The definite integral

The first four lessons were about rates: given a quantity, how fast is it changing? This lesson turns the question around. Given a rate, how much has accumulated? If you know a pump's flow rate over time, how much water has it moved? If you know a vehicle's velocity, how far has it gone? If you know the power a machine draws, how much energy has it used?

The tool for these questions is the definite integral. You will see how it arises from adding up many small pieces, how to compute it exactly with antiderivatives using the Fundamental Theorem of Calculus, and how to apply it to distance, energy, volume and averages. The theorem is the bridge between the two halves of calculus, and it is one of the great ideas of mathematics.

The accumulation problem

If a rate is constant, accumulation is just multiplication. A pump delivering a steady 2 L/s for 30 s moves \(2 \times 30 = 60\) L. On a graph of rate against time, that product is the area of a rectangle under the horizontal line at height 2.

Real rates change. A pump starts up, a vehicle accelerates, a heater cycles on and off. The idea is to split the time into many short intervals, treat the rate as nearly constant on each one, and add up the small products. On the graph, each product is a thin rectangle, and the total approximates the area under the rate curve.

So two questions turn out to be the same question: "how much has accumulated?" and "what is the area under the graph of the rate?" The definite integral answers both.

Riemann sums

To estimate the area under \(y = f(x)\) from \(x = a\) to \(x = b\):

• Split \([a, b]\) into \(n\) equal subintervals of width \(\Delta x = \dfrac{b - a}{n}\).
• In each subinterval, choose a sample point \(x_i^*\): the left end, the right end, or the midpoint.
• Build a rectangle of height \(f(x_i^*)\) and width \(\Delta x\), and add up the areas.

This total is called a Riemann sum. The Greek letter \(\Sigma\) (sigma) means "add up": the expression is read "the sum, for \(i\) from 1 to \(n\), of \(f(x_i^*)\Delta x\)."

For an increasing function, left sums underestimate and right sums overestimate. The midpoint rule is usually much better than either for the same amount of work. With more rectangles, every version gets closer to the true value.

The definite integral

The definite integral is the limit of Riemann sums as the rectangles become infinitely thin:

The notation mirrors the sum. The integral sign \(\int\) is a stretched S for "sum", \(f(x)\) is the height, and \(dx\) is the infinitely thin width. The numbers \(a\) and \(b\) are the lower and upper limits of integration. For any continuous function the limit exists and does not depend on which sample points you choose.

Signed area

When \(f(x)\) is negative, \(f(x_i^*)\Delta x\) is negative, so regions below the x-axis count as negative area. The definite integral gives the net signed area: area above the axis minus area below. This is exactly what you want physically. A negative velocity means moving backward, and integrating velocity should subtract that backward motion.

Properties of the definite integral

• Constants factor out and integrals of sums split: \(\int_a^b \big[cf(x) + g(x)\big]dx = c\int_a^b f(x)\,dx + \int_a^b g(x)\,dx\).
• Intervals join: \(\int_a^b f\,dx + \int_b^c f\,dx = \int_a^c f\,dx\).
• Reversing the limits changes the sign: \(\int_b^a f\,dx = -\int_a^b f\,dx\), and \(\int_a^a f\,dx = 0\).
• Comparison: if \(f(x) \le g(x)\) on \([a, b]\), then \(\int_a^b f\,dx \le \int_a^b g\,dx\).

Antiderivatives

Computing limits of sums every time would be impractical. The shortcut runs through antiderivatives. A function \(F\) is an antiderivative of \(f\) if \(F'(x) = f(x)\). For example, \(x^3\) is an antiderivative of \(3x^2\), and so are \(x^3 + 5\) and \(x^3 - 100\): adding a constant changes nothing in the derivative.

The family of all antiderivatives is the indefinite integral, written with a constant \(C\):

Every derivative rule from Lesson 3, read backward, gives an antiderivative:

• \(\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C\) for \(n \ne -1\), and \(\int \dfrac{1}{x}\,dx = \ln|x| + C\).
• \(\int e^{x}\,dx = e^{x} + C\), and \(\int e^{kx}\,dx = \dfrac{1}{k}e^{kx} + C\).
• \(\int \cos x\,dx = \sin x + C\) and \(\int \sin x\,dx = -\cos x + C\).
• \(\int \sec^2 x\,dx = \tan x + C\) and \(\int \dfrac{1}{1 + x^2}\,dx = \arctan x + C\).

Constants factor out and sums split, just as with derivatives. Products and compositions are harder; Calculus II is largely about techniques for them.

Initial value problems

If you know a rate and one value of the quantity, the constant \(C\) is no longer free. This is how motion is predicted from acceleration.

The Fundamental Theorem of Calculus

The Fundamental Theorem connects the area problem (integrals) to the rate problem (derivatives). It has two parts.

Part 1: accumulation functions

Define a function that measures the accumulated area from a fixed start \(a\) up to a variable end \(x\):

Part 1 says that if \(f\) is continuous, then \(A'(x) = f(x)\). The rate at which area accumulates equals the height of the curve at the moving edge. That is intuitive: when you push the right edge a little further, the area grows by a thin strip whose height is \(f(x)\).

Combined with the chain rule, it handles variable upper limits: \(\dfrac{d}{dx}\int_0^{x^2}\cos t\,dt = \cos(x^2)\cdot 2x\).

Part 2: evaluating integrals

Part 2 is the practical one. If \(F\) is any antiderivative of a continuous function \(f\), then

To find an area or an accumulated total, find any antiderivative, evaluate it at the two limits, and subtract. The constant \(C\) cancels, so you can always leave it out of definite integrals.

Look back at the example from the definition: the limit of sums gave \(\int_0^2 x^2\,dx = 8/3\). With the theorem it is just \(\big[\tfrac13x^3\big]_0^2 = \tfrac83\). The same answer, in one line.

In words, the theorem says that integrating a rate of change gives the total change: \(\int_a^b F'(x)\,dx = F(b) - F(a)\). Integrate a flow rate to get a volume, a velocity to get a displacement, a power to get an energy, a current to get a charge.

Substitution

Substitution reverses the chain rule, just as antiderivatives reverse the basic derivative rules. If the integrand contains a function \(u = g(x)\) together with its derivative \(g'(x)\), replace them by \(u\) and \(du = g'(x)\,dx\).

For a definite integral, change the limits to \(u\)-values as you substitute. Then you never need to substitute back.

Calculus II develops substitution further, along with integration by parts and other techniques.

Applications

Net change versus total distance

Integrating velocity gives displacement: where the object ends up relative to where it started. To find the total distance traveled, integrate the speed \(|v(t)|\). In practice, split the interval where \(v\) changes sign and add the sizes of the pieces.

Energy from power, volume from flow

Average value

The average value of a function on \([a, b]\) is the height of the rectangle with the same area:

Average values matter in design. Electrical equipment is sized for average (RMS) current, HVAC systems for average and peak loads, and fatigue calculations for average and alternating stresses.

Work and fluid force

Two more applications show the same idea at work: slice the problem into thin pieces, write the small contribution of one piece, and integrate. Both appear constantly in mechanical and civil engineering.

Work done by a varying force

If a constant force \(F\) moves an object a distance \(d\), the work done is \(W = Fd\). If the force changes with position, each short step \(dx\) needs work \(F(x)\,dx\), and the total is

For a spring, Hooke's law gives \(F = kx\), so stretching from 0 to \(x\) takes \(W = \int_0^x kx\,dx = \tfrac12 kx^2\). That is the elastic energy stored in the spring, the same formula you will meet in mechanics.

Notice that each slice needs a different lift. That is why the problem needs an integral rather than a single multiplication: the water at the bottom has further to go than the water at the top.

Force of water on a gate

Water pressure increases with depth: at depth \(y\) it is \(p = \rho g y\). A vertical gate or wall therefore feels more force near the bottom than near the top. Slice the gate into horizontal strips of height \(dy\): a strip of width \(w\) at depth \(y\) has area \(w\,dy\) and feels a force of \(\rho g y\,w\,dy\).

The same method gives the force on dam faces, tank walls, ship hulls and retaining walls holding back soil, which you will meet in Fluid Mechanics and Geotechnical courses.

Numerical integration with data

Engineers often have measurements, not formulas. With values at equal spacing \(h\), the trapezoid rule joins neighboring points with straight lines and adds the trapezoid areas:

The midpoint rule, the trapezoid rule and Simpson's rule (which fits parabolas instead of lines) are the basis of the numerical integration built into spreadsheets, data loggers and simulation software. ENCM-100 shows how to code the trapezoid rule in a few lines of Python.