Definite Integrals and Area
This topic covers the use of definite integration to calculate the area between a function's curve and the x-axis. It is crucial to distinguish between a definite integral, which can be negative and represents 'net area', and the total geometric area, which must always be positive.
Part of the ESAT Mathematics 2 syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- A definite integral calculates the 'signed area': area above the x-axis is positive, and area below is negative.
- If a curve crosses the x-axis within the integration interval, a single definite integral will give you the net result (sum of positive areas minus sum of areas below), not the total physical area.
- To find the total area, first find the x-intercepts (roots) of the function within the given interval.
- Split the calculation into separate integrals for each section defined by the roots and the interval boundaries.
- Calculate each definite integral, take the absolute value of any negative results, and then sum all the values to get the total area.
Diagram
Formulae
∫[a, b] f(x) dx = F(b) - F(a) Use the Fundamental Theorem of Calculus to evaluate a definite integral, where F(x) is the antiderivative of f(x).
Total Area = |∫[a, c] f(x) dx| + |∫[c, b] f(x) dx| To find the total area from a to b when the function f(x) crosses the x-axis at x=c. The integral must be split at the root.
Definitions
- Definite Integral
- The value of ∫[a, b] f(x) dx, representing the net accumulation of signed area between the curve y=f(x) and the x-axis from x=a to x=b.
- Total Area
- The sum of the geometric areas of all regions enclosed by a curve and the x-axis over an interval. This value is always non-negative.
Worked example
Find the total area enclosed between the curve y = x2 - 2x, the x-axis, and the lines x = -1 and x = 3.
- 1
Step 1:
Identify that the question asks for 'total area', not just the definite integral.
This means we must check for regions below the x-axis.
- 2
Step 2:
Find the x-intercepts of y = x2 - 2x by setting y=0.
Factoring gives x(x - 2) = 0, so the roots are at x = 0 and x = 2.
- 3
Step 3:
The roots 0 and 2 are within our interval [-1, 3].
We must split the integration at these points.
The intervals are [-1, 0], [0, 2], and [2, 3].
- 4
Step 4:
A quick sketch of the U-shaped parabola shows the curve is above the axis for [-1, 0], below for [0, 2], and above for [2, 3].
We expect the middle integral to be negative.
- 5
Step 5:
Integrate f(x) = x2 - 2x to get F(x) = (x3)/3 - x2.
- 6
Step 6:
Evaluate the integral for each section:
• For [-1, 0]:
F(0) - F(-1) = (0) - ((-1)3/3 - (-1)2) = 0 - (-1/3 - 1) = 4/3.
• For [0, 2]:
F(2) - F(0) = ((23)/3 - 22) - (0) = (8/3 - 4) = -4/3• For [2, 3]:
F(3) - F(2) = ((33)/3 - 32) - ((23)/3 - 22) = (9 - 9) - (8/3 - 4) = 0 - (-4/3) = 4/3 - 7
Step 7:
Sum the absolute values of these results:
|4/3| + |-4/3| + |4/3| = 4/3 + 4/3 + 4/3 = 12/3 = 4.
Answer: 4
Common mistakes
- ×Calculating a single integral from the start to the end of the interval (e.g., from -1 to 3) without splitting it at the x-intercepts. This calculates the net signed area, not the total area.
- ×Making sign errors during the subtraction F(b) - F(a), especially when F(a) is negative. For instance, `10 - (-2)` is `12`, not `8`.
- ×Arithmetic errors with fractions when evaluating the antiderivative at the limits. This is a common source of lost marks in a non-calculator exam.
No-calculator tips
- ✓Always sketch the graph. It visually confirms where the function is positive or negative, preventing the main conceptual error of this topic.
- ✓To simplify fraction arithmetic, evaluate the polynomial terms at the limits first as whole numbers where possible, then combine the fractions at the end.
- ✓For symmetric odd functions (e.g., y=x3 or y=sin(x)) integrated over a symmetric interval (e.g., -a to a), the definite integral is always zero. This can save significant calculation time.