Techniques of Integration

The Antiderivative #

Which function would differentiate to this?

Integrating is asking which function would have differentiated to this one

Differentiating took one off the power, so going back puts one on.

Then undo the multiply by dividing by that new power. It leaves x cubed.

The c matters: sliding a curve up changes none of its gradients, so differentiating loses c.

Integrating Powers #

Add one to the power and divide by the new power.

Add one to the power and divide by the new power

Add one to the power first. That is the move differentiating undid.

Then divide by the new power and add c, the constant of integration.

The rule fails only at n = −1, where dividing by n+1 would divide by zero.

Finding the Constant of Integration #

One known point pins the whole curve down.

One known point pins down C, collapsing the family to a single curve

f'(x) = 2x integrates to $x^2+C$x² + C — a whole family of curves, one per C.

One fact picks one curve: this one passes through the point (1, 4).

Substitute the point in: $4=1^2+C$4 = 1² + C, so C = 3.

So $f\left(x\right)=x^2+3$f(x) = x² + 3 — the family collapses to the member that fits.

Reading the Integral Sign #

A stretched S that adds up thin slivers.

The stretched S is an instruction to add up slivers of height f(x) and width dx

The sign is a letter S stretched tall — S for sum. It adds up slivers.

One sliver is f(x) dx: height f(x), times a width dx too small to see.

The numbers top and bottom of the S are the limits of integration: the sweep starts at a and stops at b.

Read aloud: "the integral from a to b of f of x, dx". The function f(x) inside is called the integrand.

Differentials #

dy and dx come apart as tiny amounts.

A derivative comes apart: the tiny rise dy is the gradient times the tiny run dx

$\frac{dy}{dx}$dy/dx looks like a fraction because it began as one: a rise over a run.

Treat the pieces separately: if $y=x^2$y = x², the tiny rise dy is 2x times dx.

The same rule works for any letter: $u=x^3+1$u = x³ + 1 gives $du=3x^{2}dx$du = 3x² dx.

This is allowed because dy and dx shrink together, and their ratio stays the gradient.

The Definite Integral #

Ever thinner slices give the exact area.

Slicing a region into ever thinner rectangles gives its exact area in the limit

Slice the region into rectangles and add them. The total is close, but visibly coarse.

Each rectangle is a height times a width: f(x) times $\Delta x$Δx, added across.

Take that to its limit and the area is exact. That limit is the definite integral.

Left, Right and Midpoint Riemann Sums #

Three places to read each rectangle’s height.

Rectangles measured at the left, the right or the middle of each strip estimate an area

Divide the interval into strips. A left sum takes each height at the strip’s left edge.

A right sum reads the height at the right edge instead, and every rectangle grows.

A midpoint sum reads the height at the middle of each strip, and lands between the other two.

On a rising curve every left rectangle falls short, so the left sum underestimates.

A falling curve reverses it: the left edge is now the tall side, so the left sum overestimates.

For $y=\frac{x^2}{4}$y = x²/4 on [0, 4] in 4 strips, the three sums are 3.5, 5.25 and 7.5; the exact area is $\frac{16}{3}$16/3.

The Riemann Sum in Sigma Notation #

$\sum f\left(x_i\right)\Delta x$Σ f(xᵢ) Δx, shrunk until it becomes $\int$∫.

Writing the rectangle sum with sigma and shrinking the strips produces the integral

Each rectangle is $\Delta x$Δx wide and $f\left(x_i\right)$f(xᵢ) tall, so its area is $f\left(x_i\right)\Delta x$f(xᵢ) Δx.

Sigma adds the whole list: the sum from i = 1 to n of $f\left(x_i\right)\Delta x$f(xᵢ) Δx.

The width is $\left(b-a\right)/n$(b − a) / n, and the strip numbered i sits at $a+i\Delta x$a + iΔx.

Let n run to infinity and the sum becomes the integral: $\sum$Σ turns into $\int$∫, $\Delta x$Δx into dx.

That is where the integral sign comes from: a stretched S, standing for this sum.

The Fundamental Theorem of Calculus #

Area and antiderivative are the same thing.

Areas and antiderivatives turn out to be the same question asked twice

Let A(x) be the area under the curve up to x. Increase x a little and the area grows.

That growth is a thin strip of height f(x), so the area function differentiates to f.

So an area can be found by antidifferentiating, and evaluating at both ends.

Area Below the Axis #

The integral keeps score with signs.

The integral counts area below the axis as negative

The arch of sin x above the axis, from 0 to $\pi$π, has area 2, and the integral is 2.

From $\pi$π to $2\pi$2π every sliver height sin x is negative, so the integral counts −2.

Integrate over the whole cycle and the two halves cancel to 0 — the integral is a signed count.

Asked for the area itself, split at the crossings and add the sizes: 4.

Properties of the Definite Integral #

Flip the limits, and the sign flips too.

Reversing the limits flips the sign, equal limits give zero, and constants and sums come apart

Swap the limits and the sign flips: each dx has become a step backwards.

With both limits the same there is no width, so the integral is exactly 0.

A constant factor comes straight out: scaling every height scales the whole area.

And a sum splits: at each x the heights add, so the two areas add as well.

Two adjacent intervals join: a to b, plus b to c, makes a to c in one integral.

Integration by Substitution #

Substitution is the chain rule read backwards.

Substitution is the chain rule read backwards

Differentiating a composite leaves the inside derivative multiplying the result.

Spot that factor in an integral, then name the inside u and rewrite the bracket in terms of it.

Here 2x is the derivative of $x^2+1$x² + 1, so the substitution $u=x^2+1$u = x² + 1 works.

Replace 2x dx with du, integrate the power, then write $x^2+1$x² + 1 back in place of u.

Integrating f(ax + b) #

A stretched input costs a divide.

Integrating f(ax + b) keeps the outer shape but divides by the inside coefficient

Differentiating sin 2x brought out a factor of 2, so integrating cos 2x must divide by 2.

The rule: integrate as if the inside were plain x, then divide by its coefficient a.

Raise the power and divide by the new power 5, then by the inside 3. That makes 15.

The shortcut only works for a linear inside ax + b, whose derivative is the constant a; a curved inside has a derivative that changes with x.

Integration by Parts #

Trade the integral for an easier one.

The product rule read backwards trades one integral for an easier one

Integrate the product rule and rearrange: your integral swaps for a different one.

It only helps if the new integral is easier, so choose u as the part that simplifies.

Integrating Trigonometric Functions #

The trig derivatives, run in reverse.

The trigonometric derivatives run backwards, with the signs swapping over

Sine differentiates to cosine, so cosine integrates back to sine.

Walk the ring backwards to integrate, and that is where the minus sign comes from.

The area under one arch of sine works out to exactly 2.

Evaluate it: −cos at $\pi$π is 1, at 0 it is −1, and the difference is exactly 2.

Integrating sin² and cos² #

The double angle tames the square.

The double angle formula rewrites ${\sin }^2$sin² and ${\cos }^2$cos² into pieces that integrate directly

${\sin }^{2}x$sin²x is not on the list: nothing on the sine–cosine ring differentiates to ${\sin }^2$sin².

The double angle formula rearranges: ${\sin }^{2}x=\left(1-\cos 2x\right)/2$sin²x = (1 − cos 2x) / 2 — the square is gone.

Now each piece is easy: the $\frac{1}{2}$1/2 gives $\frac{x}{2}$x/2, and the cos 2x piece uses the linear shortcut.

${\cos }^{2}x$cos²x rewrites the same way, with a plus: $\left(1+\cos 2x\right)/2$(1 + cos 2x) / 2 — and each piece integrates as before.

Integrating Exponentials and Logarithms #

One over x integrates to the natural log.

One over x integrates to the natural log, which fills the gap the power rule left

e to the x is its own derivative, so it is its own integral too.

The power rule fails for 1 over x: adding one to −1 gives a zero to divide by.

But ln x differentiates to 1 over x, so it fills exactly that gap.

Integration by Partial Fractions #

Split one awkward fraction into two easy ones.

Splitting one awkward fraction into two simple ones makes each integrable

A product in the denominator matches none of the standard integrals.

Write it as two fractions over the factors, then clear the denominators.

Put x = 1 so the B term vanishes, then x = −1 to remove A. That gives both constants.

Each piece is a constant over a linear bracket, so each integrates to a logarithm.

A repeated factor needs a fraction for each power, or the pieces cannot rebuild it.

A quadratic that will not factor keeps a linear numerator: Ax + B, not a bare constant.

Integrals of f′ over f #

A logarithm hiding in a fraction.

When the numerator of a fraction is the derivative of the denominator, the integral is the log of the denominator

Differentiate ln f and the chain rule gives f' over f — a fraction with a special shape.

Read backwards: when the numerator is the derivative of the denominator, the integral is ln of the denominator.

Check the numerator: 2x is the derivative of $x^2+1$x² + 1, so the integral is $\ln \left(x^2+1\right)+c$ln(x² + 1) + c.

The skill is the check: differentiate the denominator first and see if the numerator matches.

Integrating sec²x, sec x tan x and tan x #

Three tan-family facts, run in reverse.

The tan-family derivatives read backwards, and tan x itself integrates to a logarithm

Every derivative fact reads backwards: ${\sec }^{2}x$sec²x integrates straight to tan x.

The sec derivative reverses the same way: sec x tan x integrates to sec x.

tan x itself is sin over cos — an f' over f logarithm that carries one minus sign.

Minus ln cos is ln sec — the two answers the books give are one answer.

Integrals That Give tan⁻¹ and sin⁻¹ #

Fractions whose antiderivatives are angles.

Fractions built from $1+x^2$1 + x² and $1-x^2$1 − x² integrate to inverse trig functions

None of the standard rules fits 1 over $1+x^2$1 + x², yet it integrates to the angle ${\tan }^{-1}x$tan⁻¹x.

The shaded area from 0 to 1 is $\frac{\pi }{4}$π/4 — the angle whose tangent is 1.

In the same way, 1 over the root of $1-x^2$1 − x² is the derivative of ${\sin }^{-1}x$sin⁻¹ x, so it integrates to ${\sin }^{-1}x$sin⁻¹ x.

With $a^2$a² in place of the 1, the result is $\left(\frac{1}{a}\right){\tan }^{-1}\left(\frac{x}{a}\right)$(1/a) tan⁻¹(x/a): 1 over a multiplies outside, and x over a replaces x inside.

The sign decides the method: $x^2-1$x² − 1 factors, so use partial fractions; $x^2+1$x² + 1 does not factor, and integrates to ${\tan }^{-1}$tan⁻¹.

Choosing a Trigonometric Substitution #

The sign inside the root picks sine or tangent.

A root of $a^2$a² minus $x^2$x² becomes a cosine once x is written as $a\sin \theta$a sin θ

There is no factor of x beside the root, so the substitution $u=4-x^2$u = 4 − x² has nothing to cancel.

Substitute $x=2\sin \theta$x = 2 sin θ: the identity turns the root of $4-x^2$4 − x² into a cosine.

Differentiate the substitution for dx, and what is left is a cosine squared.

The sign inside decides the substitution: a minus takes sine, a plus takes tangent.

With a plus inside, $x=3\tan \theta$x = 3 tan θ cancels the whole denominator and leaves a constant.

Integrals That Give arsinh and arcosh #

A plus inside the root turns the answer over.

A root of x squared plus a squared integrates to an inverse hyperbolic function

arsinh x differentiates to 1 over the root of $x^2+1$x² + 1, using the identity ${\cosh }^{2}y=1+{\sinh }^{2}y$cosh²y = 1 + sinh²y.

Read those derivatives backwards: 1 over each root integrates to an inverse hyperbolic function.

With a general a the answer is arsinh of x over a: the a only scales the input.

The sign inside the root decides the family: a minus is circular, a plus is hyperbolic.

Because arsinh is a logarithm, the answer may be written either way and still be right.