiliary equation, and its discriminant decides whether the motion still swings

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Resistance grows with speed, so a kẋ term appears and the auxiliary equation gains a middle term.

Light damping is $k^2<4{\omega }^2$k² < 4ω²: the roots are complex, so the motion oscillates inside a shrinking envelope.

Heavy damping is $k^2>4{\omega }^2$k² > 4ω²: two negative real roots, so the particle returns slowly with no oscillation at all.

$k^2=4{\omega }^2$k² = 4ω² is critical damping: the fastest return that never crosses the center.

The three cases of the discriminant are the three kinds of damping, each with its own name.