A definite integral combines the signed contributions of a function over a stated interval. Geometric area counts the size of a region and cannot be negative. These match when the graph is on or above the horizontal axis, but differ when a curve runs below it. Before integrating, identify the boundaries and decide exactly which quantity the question requests.
The examples here use simple polynomials and original regions. Coordinates are in generic units unless a question specifies lengths. “Square units” is appropriate for geometric coordinate area; if both axes represent centimetres, the area is in cm². An integral in another context can have different units, so the word integral does not automatically imply physical area.
Write an antiderivative, then apply the bounds
For the nonnegative integer powers used here, an antiderivative of axⁿ is axⁿ⁺¹/(n + 1). Increase the power and divide by the new power. Integrate each term separately, retaining its sign. For example, an antiderivative of 6x² − 4x + 3 is 2x³ − 2x² + 3x. Differentiating this expression returns the original polynomial.
An indefinite integral describes a family of antiderivatives, so it includes an arbitrary constant C. For a definite integral from a to b, choose an antiderivative F and calculate F(b) − F(a). If C were included at both ends it would cancel. Do not add an extra unexplained “+ C” to a completed numerical definite integral.
Keep the lower substitution in brackets. The instruction is upper value minus the entire lower value, not upper value minus whichever term is most convenient. Reversing the limits reverses the sign of the integral, although the geometric region between the same two vertical boundaries has unchanged area.
Subject references: OpenStax: The definite integral; OpenStax: Areas between curves. These support the concepts; the questions, datasets, calculations and diagrams here are original.
Worked analysis 1: a region entirely above the axis
Find the area bounded by y = 3x² + 2x, the x-axis, x = 1 and x = 3. On this interval x is positive, so y is positive. The signed integral therefore equals the area.
An antiderivative is F(x) = x³ + x². At the upper bound, F(3) = 27 + 9 = 36. At the lower bound, F(1) = 1 + 1 = 2. Thus the area is 36 − 2 = 34 square units.
A size check helps: the curve rises from height 5 to height 33 across a width of 2. Its area must lie between the areas of rectangles of these two heights, namely 10 and 66 square units. The result 34 passes that broad check. This does not prove the arithmetic, but it would immediately challenge an answer such as 340.
The shaded region lies above y = 0 and below y = 3x² + 2x, between the vertical lines x = 1 and x = 3. Its endpoint heights are 5 and 33. These equations and bounds completely define the region even without the drawing.
Worked analysis 2: a region wholly below the axis
For y = x − 4 between x = 0 and x = 3, calculate both the definite integral and the geometric area. The graph has heights −4 and −1 at the endpoints, and stays below zero throughout the interval.
Take F(x) = x²/2 − 4x. The definite integral is F(3) − F(0) = 9/2 − 12 = −15/2. Its negative sign records that the function lies below the axis. The area is instead 0 minus the function integrated over the interval: 15/2 = 7.5 square units.
Here a trapezium check is available without calculus. The vertical distances from the axis are 4 and 1, and the horizontal width is 3. Area = ½(4 + 1) × 3 = 7.5 square units. Calling the area −7.5 would describe the signed integral rather than the size of the region.
Worked analysis 3: split at an axis crossing
Find the total area between y = 2x − 2 and the x-axis from x = 0 to x = 3. First solve 2x − 2 = 0, giving x = 1. The graph is negative on the first interval and positive on the second.
Using F(x) = x² − 2x, the integral from 0 to 1 is −1, and the integral from 1 to 3 is 3 − (−1) = 4. The complete signed integral is −1 + 4 = 3. The total geometric area is instead 1 + 4 = 5 square units.
The diagram shades the triangle below the axis between (0, −2), (0, 0) and (1, 0), and the triangle above it between (1, 0), (3, 0) and (3, 4). Their areas are 1 and 4 square units. Different hatching distinguishes the two signs without relying only on colour.
Taking the absolute value of the final integral would give |3| = 3, still wrong for the total area. Negative and positive contributions have already cancelled before that absolute value is taken. Split wherever the sign changes, make each region’s area positive, and then add.
Worked analysis 4: the area between a line and a curve
Find the enclosed area between y = x + 2 and y = x². The intersections satisfy x² = x + 2, so x² − x − 2 = (x − 2)(x + 1) = 0. The bounds are x = −1 and x = 2, and the intersection points are (−1, 1) and (2, 4).
At x = 0 the line has height 2 while the curve has height 0. The line is the upper boundary between these intersections. The vertical thickness is therefore (x + 2) − x². Integrate that difference, not the sum of the two functions.
An antiderivative of the thickness is F(x) = x²/2 + 2x − x³/3. At x = 2 its value is 10/3. At x = −1 its value is −7/6. Subtracting gives 10/3 − (−7/6) = 9/2 = 4.5 square units.
The enclosed region is above y = x² and below y = x + 2 for −1 ≤ x ≤ 2. The boundary curves meet at (−1, 1) and (2, 4); the vertical gap at x = 0 is 2 units. This complete description replaces any need to infer a missing shaded region.
Checking which graph is higher matters even when both graphs lie above the axis. The area is based on their separation. If their order changes inside a longer requested interval, split at each relevant intersection and use upper minus lower on each part.
A short audit before accepting an area answer
Ask four questions. Have all bounds been stated or found? Have all relevant roots or intersections been considered? Is the expression being integrated the required height or gap? Is the final answer an integral with a sign, or a nonnegative area? These checks address different errors and cannot replace one another.
Do not rely on a sketch alone to discover exact intersections. A drawing can suggest x = 2, but solving the equations establishes it. Equally, solving for every root does not mean every root is a boundary: retain those that lie in the requested interval and actually divide the region in question.
Three solved practice tasks
Practice 1: use two nonzero bounds
Find the area below y = 2x + 1 and above the x-axis from x = 1 to x = 4. The graph is positive. With F(x) = x² + x, area = (16 + 4) − (1 + 1) = 18 square units. The trapezium check is ½(3 + 9) × 3 = 18.
Practice 2: two crossings require three pieces
Find the total area between y = x² − 1 and the x-axis from x = −2 to x = 2. The roots are −1 and 1. An antiderivative is x³/3 − x. The left positive region has area 4/3, the middle negative region has area 4/3, and the right positive region has area 4/3. Total area is 4 square units; the signed integral is only 4/3.
Practice 3: a horizontal line above a parabola
Find the area enclosed by y = 9 and y = x². Intersections occur at x = −3 and x = 3. The upper-minus-lower expression is 9 − x², with antiderivative 9x − x³/3. The area is 18 − (−18) = 36 square units. Symmetry gives the same result by doubling the integral from 0 to 3.
Frequently asked questions
Why does an indefinite integral include C?
Different antiderivatives can differ by a constant, because differentiating any constant gives zero. C represents that family rather than an extra measurable area.
Why is there no C in a numerical definite integral?
The same constant appears at both bounds and cancels in F(b) − F(a). The stated bounds determine the numerical result.
Can a definite integral be negative?
Yes. Contributions below the axis are negative. A negative result can be correct for the signed integral but cannot be a negative geometric area.
Can I take the absolute value only at the end?
Only when that treatment matches the region, such as a function with one sign throughout. Across sign changes, cancellation can hide area, so split first.
How do I find bounds for an enclosed line-and-curve region?
Solve the two equations together and check which intersections bound the requested region. Do not assume an approximate drawing supplies exact coordinates.
What if the upper graph changes?
Divide the interval at the relevant intersections. On each piece, integrate the graph that is higher minus the graph that is lower.
Are the units always square centimetres?
No. Use square units for abstract coordinate area, or squared length units when both axes measure the same length unit. Other integral applications can have other units.
Continue with targeted practice
Use the existing Mathematics topical revision collection to find further resources. Some contents may require membership. This link does not promise a specific downloadable file or an official answer key for this lesson.
WhatsApp Us Now