Part IV · Series, Exponentials, and Integral Calculus

Chapter 13 — Integration Techniques and Geometric Applications

Why This Matters

Once an integral represents an accumulation, the remaining challenge is choosing a method that exposes an antiderivative or a geometry. Substitution handles a hidden chain rule; integration by parts handles a product. Applications require careful intervals and diagrams, not formula substitution alone.

13.1 Integration by parts

The product rule says $$(uv)'=u'v+uv'$$ . Rearranging and integrating gives

\int u\,dv=uv-\int v\,du.

Choose $$u$$ so differentiation simplifies it, and choose $$dv$$ so integration is available.

Example — polynomial times exponential

\int xe^x\,dx.

Take $$u=x$$ , $dv=e^x\,dx$ . Then $$du=dx$$ , $$v=e^x$$ , so

\int xe^x\,dx=xe^x-\int e^x\,dx=e^x(x-1)+C.

Differentiate the final answer to check it.

Integration by parts is the product rule read backwards. Differentiate a product, rearrange the two derivative terms, and integrate. The useful choice is the one that leaves an integral simpler than the one you started with; the formula itself does not decide that choice.

\int u\,dv=uv-\int v\,du

Choose u so differentiating it simplifies it.
Choose dv so integrating it is direct.
Write du and v explicitly before substituting into the formula.

13.2 Area between curves

If $f(x)\ge g(x)$ on $$[a,b]$$ , area between the curves is

\int_a^b[f(x)-g(x)]\,dx.

If the curves cross, first solve $$f(x)=g(x)$$ and split the interval at each intersection. Otherwise positive and negative pieces cancel and yield signed, not physical, area.

Why this works. A thin vertical strip has height equal to top curve minus bottom curve. Adding strips across the interval gives area only after the intersection points establish where that ordering is true.

A=\int_a^b(\text{top}-\text{bottom})\,dx

The region between y equals x and y equals x squared on zero to one.

13.3 Arc length and volumes

For a differentiable curve $$y=f(x)$$ , approximate a small arc by a straight segment:

ds=\sqrt{dx^2+dy^2}=\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx.

Thus

L=\int_a^b\sqrt{1+[f'(x)]^2}\,dx.

When a region is rotated about an axis, thin disks or washers produce volume. For rotation about the $$x$$ -axis with outer radius $$R(x)$$ and inner radius $$r(x)$$ ,

V=\pi\int_a^b\left[R(x)^2-r(x)^2\right]\,dx.

Draw the region and a representative slice before choosing bounds.

Length begins with local straightness. Over a tiny horizontal change, a smooth curve is nearly its tangent. The Pythagorean length of that tiny segment combines horizontal and vertical changes; adding the segments and taking the limiting refinement produces the arc-length integral.

A smooth curve approximated by short tangent-like straight segments.

13.4 Improper integrals

An integral over an infinite interval or with an unbounded integrand is defined by a limit. For example,

\int_1^\infty\frac1{x^p}\,dx =\lim_{b\to\infty}\int_1^b\frac1{x^p}\,dx.

It converges precisely when $$p>1$$ . The infinity symbol is a limiting process, not a permitted endpoint at which an antiderivative is directly evaluated.

Circle formulas as an integration check

The upper unit semicircle is $y=\sqrt{1-x^2}$ for $-1\le x\le1$ . Its derivative is $y'=-x/\sqrt{1-x^2}$ , so the arc-length integrand becomes

\sqrt{1+(y')^2}=\frac1{\sqrt{1-x^2}}.

The endpoints are unbounded, so the semicircle length is an improper integral, not an ordinary substitution at $x=\pm1$ :

\pi=\int_{-1}^{1}\frac{dx}{\sqrt{1-x^2}} =2\int_0^1\frac{dx}{\sqrt{1-x^2}}.

Scaling every length by a radius $$R$$ gives circumference $2\pi R$ . Likewise, the unit-circle area is four equal quarter-circle areas,

\pi=4\int_0^1\sqrt{1-x^2}\,dx,

so a radius- $$R$$ circle has area $\pi R^2$ . These formulas are familiar, but their derivation is a useful lesson in distinguishing a convergent endpoint singularity from an invalid direct evaluation.

Improper integrals are evaluated strictly as limits. Infinite bounds or unbounded integrands cannot be evaluated by direct substitution — a limit must be taken.

Improper integral of 1/x^2 on [1, infinity)

Starting point. Evaluate $\int_1^\infty \frac{1}{x^2} dx$ .

\[\int_1^\infty \frac{dx}{x^2} = \lim_{b\to\infty} \int_1^b x^{-2} dx\]
Replace the infinite upper bound with a finite parameter $$b$$ and take a limit.reversible
\[\int_1^b x^{-2} dx = \left[ -\frac{1}{x} \right]_1^b = -\frac{1}{b} - (-1) = 1 - \frac{1}{b}\]
Evaluate the definite integral using the Fundamental Theorem.reversible
\[\lim_{b\to\infty} \left(1 - \frac{1}{b}\right) = 1 - 0 = 1\]
Take the limit as $b \to \infty$ .reversible

Conclusion. $\int_1^\infty \frac{dx}{x^2} = 1$ (converges). In contrast, $\int_1^\infty \frac{dx}{x} = \lim_{b\to\infty} \ln b = \infty$ (diverges).

Identify the source of impropriety: infinite bound (Type I) or vertical asymptote / discontinuity (Type II).
Replace each improper point with a limit variable (e.g. $\lim_{b\to\infty} \int_a^b$ or $\lim_{c\to 0^+} \int_c^b$ ).
Evaluate the definite integral in terms of the parameter.
Take the limit: if the limit is a finite real number, the integral converges; if the limit is $\pm\infty$ or does not exist, it diverges.

13.5 Choosing integration by parts deliberately

Integration by parts transfers a derivative from one factor to the other. It is most useful when that transfer makes the remaining integral simpler. A practical preference order for choosing $$u$$ is often: logarithmic functions, inverse trigonometric functions, algebraic polynomials, trigonometric functions, then exponentials. This is a guide, not a theorem. The decisive question is whether $\int v\,du$ is easier than the original integral.

Example — a logarithm needs an implicit second factor

Find

\int\ln x\,dx \qquad (x>0).

There is no visible product, but write $\ln x\cdot1$ . Choose

u=\ln x,\quad dv=dx, \qquad du=\frac{dx}{x},\quad v=x.

Then

\int\ln x\,dx=x\ln x-\int x\frac{dx}{x} =x\ln x-\int1\,dx =x\ln x-x+C.

The restriction $$x>0$$ comes from the real logarithm. On a negative interval one would instead work with $\ln|x|$ .

Example — repeated integration by parts

Evaluate $\int e^x\sin x\,dx$ . Let this integral be $$I$$ . First choose $u=\sin x$ , $dv=e^x\,dx$ :

I=e^x\sin x-\int e^x\cos x\,dx.

Call the remaining integral $$J$$ . Applying integration by parts to $$J$$ with $u=\cos x$ gives

J=e^x\cos x+\int e^x\sin x\,dx=e^x\cos x+I.

Substitute this into the first equation:

I=e^x\sin x-e^x\cos x-I.

Therefore $2I=e^x(\sin x-\cos x)$ and

\int e^x\sin x\,dx=\frac{e^x}{2}(\sin x-\cos x)+C.

The reappearance of the original integral is not failure; it creates a linear equation for that integral.

Strategic choice of $$u$$ and $$dv$$ . The goal of integration by parts is to make $\int v\,du$ strictly simpler than $\int u\,dv$ .

Polynomial $\times$ Exponential / Trig: set $u = \text{polynomial}$ (so $$du$$ reduces in degree) and $dv = e^{ax} dx$ or $\cos(ax) dx$ .
Logarithmic or Inverse Trig factors: set $u = \ln x$ or $u = \arctan x$ (so $$du$$ becomes an algebraic fraction) and $$dv = dx$$ or $$x^n dx$$ .
Repeated parts: for $$x^2 e^x$$ , apply parts twice to eliminate powers of $$x$$ .
Cyclic parts: for $e^{ax}\cos(bx)$ , applying parts twice reproduces the original integral; solve for the integral algebraically.

Why LIATE works

From

Functions higher in LIATE (L, I) simplify dramatically when differentiated (logs become rational; inverse trig becomes algebraic).
Functions lower in LIATE (T, E) do not become more complicated when integrated ( $e^{ax}$ and $\sin(ax)$ retain their form).

Differentiating $u = \ln x$ gives $du = \frac{1}{x} dx$ (simplifies).
Integrating $$dv = x^2 dx$$ gives $v = \frac{x^3}{3}$ (manageable).
The product $v\,du = \frac{x^2}{3} dx$ is a simple monomial.

Result. LIATE maximizes algebraic simplification in the resulting integral $\int v\,du$ .

13.6 From a drawing to an area or volume integral

An application integral begins with geometry, not an antiderivative. Mark all intersections, choose vertical or horizontal slices, label the slice dimensions, and only then write the integral. For area with vertical slices, the height is “top minus bottom.” For a washer about the $$x$$ -axis, the cross-sectional area is $\pi(R^2-r^2)$ , not $\pi(R-r)^2$ .

[Figure: The region between

\(y=x\)

and

\(y=x^2\)

\([0,1]\)

, with a vertical sample

slice labelled top $$x$$ , bottom $$x^2$$ , height $$x-x^2$$ . A companion washer diagram labels outer radius $$R(x)$$ and inner radius $$r(x)$$ .]

Example — area after finding the intersections

Find the area enclosed by $$y=2x$$ and $$y=x^2$$ . Solve

x^2=2x\quad\Longrightarrow\quad x(x-2)=0,

so the intersections are at $$x=0$$ and $$x=2$$ . A test value such as $$x=1$$ gives $$2x=2$$ and $$x^2=1$$ , so the line is above the parabola. Hence

A=\int_0^2(2x-x^2)\,dx =\left[x^2-\frac{x^3}{3}\right]_0^2 =4-\frac83=\frac43.

The limits came from intersections, not from an arbitrary visible portion of a sketch.

Example — a disk volume

Rotate the region under $y=\sqrt{x}$ from $$x=0$$ to $$x=4$$ about the $$x$$ -axis. At position $$x$$ , the cross-section is a disk with radius $\sqrt{x}$ , so its area is $\pi(\sqrt{x})^2=\pi x$ . Therefore

V=\pi\int_0^4x\,dx =\pi\left[\frac{x^2}{2}\right]_0^4=8\pi.

Squaring the radius is why the radical disappears. A volume must have cubic units; the factor $$dx$$ supplies the slice thickness.

Integration by parts is the product rule in reverse. It trades an intractable integral $\int u\,dv$ for a manageable integral $\int v\,du$ .

Derivation of integration by parts

Starting point. By the product rule for differentiation: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$ .

\[u(x)v(x) = \int u'(x)v(x) dx + \int u(x)v'(x) dx\]
Integrate both sides with respect to $$x$$ .reversible
\[u v = \int v\,du + \int u\,dv\]
Rewrite using differential notation $$du = u'(x)dx$$ and $$dv = v'(x)dx$$ .reversible
\[\int u\,dv = u v - \int v\,du\]
Rearrange to isolate $\int u\,dv$ .reversible

Conclusion. $\int u\,dv = u v - \int v\,du$ . For definite integrals: $\int_a^b u\,dv = [u v]_a^b - \int_a^b v\,du$ .

Choose $$u$$ and $$dv$$ using the LIATE priority rule (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential).
Differentiate $$u$$ to get $$du = u' dx$$ ; integrate $$dv$$ to get $v = \int dv$ (choose $$C = 0$$ ).
Apply formula: $\int u\,dv = u v - \int v\,du$ .
Check that the new integral $\int v\,du$ is simpler than the original.

13.7 Arc length: local straightness, global curvature

Over a very short horizontal displacement $$dx$$ , a smooth curve changes vertically by approximately $$dy=f'(x)dx$$ . Pythagoras gives the short length

ds\approx\sqrt{(dx)^2+(dy)^2} =\sqrt{1+[f'(x)]^2}\,dx.

Adding these short lengths and taking a limit gives the arc-length formula from Section 13.3. The formula is not “height plus width”: a curved path is the hypotenuse of many infinitesimal right triangles.

Example — arc length of a straight line verifies the formula

For $$y=3x$$ on $$[0,2]$$ , $$f'(x)=3$$ . Thus

L=\int_0^2\sqrt{1+9}\,dx=2\sqrt{10}.

This agrees with distance between endpoints $$(0,0)$$ and $$(2,6)$$ : $\sqrt{2^2+6^2}=2\sqrt{10}$ . A general formula earns trust when it returns a known geometric result in a simple case.

Arc length sums infinitesimal Pythagorean hypotenuses. A smooth curve is locally straight; the differential element of arc length is $ds = \sqrt{dx^2 + dy^2}$ .

Arc length formula derivation

Starting point. Consider a curve segment with endpoints $$(x, y)$$ and $$(x+dx, y+dy)$$ .

\[(ds)^2 = (dx)^2 + (dy)^2\]
Apply the Pythagorean theorem to an infinitesimal step.reversible
\[ds = \sqrt{(dx)^2 + (dy)^2} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx\]
Factor out $dx = \sqrt{(dx)^2}$ from under the radical.reversible
\[L = \int_a^b ds = \int_a^b \sqrt{1 + [f'(x)]^2}\,dx\]
Integrate all infinitesimal arc length elements from $$x = a$$ to $$x = b$$ .reversible

Conclusion. The arc length of $$y = f(x)$$ on $$[a, b]$$ is $L = \int_a^b \sqrt{1 + [f'(x)]^2}\,dx$ . Because $\sqrt{1 + (f')^2} \ge 1$ , $L \ge b-a$ (the straight-line minimum).

13.8 Improper integrals and convergence checks

When an endpoint is infinite, replace it by a finite variable before integrating. For $p\ne1$ ,

\int_1^b x^{-p}\,dx=\frac{b^{1-p}-1}{1-p}.

If $$p>1$$ , then $$1-p<0$$ and $b^{1-p}\to0$ , so the limiting integral is $$1/(p-1)$$ . If $$p<1$$ , the power grows without bound. At $$p=1$$ the integral is $\ln b$ , which also grows without bound. This establishes the threshold $$p>1$$ rather than merely memorising it.

Example — a finite integral with an infinite endpoint value

Consider

\int_0^1\frac{dx}{\sqrt{x}}.

The integrand is unbounded at zero, so write

\lim_{a\to0^+}\int_a^1x^{-1/2}\,dx =\lim_{a\to0^+}\left[2\sqrt{x}\right]_a^1 =2.

An unbounded height does not by itself imply infinite accumulated area. The limit decides; directly inserting $$x=0$$ into $1/\sqrt{x}$ would not.

Area between curves requires finding intersections first. Area is always non-negative: $\int (\text{upper} - \text{lower}) dx$ .

Area between y = x and y = x^2

Starting point. Find the area bounded by $$y_1 = x$$ and $$y_2 = x^2$$ .

\[x = x^2 \implies x(x-1) = 0 \implies x = 0, \quad x = 1\]
Find intersection points to determine integration limits $$[0, 1]$$ .reversible
\[\text{On } [0, 1]: \quad x \ge x^2 \implies \text{upper} = x, \quad \text{lower} = x^2\]
Test a point in $$(0, 1)$$ (e.g. $$x = 0.5$$ : $$0.5 > 0.25$$ ) to confirm curve ordering.reversible
\[A = \int_0^1 (x - x^2) dx = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}\]
Integrate $\text{upper} - \text{lower}$ over $$[0, 1]$$ .reversible

Conclusion. The bounded area is exactly $\frac{1}{6}$ . If curves cross multiple times, split the integral at each intersection so the integrand is always $$|y_1 - y_2|$$ .

Set $$f(x) = g(x)$$ and solve for all intersection points $x = a, b, c, \dots$
Determine which function is on top on each subinterval $$[a, b]$$ by testing a sample point.
Set up $\int_a^b (f(x) - g(x)) dx$ where $f \ge g$ .
Evaluate and sum the areas of all enclosed regions.

Common Mistakes

Choosing $$u$$ and $$dv$$ for integration by parts without checking that the new integral is simpler.
Mixing $$u$$ -limits with an antiderivative returned to $$x$$ .
Subtracting curves in the wrong order or failing to split at crossings.
Treating an improper integral as ordinary finite-bound evaluation.

Exercises

Evaluate $\int x\cos x\,dx$ .
Find the area between $$y=x$$ and $$y=x^2$$ on $$[0,1]$$ .
Find the volume generated by rotating $$y=x$$ on $$[0,2]$$ about the $$x$$ -axis.
Determine whether $\int_1^\infty1/x^{3/2}\,dx$ converges.
Derive the circumference of a circle from a parametric or geometric argument and explain why radians are natural for arc length.
Evaluate $\int x\ln x\,dx$ by integration by parts.
Show all steps in evaluating $\int e^x\cos x\,dx$ .
Find the area enclosed by $$y=4-x^2$$ and the $$x$$ -axis.
Find the volume obtained by rotating the region under $$y=2x$$ on $$[0,1]$$ about the $$x$$ -axis.
Determine whether $\int_0^1x^{-2/3}\,dx$ converges and evaluate it if it does.