01 · An above-axis polynomial
Find the area under y = x² + 1 from x = 0 to x = 2.
Hint
The curve is always positive.
Worked solution
[x³/3 + x]₀² = 14/3 square units.
Understand · explore · practise
Areas and signed integrals
Distinguish signed integrals from total area. Split at axis crossings, use exact polynomial, exponential and trigonometric integrals, and compare absolute function values with absolute inputs.
Before you startDefinite integrals, solving equations, graph signs and modulus transformations.
01 / Separate the integral from geometric area
Below-axis regions subtract from an integral but add to total area.
Signed integral = area above − area below
Total geometric area = area above + area below
Both component areas are nonnegative.
A definite integral keeps the sign of the function. Geometric area counts every region positively. Before calculating, sketch the curve, locate its zeros and decide whether the question asks for a signed integral or a total area.
Separate positive and negative regionsExplore
Area above the axis = 3.666667.
Area below the axis = 4.5.
Signed integral = −0.833333.
Total geometric area = 8.166667.
Split points inside the interval: −1, 2.
Green marks regions above the axis; amber marks regions below it. Values come from exact polynomial antiderivatives, displayed as decimals. The shaded drawing uses sampled curves.
02 / Use an integral directly above the axis
First establish that the curve stays nonnegative.
Find the area under y = eˣ from x = 0 to x = ln 3.Worked example
eˣ > 0 throughout the interval
The integral equals the geometric area.
∫₀^(ln 3) eˣ dx = [eˣ]₀^(ln 3)
Use a standard primitive.
3 − 1 = 2 square units
The area is positive.
03 / Negate an integral below the axis
Do not report a negative geometric area.
Find the area between y = x² − 4 and the x-axis for 0 ≤ x ≤ 1.Worked example
x² − 4 < 0 on [0,1]
The entire curve segment lies below the axis.
Signed integral = [x³/3 − 4x]₀¹ = −11/3
This negative value is not the geometric area.
Area = 11/3 square units
Negate the below-axis integral.
02 · Below-axis exponential
Find the area between y = −eˣ and the x-axis from 0 to ln 2.
Hint
The signed integral is negative.
Worked solution
Area = ∫₀^(ln 2) eˣ dx = 1 square unit.
03 · Identify what is asked
If a below-axis region has geometric area 7, what is its signed integral with increasing x bounds?
Hint
The function is negative there.
Worked solution
−7.
04 / Split at each sign change
One absolute value around the final answer is not enough.
Find the total area between y = x² − 1 and the axis from x = 0 to x = 2.Worked example
x² − 1 = 0 gives x = 1 inside the interval
The curve is negative before 1 and positive after 1.
Below-axis area = −∫₀¹ (x² − 1) dx = 2/3
Negate just this piece.
Above-axis area = ∫₁² (x² − 1) dx = 4/3
Keep this piece positive.
Total area = 2; signed integral = 2/3
Adding the areas differs from subtracting them.
Watch: separate positive and negative regions
Pause, replay or seek freely. The notes explain the same idea and stay in view.
04 · A failed absolute-value shortcut
For the worked example, is |∫₀² (x² − 1) dx| the total area?
Hint
Compare 2/3 with the sum of the two region areas.
Worked solution
No. It is 2/3, whereas the total area is 2.
05 · Three regions
For f(x) = x² − x − 2 on [−2,3], where must you split to compute total area?
Hint
Factor f(x).
Worked solution
f(x) = (x + 1)(x − 2); split at −1 and 2.
05 / Use exact logarithmic crossing points
Solve the axis crossing before integrating.
Find total area between y = eˣ − 2 and the axis from 0 to ln 4.Worked example
eˣ − 2 = 0 gives x = ln 2
The curve increases through this crossing.
Area below = ∫₀^(ln 2) (2 − eˣ) dx = 2ln 2 − 1
Keep the exact crossing value.
Area above = ∫_(ln 2)^(ln 4) (eˣ − 2) dx = 2 − 2ln 2
Subtract the correct endpoint values.
Total area = 1 square unit
The logarithmic terms cancel in the sum.
06 · The signed version
What is the signed integral of eˣ − 2 from 0 to ln 4?
Hint
Integrate once without changing signs.
Worked solution
3 − 4ln 2. This differs from the total area 1.
07 · Locate another exponential crossing
Where does e²ˣ − 5 cross the axis?
Hint
Solve e²ˣ = 5.
Worked solution
x = (ln 5)/2.
06 / Split trigonometric regions using exact angles
Use radians and all crossings inside the interval.
Find the signed integral and total area for y = sin x on [0,2π].Worked example
sin x changes sign at x = π
It is positive on (0,π) and negative on (π,2π).
Above-axis area = 2; below-axis area = 2
Use the primitive −cos x on each piece.
Signed integral = 0; total area = 4
Equal regions cancel only in the signed result.
08 · A shifted cosine crossing
Where does cos x − 1/2 cross the axis on [0,π]?
Hint
Use the exact cosine value.
Worked solution
x = π/3.
09 · Shifted cosine total area
Find the total area between cos x − 1/2 and the axis on [0,π].
Hint
Split at π/3 and use sin x − x/2.
Worked solution
The positive part is √3/2 − π/6; the negative-region area is √3/2 + π/3. Total = √3 + π/6.
07 / Reflect negative outputs for y = |f(x)|
The reflected graph’s integral gives total area.
Total area on [a,b] = ∫ₐᵇ |f(x)| dx
For a continuous real function on a finite interval.
Taking |f(x)| leaves above-axis pieces unchanged and reflects below-axis pieces upwards. It does not move the x-coordinates of the zeros. Integrate piecewise using f or −f according to the sign of f.
Integrate |x − 1| from x = −2 to x = 2.Worked example
|x − 1| = 1 − x for x < 1; x − 1 for x ≥ 1
Split at x = 1.
∫₋₂¹ (1 − x) dx = 9/2
The left piece is a positive triangle.
∫₁² (x − 1) dx = 1/2
The right piece is another positive triangle.
Total = 5
Both pieces add.
10 · Output modulus and zeros
Does replacing f(x) by |f(x)| change its zeros?
Hint
|f(x)| is zero exactly when f(x) is zero.
Worked solution
No. It changes the signs of negative outputs, not where the function vanishes.
08 / Keep f(|x|) separate from |f(x)|
An input modulus reflects the right half of the graph leftwards.
For f(x) = x − 1, compare f(|x|) and |f(x)|.Worked example
f(|x|) = |x| − 1
Replace the input x before evaluating f.
|f(x)| = |x − 1|
Take the modulus after evaluating f.
At x = −2, the values are 1 and 3
The two functions are different.
f(|x|) can still be negative
For example, its value at x = 0 is −1.
For a function defined on the needed nonnegative inputs, f(|x|) is even. It is not necessarily nonnegative. If the question asks for total area, you may still need to split at its zeros.
11 · Signed area of an input modulus
Evaluate ∫₋₂² (|x| − 1) dx.
Hint
The function is even: use twice the integral on [0,2].
Worked solution
2∫₀² (x − 1) dx = 0.
12 · Total area of that graph
Find the total area between y = |x| − 1 and the axis on [−2,2].
Hint
Use even symmetry and split the positive half at x = 1.
Worked solution
2[∫₀¹ (1 − x) dx + ∫₁² (x − 1) dx] = 2 square units.
09 / Use symmetry without losing the question’s meaning
An odd function’s zero integral does not imply zero area.
For an odd continuous function, equal opposite regions on [−a,a] cancel in the signed integral. For an even function, its signed integral is twice the integral on [0,a]. Total area can use symmetry too, but negative pieces must still count positively.
A function touching the axis without changing sign does not require a sign reversal in the area calculation. Find the sign on each interval instead of assuming every zero is a crossing.
13 · A touching zero
Does y = (x − 1)² change sign at x = 1?
Hint
A square is nonnegative.
Worked solution
No. Its integral over an increasing interval equals its geometric area even though it touches the axis.
10 / Use a sketch and a sign table
Keep geometry, algebra and the final units consistent.
Mark the bounds and roots, test one point in each interval, and write an area sum before evaluating. A total area cannot be negative. The magnitude of a signed integral cannot exceed the total area of the same continuous function on the same interval.
If an integrand has a singularity inside the interval, inspect it separately; finite endpoint substitution alone is not a valid area calculation. The smooth examples here have no such singularities.
14 · Compare the two totals
If the area above is 8 and the area below is 3, give the signed integral and total area.
Hint
Subtract for the signed value; add for total area.
Worked solution
Signed integral = 5; total area = 11.
15 · Recover a missing component
The total area is 13 and the signed integral is −5. Find the areas above and below.
Hint
Solve P + N = 13 and P − N = −5.
Worked solution
Area above P = 4; area below N = 9.
11 / Find signs before finding area
Split only where needed, then add positive region areas.
- Distinguish a signed integral from a total geometric area.
- Find the relevant roots and inspect interval signs.
- Negate below-axis integrals when adding total area.
- Do not confuse |f(x)| with f(|x|).
- Use exact crossings and symmetry carefully.
16 · A final polynomial area
Find the total area between y = x and the axis on [−2,3].
Hint
Split at zero; the two triangles have base and height 2, and base and height 3.
Worked solution
2 + 9/2 = 13/2 square units. The signed integral is 5/2.
Section 1 of 11 · Separate the integral from geometric area