Difficulty: Intermediate | Prerequisites: Properties of definite integrals, basic area formulas (triangle, trapezoid, circle).
Before you learn the Fundamental Theorem of Calculus, there is a class of definite integrals you can evaluate by hand using nothing more than geometry. When the graph of f(x) is made up of straight lines and semicircles, the definite integral is the net signed area enclosed between the curve and the x-axis. This topic trains you to decompose that region into triangles, rectangles, trapezoids and semicircles, compute their areas, and assign the correct signs. It sits between "properties of integrals" (the algebraic rules) and "evaluating integrals with antiderivatives" (the FTC).
When a graph is built from line segments and semicircles, you evaluate the definite integral by computing the areas of simple geometric shapes and adding or subtracting them depending on whether the region sits above or below the x-axis. Area above is positive, area below is negative.
Net signed area
The value returned by a definite integral: area above the x-axis counts as positive and area below counts as negative. Think of it as "area with a direction."
Total (unsigned) area
The sum of all enclosed areas treated as positive regardless of whether the region is above or below the x-axis. Computed by integrating |f(x)| instead of f(x). In simple terms, you ignore the sign and just add up all the regions.
Piecewise linear function
A function whose graph is made up of straight line segments joined end to end. Each segment forms the side of a triangle, rectangle or trapezoid, which is what makes geometric evaluation possible.
Semicircle (in integration contexts)
When part of a graph follows the equation y = √(r² − x²) or y = −√(r² − x²), it traces a semicircle of radius r. The enclosed area is ½πr². In simple terms, it is half a circle, and you use half the circle area formula.
Absolute value integral
An integral of the form ∫ₐᵇ |f(x)| dx, which gives the total (unsigned) area between the curve and the x-axis. To evaluate it, you must split the interval at every point where f(x) = 0 and take the positive version of each piece.
Identify the geometric shapes formed between the graph and the x-axis over each sub-interval.
Common shapes: triangles (area = ½ × base × height), rectangles (area = base × height), trapezoids (area = ½ × (b₁ + b₂) × height).
Assign signs: regions above the x-axis contribute positive area, regions below contribute negative area.
Add the signed areas across the full interval to get the integral's value.
Example from the worksheet (problems 11 to 16): the graph of f(x) passes through (−4, 3), (−2, 0), (1, 3), (3, 3), (5, 0), (7, −3). Between x = −4 and x = −2 the region is a triangle with base 2 and height 3, giving area = ½(2)(3) = 3. Between x = −2 and x = 1 there is a triangle with base 3 and height 3, area = ½(3)(3) = 9/2. And so on for each segment.
To evaluate ∫ over a wide interval, split it at the graph's vertices and compute each piece.
Example: ∫₋₄⁷ f(x) dx = (triangle from −4 to −2) + (triangle from −2 to 1) + (rectangle from 1 to 3) + (triangle from 3 to 5) + (triangle from 5 to 7, below x-axis).
The pieces from 5 to 7 give negative area because f(x) < 0 there.
When the graph includes a semicircular arc, the area enclosed is ½πr².
If the semicircle sits below the x-axis, its contribution is −½πr².
Example from problems 17 to 24: the graph has a semicircle of radius 1 centred below the x-axis from x = −2 to x = 0. Its area contribution is −½π(1)² = −π.
Triangular portions in the same graph are handled the usual way, then all pieces are summed.
Taking the absolute value of f(x) reflects any portion below the x-axis upward, so all area becomes positive.
To compute: find where f(x) = 0, split the integral at those zeros, and negate the parts where f(x) was negative.
Result: ∫ₐᵇ |f(x)| dx = (sum of all geometric areas, all taken as positive).
Example: if ∫₋₄⁶ f(x) dx = 3 − 2π (net signed), then ∫₋₄⁶ |f(x)| dx = 5 + 2π (every piece made positive).
∫ₐᵇ [f(x) + c] dx = ∫ₐᵇ f(x) dx + c(b − a).
Adding a constant c shifts the entire graph up by c units. The extra area is simply a rectangle of width (b − a) and height c.
Example: ∫₋₄⁶ [f(x) + 2] dx = (3 − 2π) + 2(6 − (−4)) = 3 − 2π + 20 = 23 − 2π.
For integrands like (6 − x), (2x − 3), or (2 − |x|), sketch the line on the given interval, identify the shapes, and compute.
Watch for lines that cross the x-axis within the interval, which split the region into a positive and a negative part.
Example: ∫₋₁⁵ (2x − 3) dx. The line 2x − 3 = 0 at x = 3/2. Below the axis from −1 to 3/2 (a triangle), above from 3/2 to 5 (a triangle). Compute each and sum: result is 6.
Example: ∫₀ᵃ √(a² − x²) dx represents a quarter-circle of radius a, so the answer is ¼πa².
\text{Triangle: } A = \tfrac{1}{2} \times \text{base} \times \text{height}\text{Rectangle: } A = \text{base} \times \text{height}\text{Trapezoid: } A = \tfrac{1}{2}(b_1 + b_2) \times h\text{Semicircle: } A = \tfrac{1}{2}\pi r^2\text{Quarter-circle: } A = \tfrac{1}{4}\pi r^2\int_a^b [f(x) + c]\,dx = \int_a^b f(x)\,dx + c(b - a)This technique is how you estimate total distance from a velocity-time graph when you do not have a formula for velocity. Plot the data, connect the dots with line segments, and the area under the curve gives total displacement. Speedometer readings on a car trip work exactly this way.
Students often forget to make area negative when the graph sits below the x-axis. The definite integral is net signed area, not total area.
Students confuse ∫ f(x) dx with ∫ |f(x)| dx. The first gives net signed area; the second gives total area. They are equal only when f(x) ≥ 0 everywhere on the interval.
When a semicircle appears, students sometimes use the full circle formula πr² instead of the half-circle formula ½πr².
Students sometimes miscalculate the base of a triangle by forgetting that the base is the horizontal distance along the x-axis, not the length of the sloped side.
⚠️ Graph-based integral problems are a staple of Brief Calculus exams. You will almost certainly see a piecewise linear graph and be asked to evaluate integrals over various sub-intervals.
⚠️ Problems mixing semicircles with line segments test whether you can identify each shape and sum correctly.
⚠️ Absolute value integrals (∫ |f(x)| dx) frequently appear as a follow-up to a net-area problem. Know how to convert.
⚠️ "Sketch the region" problems (like 25 to 30) require you to draw the function, identify where it crosses the x-axis, and decompose into shapes. The sketch is not optional; it prevents sign errors.
True or False: If f(x) < 0 on [a, b], then ∫ₐᵇ f(x) dx is negative. (True)
Fill in the blank: The area of a semicircle with radius 3 is ___. (9π/2)
True or False: ∫ₐᵇ |f(x)| dx is always greater than or equal to |∫ₐᵇ f(x) dx|. (True)
Fill in the blank: A triangle with base 4 and height 3 has area ___. (6)
True or False: ∫ₐᵇ [f(x) + 2] dx = ∫ₐᵇ f(x) dx + 2. (False, it equals ∫ₐᵇ f(x) dx + 2(b − a))
Q: A graph of f(x) forms a triangle above the x-axis with vertices at (0, 0), (4, 0) and (2, 5). What is ∫₀⁴ f(x) dx?
A: The triangle has base 4 and height 5, so the area is ½(4)(5) = 10.
Q: The graph of f(x) is a semicircle of radius 2 sitting below the x-axis from x = 0 to x = 4. What is ∫₀⁴ f(x) dx?
A: The semicircle area is ½π(2)² = 2π, and since it is below the axis the integral is −2π.
Q: Given ∫₋₄⁶ f(x) dx = 3 − 2π (from a graph with line segments and a semicircle), find ∫₋₄⁶ |f(x)| dx if the regions below the axis have total area 1 + 2π.
A: The net integral = (positive area) − (negative area) = 3 − 2π. So positive area = (3 − 2π) + (1 + 2π) = 4. Then ∫ |f(x)| dx = 4 + (1 + 2π) = 5 + 2π.
Q: Evaluate ∫₋₄⁶ [f(x) + 2] dx given that ∫₋₄⁶ f(x) dx = 3 − 2π.
A: (3 − 2π) + 2(6 − (−4)) = 3 − 2π + 20 = 23 − 2π.
Q: Evaluate ∫₁⁵ [6 − x] dx by sketching the region.
A: The line y = 6 − x goes from (1, 5) to (5, 1), all above the x-axis. The region is a trapezoid with parallel sides 5 and 1 and width 4: area = ½(5 + 1)(4) = 12.
Q: Evaluate ∫₀ᵃ √(a² − x²) dx for a > 0.
A: This is a quarter-circle of radius a in the first quadrant. Area = ¼πa².
This material connects directly to Riemann sums, where the rectangles used to approximate area become exact as their width tends to zero. Geometric evaluation is the "exact" version of what Riemann sums approximate.
It also connects to the Fundamental Theorem of Calculus: once you learn to find antiderivatives, you will have a formula-based way to compute the same values you are here getting from shapes. The geometric approach remains useful as a check.
Absolute value integrals reappear in total distance versus displacement problems in applied calculus.
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