Applications of Integration — Study Notes

Applications of Integration — Study Notes

The big idea: a definite integral is a limit of a sum

Take a continuous function on \([a,b]\), chop the interval into \(n\) thin pieces, multiply the function's height in each piece by that piece's width, and add everything up. As the pieces are made infinitely thin, this running total settles onto a single number — the definite integral \(\int_a^b f(x)\,dx\). This "limit of a Riemann sum" is the rigorous meaning behind the integral sign; the familiar area picture is simply the most common example of it.

What the integral really measures: signed area

When \(f(x)\ge 0\) on \([a,b]\), the integral equals the geometric area between the curve and the \(x\)-axis. When \(f(x)\le 0\), the integral comes out negative and equals the negative of that area. If the curve crosses the axis, the positive and negative pieces partly cancel. This is the single most common trap in the chapter: a definite integral gives signed area, not geometric area. To get true (geometric) area you must split the interval at every point where \(f\) changes sign and add the absolute values of the pieces. Classic case: \(\int_0^{2\pi}\sin x\,dx=0\), yet the area trapped between \(y=\sin x\) and the axis over \([0,2\pi]\) is \(4\).

Evaluating integrals the easy way: the Fundamental Theorem

Computing integrals straight from the limit-of-a-sum definition is tedious. The Fundamental Theorem of Calculus rescues us: find any antiderivative \(F\) (so \(F'=f\)), then \(\int_a^b f(x)\,dx=F(b)-F(a)\). No constant of integration is needed — it cancels in the subtraction. The first form of the theorem adds that differentiating an integral with a variable upper limit returns the integrand, which is exactly why differentiation and integration are inverse operations.

Properties that save time

A handful of properties turn ugly integrals into easy ones. The ones examiners lean on:

• Odd / even shortcut. Over a symmetric interval \([-a,a]\) an odd integrand integrates to \(0\) and an even one to twice the half-integral. Spotting that an integrand is odd can finish a question in a single line.
• The "king" property \(\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx\). Adding the original integral to its reflected copy often makes the awkward part cancel, leaving something trivial.
• The \([0,2a]\) split. \(\int_0^{2a}f\,dx=2\int_0^a f\,dx\) when \(f(2a-x)=f(x)\), and \(0\) when \(f(2a-x)=-f(x)\).

Train yourself to glance at the limits and the symmetry before reaching for a substitution — a large share of the book-back questions are built to collapse under one of these properties.

Special techniques

Bernoulli's formula is a shortcut for \(\int(\text{polynomial})\times(\text{easily-integrable function})\,dx\). Rather than integrating by parts over and over, you write one alternating series \(uv_1-u'v_2+u''v_3-\cdots\) that stops as soon as the polynomial's derivatives run out. It is the same idea as repeated integration by parts, packaged so the answer can be written in a single step.

Improper integrals extend integration to infinite intervals (or unbounded integrands) and are handled as limits. The star of this section is the Gamma function \(\Gamma(n)=\int_0^\infty e^{-x}x^{\,n-1}\,dx\), a continuous generalisation of the factorial: \(\Gamma(n+1)=n\,\Gamma(n)\), \(\Gamma(1)=1\), and \(\Gamma(n+1)=n!\) for whole numbers. Many fearsome-looking exponential integrals reduce instantly to a Gamma value after a substitution.

Reduction (Wallis) formulae give clean closed forms for \(\int_0^{\pi/2}\sin^n x\,dx\) and \(\int_0^{\pi/2}\cos^n x\,dx\), with the answer depending only on whether \(n\) is odd or even (the even case carries an extra factor of \(\tfrac{\pi}{2}\)). Memorise the pattern; it shows up every year.

Area problems: a reliable method

Sketch the region first — guessing is where sign errors creep in. Then pick a strip direction. Vertical strips (integrate in \(dx\)) suit regions bounded above and below by curves; horizontal strips (integrate in \(dy\)) suit regions bounded left and right. For the area between two curves, integrate (upper − lower) for vertical strips, or (right − left) for horizontal strips, between the curves' points of intersection. And split with absolute values wherever the boundary crosses the axis.

Volume of solids of revolution

Spin a region about an axis and it sweeps out a solid; slicing perpendicular to the axis produces thin discs, and summing their volumes gives the disc method. Revolving about the \(x\)-axis makes the radius \(y\), so \(V=\pi\int_a^b y^2\,dx\); revolving about the \(y\)-axis makes the radius \(x\), so \(V=\pi\int_c^d x^2\,dy\). The two dependable mistakes are (i) using the wrong variable of integration for the chosen axis, and (ii) forgetting to square the radius before integrating.

Common exam traps — quick checklist

• Signed area vs geometric area: split at sign changes for true area.
• Symmetric limits with an odd integrand \(\Rightarrow\) the answer is \(0\); don't integrate the long way.
• Match the strip and variable to the axis: \(y\,dx\) on the \(x\)-axis side, \(x\,dy\) on the \(y\)-axis side.
• Revolution: square the radius — \(y^2\,dx\) about the \(x\)-axis, \(x^2\,dy\) about the \(y\)-axis.
• Gamma: \(\Gamma(n+1)=n!\) holds only for whole numbers; keep the shift in \(n\) straight.