Properties of the Definite Integral

Reading an integral as an area, evaluating three of them from a rectangle, a trapezoid and a quarter circle, and the properties that take an integral apart.

The Riemann sums lesson evaluated one integral straight from the definition. It took a page of algebra for a single parabola. This lesson collects the properties that evaluate an integral with no sum at all.
Start with what the integral means as a picture. When ff is continuous and never negative on [a,b][a, b], every rectangle in every Riemann sum stands on the axis and rises to the curve. The limit of those sums is then the area of the region between the graph and the axis.
An integral whose region is a rectangle, a triangle, a trapezoid or a piece of a circle therefore needs no sum. The area formula gives it.
Where ff dips below the axis the rectangles have negative height, and the integral counts that piece of the region negatively. Every integrand in this lesson stays at or above the axis, so area and integral agree throughout.

Three regions

Sketch each region and evaluate the integral from an area formula.
∫253 dx,∫04(x+1) dx,∫039−x2 dx.\int_{2}^{5} 3\,dx, \qquad \int_{0}^{4} (x + 1)\,dx, \qquad \int_{0}^{3} \sqrt{9 - x^2}\,dx.

A rectangle

A constant integrand: the region from x = 2 to x = 5 is a rectangle.
The integrand is constant, so the region is a rectangle of height 33 standing on a base of width 5−2=35 - 2 = 3.
∫253 dx=3(3)=9.\int_{2}^{5} 3\,dx = 3(3) = 9.

A trapezoid

A linear integrand: the region from x = 0 to x = 4 is a trapezoid lying on its side.
The graph is a line, so the region is a trapezoid resting on its side. Its two parallel sides are the vertical segments at x=0x = 0 and x=4x = 4, of lengths f(0)=1f(0) = 1 and f(4)=5f(4) = 5, and the distance between them is 44.
∫04(x+1) dx=12(4)(1+5)=12.\int_{0}^{4} (x + 1)\,dx = \tfrac{1}{2}(4)(1 + 5) = 12.

A quarter circle

A circular integrand: the region from x = 0 to x = 3 is a quarter of a disc of radius 3.
The graph of ff is the upper half of the circle of radius 33 centred at the origin. Keeping only 0≤x≤30 \le x \le 3 leaves the quarter of that disc lying in the first quadrant.
∫039−x2 dx=14π(3)2=9π4.\int_{0}^{3} \sqrt{9 - x^2}\,dx = \tfrac{1}{4}\pi(3)^2 = \frac{9\pi}{4}.

The variable is a dummy

Nothing in the definition depends on the letter used for the variable of integration. The subintervals, the sample points and the limit are the same whatever it is called, so these two integrals are one number.
∫04(x+1) dx=∫04(t+1) dt=12.\int_{0}^{4} (x + 1)\,dx = \int_{0}^{4} (t + 1)\,dt = 12.
A variable that can be renamed without changing the value is a dummy variable. The letter is a matter of readability. One exception arrives later: when a limit of integration is itself a variable, it and the variable of integration have to be kept apart.

Two special integrals

Two cases fall straight out of the definition. Both are worth recording before they are needed.
Both are read off directly. The first integral below has equal limits. The second is the trapezoid from earlier with its limits interchanged.
∫44cos⁡x dx=0,∫40(x+1) dx=−12.\int_{4}^{4} \cos x\,dx = 0, \qquad \int_{4}^{0} (x + 1)\,dx = -12.

Splitting the interval

A region can be cut into pieces and the areas added. Integrals behave the same way, and the cut point is not required to lie between the two limits.
An integrand defined in pieces is the usual reason to split. The graph of f(x)=∣x∣f(x) = |x| is two straight lines meeting at the origin, so the cut belongs at x=0x = 0.
The region under f(x) = |x| from x = −1 to x = 1 is two triangles meeting at the origin.
On [−1,0][-1, 0] the graph is the line y=−xy = -x, and on [0,1][0, 1] it is the line y=xy = x. Each piece bounds a triangle of base 11 and height 11.
∫−11∣x∣ dx=∫−10(−x) dx+∫01x dx=12+12=1.\int_{-1}^{1} |x|\,dx = \int_{-1}^{0} (-x)\,dx + \int_{0}^{1} x\,dx = \tfrac{1}{2} + \tfrac{1}{2} = 1.

Constant multiples and sums

Two more properties take an integral apart along the algebra of its integrand.
A constant factor passes through the integral sign, and a sum or difference splits term by term. Between them they reduce a polynomial integrand to the integrals of its separate powers.
Evaluate the integral of 2x2−3x+12x^2 - 3x + 1 from 11 to 33, given these three values on the same interval:
∫13x2 dx=263,∫13x dx=4,∫13dx=2.\int_{1}^{3} x^2\,dx = \frac{26}{3}, \qquad \int_{1}^{3} x\,dx = 4, \qquad \int_{1}^{3} dx = 2.
Split the integrand term by term, then pull each coefficient out in front:
∫13(2x2−3x+1) dx=2∫13x2 dx−3∫13x dx+∫13dx=2(263)−3(4)+2=523−10=223.\begin{aligned} \int_{1}^{3} (2x^2 - 3x + 1)\,dx &= 2\int_{1}^{3} x^2\,dx - 3\int_{1}^{3} x\,dx + \int_{1}^{3} dx \\[8pt] &= 2\left(\frac{26}{3}\right) - 3(4) + 2 \\[8pt] &= \frac{52}{3} - 10 = \frac{22}{3}. \end{aligned}

An integrand with a jump

Integrability does not require continuity. A function bounded on [a,b][a, b] whose only discontinuities are finitely many jumps is integrable there. Splitting the interval at those jumps evaluates it.
Evaluate the integral of ff from 00 to 55, where
f(x)={−x+6,x≤2,3,x>2.f(x) = \begin{cases} -x + 6, & x \le 2, \\[4pt] 3, & x > 2. \end{cases}
A falling line on [0, 2] and a horizontal line on [2, 5], with a jump at x = 2.
The graph falls along a line until x=2x = 2, drops from 44 to 33, and runs level from there to x=5x = 5. The function is bounded and that jump is its only discontinuity, so it is integrable on [0,5][0, 5], and the interval splits at x=2x = 2.
∫05f(x) dx=∫02f(x) dx+∫25f(x) dx=12(2)(6+4)⏟trapezoid  +  3(3)⏟rectangle=10+9=19.\begin{aligned} \int_{0}^{5} f(x)\,dx &= \int_{0}^{2} f(x)\,dx + \int_{2}^{5} f(x)\,dx \\[8pt] &= \underbrace{\tfrac{1}{2}(2)(6 + 4)}_{\text{trapezoid}} \; + \; \underbrace{3(3)}_{\text{rectangle}} \\[8pt] &= 10 + 9 = 19. \end{aligned}
The value of ff at the single point x=2x = 2 contributes nothing to either area. It makes no difference which branch of the definition is given the endpoint.