The integral as area
If
f f f is continuous and
f ( x ) ≥ 0 f(x) \ge 0 f ( x ) ≥ 0 on
[ a , b ] [a, b] [ a , b ] , then
∫ a b f ( x ) d x \int_{a}^{b} f(x)\,dx ∫ a b f ( x ) d x is the area of the region bounded by the graph of
f f f , the
x x x -axis, and the vertical lines
x = a x = a x = a and
x = b x = b x = b .
Three regions In each example below, sketch the region and evaluate the integral from an area formula.
Example 1
Evaluate the integral below.
∫ 2 5 3 d x \int_{2}^{5} 3\,dx ∫ 2 5 3 d x
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
3 3 3 standing on a base of width
5 − 2 = 3 5 - 2 = 3 5 − 2 = 3 .
∫ 2 5 3 d x = 3 ( 3 ) = 9. \int_{2}^{5} 3\,dx = 3(3) = 9. ∫ 2 5 3 d x = 3 ( 3 ) = 9. Example 2
Evaluate the integral below.
∫ 0 4 ( x + 1 ) d x \int_{0}^{4} (x + 1)\,dx ∫ 0 4 ( x + 1 ) d x
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 = 0 x = 0 x = 0 and
x = 4 x = 4 x = 4 , of lengths
f ( 0 ) = 1 f(0) = 1 f ( 0 ) = 1 and
f ( 4 ) = 5 f(4) = 5 f ( 4 ) = 5 , and the distance between them is
4 4 4 .
∫ 0 4 ( x + 1 ) d x = 1 2 ( 4 ) ( 1 + 5 ) = 12. \int_{0}^{4} (x + 1)\,dx = \tfrac{1}{2}(4)(1 + 5) = 12. ∫ 0 4 ( x + 1 ) d x = 2 1 ( 4 ) ( 1 + 5 ) = 12. Example 3
Evaluate the integral below.
∫ 0 3 9 − x 2 d x \int_{0}^{3} \sqrt{9 - x^2}\,dx ∫ 0 3 9 − x 2 d x
A circular integrand: the region from x = 0 to x = 3 is a quarter of a disk of radius 3. The graph of
f f f is the upper half of the circle of radius
3 3 3 centered at the origin. Keeping only
0 ≤ x ≤ 3 0 \le x \le 3 0 ≤ x ≤ 3 leaves the quarter of that disk lying in the first quadrant.
∫ 0 3 9 − x 2 d x = 1 4 π ( 3 ) 2 = 9 π 4 . \int_{0}^{3} \sqrt{9 - x^2}\,dx = \tfrac{1}{4}\pi(3)^2 = \frac{9\pi}{4}. ∫ 0 3 9 − x 2 d x = 4 1 π ( 3 ) 2 = 4 9 π . Regions below the axis Where the graph goes below the axis, the heights
f ( c i ) f(c_i) f ( c i ) are negative, and so is that part of every Riemann sum. The limit keeps the sign.
Signed area
Let
f f f be continuous on
[ a , b ] [a, b] [ a , b ] . The integral of
f f f from
a a a to
b b b is the area of the part of the region above the
x x x -axis minus the area of the part below it:
∫ a b f ( x ) d x = ( area above the axis ) − ( area below the axis ) . \int_{a}^{b} f(x)\,dx = (\text{area above the axis}) - (\text{area below the axis}). ∫ a b f ( x ) d x = ( area above the axis ) − ( area below the axis ) . The total area between the graph and the axis counts both parts as positive. It equals the integral only when no part of the region lies below the axis.
Example 4
The graph of
f f f on
[ 0 , 10 ] [0, 10] [ 0 , 10 ] is a semicircle of radius
2 2 2 above the axis on
[ 0 , 4 ] [0, 4] [ 0 , 4 ] . After it come two line segments that form a triangle below the axis on
[ 4 , 10 ] [4, 10] [ 4 , 10 ] , with its lowest point at
( 7 , − 3 ) (7, -3) ( 7 , − 3 ) . Evaluate the integral of
f f f from
0 0 0 to
10 10 10 , and find the total area between the graph and the axis.
Above the axis, a half-disk of radius 2; below it, a triangle with base 6 and height 3. The half-disk has area
1 2 π ( 2 ) 2 = 2 π \frac{1}{2}\pi(2)^2 = 2\pi 2 1 π ( 2 ) 2 = 2 π . The triangle has base
6 6 6 and height
3 3 3 , so its area is
9 9 9 . It lies below the axis, so the integral subtracts it:
∫ 0 10 f ( x ) d x = 2 π − 9 ≈ − 2.72. \int_{0}^{10} f(x)\,dx = 2\pi - 9 \approx -2.72. ∫ 0 10 f ( x ) d x = 2 π − 9 ≈ − 2.72. The integral is negative although part of the region lies above the axis, because the part below is larger. The total area adds the two parts:
total area = 2 π + 9 ≈ 15.28. \text{total area} = 2\pi + 9 \approx 15.28. total area = 2 π + 9 ≈ 15.28. The dummy variable 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.
∫ 0 4 ( x + 1 ) d x = ∫ 0 4 ( t + 1 ) d t = 12. \int_{0}^{4} (x + 1)\,dx = \int_{0}^{4} (t + 1)\,dt = 12. ∫ 0 4 ( x + 1 ) d x = ∫ 0 4 ( t + 1 ) d t = 12. A variable that can be renamed without changing the value is a dummy variable . The letter is a matter of readability. There's one exception later on: when a limit of integration is itself a variable, it and the variable of integration have to be kept apart.
Two special integrals The first case below falls straight out of the definition. The second extends the definition to limits written in the other order. Both are worth recording before they are needed.
Two special integrals
If
f f f is defined at
x = a x = a x = a , the interval has collapsed to a point and there is no width left to sum:
∫ a a f ( x ) d x = 0. \int_{a}^{a} f(x)\,dx = 0. ∫ a a f ( x ) d x = 0. If
f f f is integrable on
[ a , b ] [a, b] [ a , b ] , reversing the limits reverses the sign:
∫ b a f ( x ) d x = − ∫ a b f ( x ) d x . \int_{b}^{a} f(x)\,dx = -\int_{a}^{b} f(x)\,dx. ∫ b a f ( x ) d x = − ∫ a b f ( x ) d x . Example 5
Evaluate the two integrals below.
∫ 4 4 cos x d x , ∫ 4 0 ( x + 1 ) d x \int_{4}^{4} \cos x\,dx, \qquad \int_{4}^{0} (x + 1)\,dx ∫ 4 4 cos x d x , ∫ 4 0 ( x + 1 ) d x The first has equal limits. The second is the trapezoid of Example 2 with its limits interchanged, so it is the negative of
12 12 12 :
∫ 4 4 cos x d x = 0 , ∫ 4 0 ( x + 1 ) d x = − 12. \int_{4}^{4} \cos x\,dx = 0, \qquad \int_{4}^{0} (x + 1)\,dx = -12. ∫ 4 4 cos x d x = 0 , ∫ 4 0 ( x + 1 ) d x = − 12. Splitting the interval A region can be cut into pieces and the areas added. Integrals behave the same way, and the cut point doesn't have to lie between the two limits.
Splitting the interval
If
f f f is integrable on the three closed intervals determined by
a a a ,
b b b , and
c c c , then
∫ a b f ( x ) d x = ∫ a c f ( x ) d x + ∫ c b f ( x ) d x . \int_{a}^{b} f(x)\,dx = \int_{a}^{c} f(x)\,dx + \int_{c}^{b} f(x)\,dx. ∫ a b f ( x ) d x = ∫ a c f ( x ) d x + ∫ c b f ( x ) d x . An integrand defined in pieces is the usual reason to split.
Example 6
Evaluate the integral below.
∫ − 1 1 ∣ x ∣ d x \int_{-1}^{1} |x|\,dx ∫ − 1 1 ∣ x ∣ d x The graph of
f ( x ) = ∣ x ∣ f(x) = |x| f ( x ) = ∣ x ∣ is two straight lines meeting at the origin, so the cut belongs at
x = 0 x = 0 x = 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] [ − 1 , 0 ] the graph is the line
y = − x y = -x y = − x , and on
[ 0 , 1 ] [0, 1] [ 0 , 1 ] it is the line
y = x y = x y = x . Each piece bounds a triangle of base
1 1 1 and height
1 1 1 .
∫ − 1 1 ∣ x ∣ d x = ∫ − 1 0 ( − x ) d x + ∫ 0 1 x d x = 1 2 + 1 2 = 1. \int_{-1}^{1} |x|\,dx = \int_{-1}^{0} (-x)\,dx + \int_{0}^{1} x\,dx = \tfrac{1}{2} + \tfrac{1}{2} = 1. ∫ − 1 1 ∣ x ∣ d x = ∫ − 1 0 ( − x ) d x + ∫ 0 1 x d x = 2 1 + 2 1 = 1. The rule also runs the other way. Given the integral over a whole interval and over one piece of it, the integral over the other piece is their difference.
Example 7
Two integrals of
f f f are given below. Find the integral of
f f f from
0 0 0 to
5 5 5 .
∫ 0 8 f ( x ) d x = 10 , ∫ 8 5 f ( x ) d x = 3. \int_{0}^{8} f(x)\,dx = 10, \qquad \int_{8}^{5} f(x)\,dx = 3. ∫ 0 8 f ( x ) d x = 10 , ∫ 8 5 f ( x ) d x = 3. The second integral runs from right to left. Reverse it first:
∫ 5 8 f ( x ) d x = − 3. \int_{5}^{8} f(x)\,dx = -3. ∫ 5 8 f ( x ) d x = − 3. Then split
[ 0 , 8 ] [0, 8] [ 0 , 8 ] at
x = 5 x = 5 x = 5 and subtract the known piece:
∫ 0 5 f ( x ) d x = ∫ 0 8 f ( x ) d x − ∫ 5 8 f ( x ) d x = 10 − ( − 3 ) = 13. \int_{0}^{5} f(x)\,dx = \int_{0}^{8} f(x)\,dx - \int_{5}^{8} f(x)\,dx = 10 - (-3) = 13. ∫ 0 5 f ( x ) d x = ∫ 0 8 f ( x ) d x − ∫ 5 8 f ( x ) d x = 10 − ( − 3 ) = 13. A cut point outside the interval works too. Splitting at
x = 8 x = 8 x = 8 uses both given integrals as they stand:
∫ 0 5 f ( x ) d x = ∫ 0 8 f ( x ) d x + ∫ 8 5 f ( x ) d x = 10 + 3 = 13. \int_{0}^{5} f(x)\,dx = \int_{0}^{8} f(x)\,dx + \int_{8}^{5} f(x)\,dx = 10 + 3 = 13. ∫ 0 5 f ( x ) d x = ∫ 0 8 f ( x ) d x + ∫ 8 5 f ( x ) d x = 10 + 3 = 13. Subtracting the
3 3 3 as it is given gives
7 7 7 . That is the usual mistake, and it comes from skipping the reversal.
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.
Example 8
Evaluate the integral of
2 x 2 − 3 x + 1 2x^2 - 3x + 1 2 x 2 − 3 x + 1 from
1 1 1 to
3 3 3 , given these three values on the same interval:
∫ 1 3 x 2 d x = 26 3 , ∫ 1 3 x d x = 4 , ∫ 1 3 d x = 2. \int_{1}^{3} x^2\,dx = \frac{26}{3}, \qquad \int_{1}^{3} x\,dx = 4, \qquad \int_{1}^{3} dx = 2. ∫ 1 3 x 2 d x = 3 26 , ∫ 1 3 x d x = 4 , ∫ 1 3 d x = 2. Split the integrand term by term, then pull each coefficient out in front:
∫ 1 3 ( 2 x 2 − 3 x + 1 ) d x = 2 ∫ 1 3 x 2 d x − 3 ∫ 1 3 x d x + ∫ 1 3 d x = 2 ( 26 3 ) − 3 ( 4 ) + 2 = 52 3 − 10 = 22 3 . \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} ∫ 1 3 ( 2 x 2 − 3 x + 1 ) d x = 2 ∫ 1 3 x 2 d x − 3 ∫ 1 3 x d x + ∫ 1 3 d x = 2 ( 3 26 ) − 3 ( 4 ) + 2 = 3 52 − 10 = 3 22 . A constant term needs its own care. Its integral is a rectangle, so a constant
k k k integrates to
k k k times the width of the interval.
Example 9
Given the first integral below, evaluate the second.
∫ 2 9 g ( x ) d x = 4 , ∫ 9 2 [ 3 g ( x ) − 2 ] d x . \int_{2}^{9} g(x)\,dx = 4, \qquad \int_{9}^{2} \bigl[3g(x) - 2\bigr]\,dx. ∫ 2 9 g ( x ) d x = 4 , ∫ 9 2 [ 3 g ( x ) − 2 ] d x . Work over
[ 2 , 9 ] [2, 9] [ 2 , 9 ] first. The constant
2 2 2 is a rectangle of height
2 2 2 and width
7 7 7 , so its integral is
14 14 14 :
∫ 2 9 [ 3 g ( x ) − 2 ] d x = 3 ( 4 ) − 2 ( 7 ) = − 2. \int_{2}^{9} \bigl[3g(x) - 2\bigr]\,dx = 3(4) - 2(7) = -2. ∫ 2 9 [ 3 g ( x ) − 2 ] d x = 3 ( 4 ) − 2 ( 7 ) = − 2. Reversing the limits changes the sign:
∫ 9 2 [ 3 g ( x ) − 2 ] d x = 2. \int_{9}^{2} \bigl[3g(x) - 2\bigr]\,dx = 2. ∫ 9 2 [ 3 g ( x ) − 2 ] d x = 2. Example 10
Given the integral below, find the integral of
h h h from
0 0 0 to
6 6 6 .
∫ 0 6 [ 2 h ( x ) + 5 ] d x = 18 \int_{0}^{6} \bigl[2h(x) + 5\bigr]\,dx = 18 ∫ 0 6 [ 2 h ( x ) + 5 ] d x = 18 Split the integrand. The constant term is a rectangle of height
5 5 5 and width
6 6 6 , with area
30 30 30 :
2 ∫ 0 6 h ( x ) d x + 30 = 18 , so ∫ 0 6 h ( x ) d x = − 6. 2\int_{0}^{6} h(x)\,dx + 30 = 18, \qquad \text{so} \qquad \int_{0}^{6} h(x)\,dx = -6. 2 ∫ 0 6 h ( x ) d x + 30 = 18 , so ∫ 0 6 h ( x ) d x = − 6. A discontinuous integrand Integrability doesn't require continuity. A function bounded on
[ a , b ] [a, b] [ a , b ] whose only discontinuities are finitely many jumps is integrable there. Splitting the interval at those jumps evaluates it.
Example 11
Evaluate the integral of
f f f from
0 0 0 to
5 5 5 , 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} f ( x ) = { − x + 6 , 3 , x ≤ 2 , x > 2.
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 = 2 x = 2 x = 2 , drops from
4 4 4 to
3 3 3 , and runs level from there to
x = 5 x = 5 x = 5 . The function is bounded and that jump is its only discontinuity, so it is integrable on
[ 0 , 5 ] [0, 5] [ 0 , 5 ] , and the interval splits at
x = 2 x = 2 x = 2 .
∫ 0 5 f ( x ) d x = ∫ 0 2 f ( x ) d x + ∫ 2 5 f ( x ) d x = 1 2 ( 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} ∫ 0 5 f ( x ) d x = ∫ 0 2 f ( x ) d x + ∫ 2 5 f ( x ) d x = trapezoid 2 1 ( 2 ) ( 6 + 4 ) + rectangle 3 ( 3 ) = 10 + 9 = 19. The value of
f f f at the single point
x = 2 x = 2 x = 2 contributes nothing to either area. It makes no difference which branch of the definition is given the endpoint.