6.5 Definite, Improper, Double & Triple Integrals

The Definite Integral

The fundamental theorem of calculus converts integration into antidifferentiation:

$\int_{a}^{b} f(x)\,dx = F(b) - F(a), \qquad \text{where } F'(x) = f(x)$

The properties that shorten exam work:

1. $\displaystyle\int_a^b f(x),dx = -\int_b^a f(x),dx$
2. $\displaystyle\int_a^b f(x),dx = \int_a^c f(x),dx + \int_c^b f(x),dx$
3. $\displaystyle\int_0^a f(x),dx = \int_0^a f(a-x),dx$ — the king property
4. Symmetry: over $[-a, a]$,

$\int_{-a}^{a} f(x)\,dx = \begin{cases} 0 & f \text{ odd} \\ 2\displaystyle\int_0^a f(x)\,dx & f \text{ even}\end{cases}$

Property 4 alone answers many one-mark items outright. If you are asked for $\int_{-2}^{2} x^3 \cos x,dx$, note that $x^3$ is odd and $\cos x$ is even, so the product is odd and the integral is zero — no antiderivative needed.

Standard Integrals to Memorise

Integration by parts follows the ILATE ordering (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential) for choosing the first function:

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

Improper Integrals

An integral is improper when a limit is infinite or the integrand becomes unbounded inside the interval. It is defined as a limit:

$\int_{a}^{\infty} f(x)\,dx = \lim_{t \to \infty}\int_{a}^{t} f(x)\,dx$

If the limit is finite the integral converges; otherwise it diverges. The benchmark result worth memorising is the $p$-integral:

$\int_{1}^{\infty} \frac{dx}{x^{p}} \quad \text{converges when } p > 1, \quad \text{diverges when } p \leq 1$

So $\int_1^\infty dx/x^2 = 1$ converges, while $\int_1^\infty dx/x$ diverges — a distinction that appears regularly as a single objective item.

Worked example. $\displaystyle\int_0^\infty e^{-2x}dx = \lim_{t\to\infty}\left[-\tfrac{1}{2}e^{-2x}\right]_0^t = 0 + \tfrac{1}{2} = \tfrac{1}{2}$. In general $\int_0^\infty e^{-ax}dx = 1/a$ for $a > 0$, a result that reappears directly in the Laplace transform of a constant.

Double Integrals

A double integral accumulates a quantity over a two-dimensional region $R$:

$\iint_R f(x,y)\,dA = \int_{y_1}^{y_2}\!\!\int_{x_1(y)}^{x_2(y)} f(x,y)\,dx\,dy$

The non-negotiable rule of structure: the inner limits may involve the outer variable, but the outer limits must be pure constants. If your outer limits still contain a variable, the region has been set up incorrectly.

The order of integration is chosen for convenience. Sketch the region, decide whether horizontal or vertical strips describe it more simply, and integrate accordingly. When the inner integrand has no elementary antiderivative in one order, reversing the order is usually the intended trick.

Engineering uses

That last row is precisely the moment of inertia computed in Applied Mechanics; the multiple-integral topic in Section 1 is the machinery behind the centroid and inertia results in Section 2.

Worked example. Evaluate $\displaystyle\int_0^1!!\int_0^2 (x + y),dy,dx$.

Inner integral with respect to $y$: $\left[xy + \tfrac{y^2}{2}\right]_0^2 = 2x + 2$.

Outer integral with respect to $x$: $\left[x^2 + 2x\right]_0^1 = 1 + 2 = 3$.

Triple Integrals

Extending to three dimensions gives volume and mass properties of solids:

$\iiint_V f(x,y,z)\,dV$

With $f = 1$ the result is the volume; with $f = \rho$ it is the mass; with $f = \rho(y^2 + z^2)$ it is the mass moment of inertia about the $x$-axis — the quantity that governs flywheel sizing later in Paper-II.

For bodies with rotational symmetry, cylindrical coordinates are far quicker. The volume element becomes

$dV = r\,dr\,d\theta\,dz$

and the factor $r$ in the Jacobian is the single most-forgotten term in the whole topic. In spherical coordinates the element is $dV = \rho^2\sin\phi,d\rho,d\phi,d\theta$.

Worked example. The volume of a cylinder of radius $a$ and height $h$:

$V = \int_0^{h}\!\!\int_0^{2\pi}\!\!\int_0^{a} r\,dr\,d\theta\,dz = h \cdot 2\pi \cdot \frac{a^2}{2} = \pi a^2 h$

as expected — a useful sanity check that your Jacobian is right.

Changing the Order of Integration

When the inner integral has no elementary antiderivative in the given order, reversing the order is usually the intended route. The procedure is always the same: sketch the region, then re-describe it with the strips running the other way.

Worked example. Evaluate $\displaystyle\int_0^1!!\int_x^1 e^{y^2},dy,dx$.

The inner integral $\int e^{y^2}dy$ has no elementary form, so the given order is hopeless. The region is bounded by $y = x$, $y = 1$ and $x = 0$ — a triangle with vertices $(0,0)$, $(0,1)$ and $(1,1)$. Re-describing it with $y$ outer and $x$ inner, $y$ runs 0 to 1 and $x$ runs 0 to $y$:

$\int_0^1\!\!\int_0^{y} e^{y^2}\,dx\,dy = \int_0^1 y\,e^{y^2}\,dy = \left[\tfrac{1}{2}e^{y^2}\right]_0^1 = \frac{e-1}{2}$

The substitution $u = y^2$ makes the final integral immediate. Note that the limits change completely — simply swapping the differentials while keeping the old limits is the standard error.

Applications to Mechanical Engineering

Multiple integration is not an isolated topic; it is the machinery behind results used throughout Paper-II.

The last row is worth noting explicitly: the mean effective pressure of an engine cycle, computed in the IC engine section, is exactly the mean value of pressure over the swept volume — that is, the area of the indicator diagram divided by the swept volume. Recognising it as an averaging integral removes any need to memorise it separately.

Worked example. Find the second moment of area of a rectangle of width $b$ and depth $d$ about its centroidal axis.

$I_{xx} = \int_{-d/2}^{d/2}\!\!\int_{-b/2}^{b/2} y^2\,dx\,dy = b\int_{-d/2}^{d/2} y^2\,dy = b\left[\frac{y^3}{3}\right]_{-d/2}^{d/2} = \frac{bd^3}{12}$

which is the standard result quoted throughout the bending and deflection sections, here derived rather than recalled.