A definite integral of |f(x)| measures the total non-negative area between a graph and the x-axis. If f(x) is quadratic and crosses the axis, you must divide the integration interval at every root inside it, integrate f on positive sections and integrate −f on negative sections. Taking the absolute value only after integrating the whole interval usually gives a different number.
For example, let f(x)=x²−4x+3=(x−1)(x−3) on 0≤x≤4. The quadratic is positive on [0,1], negative on [1,3] and positive again on [3,4]. Its signed integral is 4/3, but ∫₀⁴|x²−4x+3|dx is 4. The negative middle region contributes its full magnitude rather than cancelling part of the positive regions. This guide gives exact working, a sign-chart method, general formulae for quadratic root pairs and parameter cases where the number of regions changes.
Signed accumulation and total area are different
A definite integral ∫_a^b f(x)dx adds values with their signs. Regions above the x-axis contribute positively, while regions below contribute negatively when a The integral ∫_a^b|f(x)|dx replaces every negative function value by its positive magnitude. Its result is the total geometric area between the graph and the x-axis over the interval. The difference matters when a curve crosses the axis. One region’s signed contribution can cancel another even though both represent real geometric area. A learner who writes |∫f| in place of ∫|f| has moved the absolute value outside the integral. In general, that changes the task, because cancellation has already occurred before the absolute value is applied. Consider f(x)=x²−4x+3. Factorise it as (x−1)(x−3). The roots are x=1 and x=3. Both lie strictly within the requested interval [0,4]. These are the only points at which the sign of this continuous quadratic can change, because its leading coefficient is positive and its roots are simple. Therefore split the domain into [0,1], [1,3] and [3,4]. There is no need to solve an absolute-value equation at every point. The two roots divide the interval into the regions whose signs must be understood. For 0 For 1 Hence |f(x)| equals f(x) on [0,1], equals −f(x) on [1,3], and equals f(x) on [3,4]. The isolated roots have value zero, so whether they are assigned to the neighbouring piece does not change the definite integral. The sign chart, not the absolute-value symbol alone, determines which expression to integrate on each piece. The correct setup is ∫₀⁴|f(x)|dx = ∫₀¹ f(x)dx − ∫₁³ f(x)dx + ∫₃⁴ f(x)dx. The minus in front of the middle integral is intentional. The signed middle integral is negative, and its opposite is the positive geometric area in that section. Another equivalent notation is ∫₀¹ f + ∫₁³(−f) + ∫₃⁴ f. This makes every piece’s non-negative integrand explicit. Do not replace the setup with |∫₀⁴ f|. That operation would allow the middle region to cancel part of the positive outer regions before taking magnitude. Integrate f(x)=x²−4x+3 term by term. An antiderivative is F(x)=x³/3−2x²+3x. Check by differentiation: F′(x)=x²−4x+3=f(x). The same antiderivative can be used on every piece. There is no need to integrate the same polynomial from scratch three times. The only change between the regions is whether we add or subtract the signed definite integral. Recording F clearly before evaluating endpoints reduces the risk of making three unrelated integration errors. Compute F(0)=0. At x=1, F(1)=1/3−2+3=4/3. At x=3, F(3)=27/3−2(9)+9=9−18+9=0. At x=4, F(4)=64/3−2(16)+12=64/3−20=4/3. Thus the key values are F(0)=0, F(1)=4/3, F(3)=0 and F(4)=4/3. A small table helps keep the interval orientation and subtraction order correct. Over [0,1], the polynomial is non-negative. The geometric area contribution is therefore ∫₀¹ f(x)dx. Using F, this equals F(1)−F(0)=4/3−0=4/3 square units. The region runs from the y-axis to the first x-intercept at x=1. Its positivity is consistent with the upward-opening quadratic being outside its two roots. This contribution can be added directly to total area without changing its sign. The same will be true of the rightmost region, but not the section between the roots. Over [1,3], f(x) is negative. The signed integral is F(3)−F(1)=0−4/3=−4/3. The geometric region below the x-axis has positive area 4/3. Thus its contribution to ∫|f| is −(−4/3)=+4/3. This is the step at which a common error occurs. A student who adds the signed value −4/3 to the two outer regions is calculating net signed area rather than total geometric area. Writing the sign decision beside the integral before substituting numerical values makes the distinction explicit and easier to check. Over [3,4], the quadratic is positive again. Its signed integral equals its geometric area. Evaluate F(4)−F(3)=4/3−0=4/3. The three regions therefore have equal positive area magnitudes in this specially chosen symmetric example, even though the middle signed contribution is negative. Their equality is not a universal feature of every quadratic interval. It follows here from the particular endpoints and symmetry around x=2. Different outer boundaries generally produce different area contributions. The total non-negative area is 4/3+4/3+4/3=12/3=4 square units. Therefore ∫₀⁴|x²−4x+3|dx=4. This is the requested result if the problem asks for total area between the quadratic and the x-axis or the integral of the absolute value. The three subinterval computations establish completeness: every permitted input is covered by one sign region, no part is counted twice except shared endpoints of zero width, and every signed contribution has been converted into positive geometric area. Without absolute values, the integral over the full interval is F(4)−F(0)=4/3. The same answer follows by summing signed pieces: +4/3−4/3+4/3=4/3. Thus ∫₀⁴ f(x)dx=4/3, while ∫₀⁴|f(x)|dx=4. Their difference is 8/3. That difference is twice the magnitude of the negative middle signed area. The absolute-value integral both removes the negative contribution and adds its positive magnitude, creating the factor of two in the correction. A tempting shortcut is to calculate ∫₀⁴ f=4/3 and then write |4/3|=4/3 as the total area. That approach is wrong because the negative middle region has already cancelled part of the positive outer regions. Taking an absolute value after the cancellation cannot reconstruct how much was lost. The correct comparison is |∫f|≤∫|f|, with equality only under suitable sign conditions, such as f keeping one sign almost everywhere on the interval. Our example makes the inequality strict: 4/3<4. This is a useful diagnostic whenever signed graphs, velocity or magnitude questions appear in calculus. Write I=∫₀⁴ f(x)dx for the signed total, and let N be the magnitude of the region on which f is negative. Then the absolute-value integral equals I+2N. One factor of N repairs the subtraction in I and the second accounts for adding that region positively. For the example, I=4/3 and N=4/3. Hence ∫₀⁴|f|=4/3+2(4/3)=4. This method can save work after the negative region is identified. It still depends on a correct sign chart and exact calculation of N; it is not a shortcut that avoids understanding where f is below zero. Let f(x)=a(x−p)(x−q), with p The magnitude of the area between the graph and x-axis over [p,q] is ∫_p^q|f(x)|dx. On this interval |f(x)|=|a|(x−p)(q−x). Substituting u=x−p and L=q−p gives |a|∫₀ᴸu(L−u)du. Integrating produces |a|L³/6, or |a|(q−p)³/6. This compact formula is valid for the between-roots section and can supplement the full piecewise calculation. In the opening polynomial, a=1, p=1 and q=3. Their separation is q−p=2. The magnitude of the negative region is |a|(q−p)³/6=1·8/6=4/3, matching the direct integral. The same formula appeared in the area between a parabola and its secant chord because the underlying vertical gap is also a quadratic with two roots. These are connected applications of the same factored expression. The formula calculates only the area between the two roots. It does not automatically include any positive regions to the left or right when the requested integration interval extends beyond them. Now integrate |f(x)| from x=0 to x=2, keeping f(x)=x²−4x+3. Only the root x=1 lies strictly inside [0,2]; the other root x=3 is outside. The function is positive on [0,1] and negative on [1,2]. The first area is F(1)−F(0)=4/3. Since F(2)=8/3−8+6=2/3, the second signed integral is F(2)−F(1)=−2/3, giving area 2/3. Therefore ∫₀²|f(x)|dx=4/3+2/3=2. Split only at roots inside the actual integration domain; extra roots outside that interval should not create unnecessary pieces. If the requested interval is [1,3], the quadratic is non-positive throughout it. Thus ∫₁³|f(x)|dx=−∫₁³ f(x)dx=4/3. There is no need to split at an interior root because no additional root lies strictly inside the interval. The integrand’s sign is known over the whole domain. This is an important efficiency principle. The number of algebraic roots of a polynomial does not determine the number of integration pieces unless those roots occur inside the stated bounds and actually mark relevant sign changes. Consider g(x)=−f(x)=−x²+4x−3. It is the reflection of f across the x-axis. For every real x, |g(x)|=|−f(x)|=|f(x)|. Therefore their absolute-value integrals over any common interval are equal. On [0,4], ∫|g| is also 4, even though the signed integral of g is −4/3 rather than +4/3. This example separates total magnitude from orientation. Reflection reverses signed area but preserves geometric area between the graph and x-axis. It also explains why absolute-value integrals naturally use |a| rather than the signed leading coefficient in between-roots formulas. Take h(x)=(x−2)² on [0,4]. The function equals zero at x=2, but it is non-negative on both sides. Therefore |h(x)|=h(x) throughout the interval. There is no need to flip the sign of the integrand after x=2. The integral is ∫₀⁴(x−2)²dx=16/3. Splitting at x=2 is allowed but not necessary. The lesson is that every zero is a candidate sign boundary, but a repeated even-multiplicity root may not change the sign. A sign chart or factor analysis is more reliable than alternating signs mechanically after every root. Consider f_k(x)=x²−4x+k=(x−2)²+k−4 on the fixed interval [0,4]. Its vertex output is k−4 at x=2, and its endpoint outputs are f_k(0)=f_k(4)=k. For k≥4, the quadratic is non-negative across the interval, so the absolute-value integral equals the ordinary integral. For k≤0, even its endpoint maxima are non-positive, so the function is non-positive on [0,4] and the absolute-value integral is the negative of its signed integral. When 0 Integrate f_k(x)=x²−4x+k on [0,4]. An antiderivative is x³/3−2x²+kx. At the upper limit, its value is 64/3−32+4k. At zero, it is zero. Therefore I(k)=∫₀⁴ f_k(x)dx=4k−32/3. This single linear expression gives signed accumulation for every real k. It does not automatically give total geometric area when the function changes sign. The next stage depends on the parameter case: retain this signed result when f_k≥0, reverse its sign when f_k≤0, and add a middle-region correction when the graph crosses the axis twice. When k≥4, f_k(x)=(x−2)²+(k−4)≥0 for every real x. Hence ∫₀⁴|f_k(x)|dx=I(k)=4k−32/3. At k=4, the graph just touches the x-axis at x=2. The formula gives 16/3, agreeing with the integral of (x−2)². The equality case belongs here because the function is non-negative, not strictly positive. Its isolated zero does not create a negative region or change the value of the integral. For k≤0, the endpoint outputs are at most zero and the upward-opening quadratic lies no higher than its larger endpoint output within [0,4]. Therefore f_k(x)≤0 throughout the interval. The absolute integrand is −f_k. The total geometric area is A(k)=−I(k)=32/3−4k. At k=0, f_0(x)=x(x−4), which is negative between the two endpoint roots and zero at x=0 and x=4. The formula gives 32/3, the magnitude of its negative signed integral. For 0 Because 0<√(4−k)<2, both roots lie strictly inside [0,4]. The quadratic is negative between them and positive in the two outer regions. The root separation is L=2√(4−k). The between-roots negative area has magnitude N(k)=L³/6=(8/6)(4−k)^(3/2). Thus N(k)=(4/3)(4−k)^(3/2). This exact expression gives the amount that must be restored when converting the signed full integral into total absolute area. For the crossing case, total area is signed integral plus twice the negative-region magnitude. Therefore A(k)=4k−32/3+(8/3)(4−k)^(3/2) for 0 Combining all cases gives A(k)=32/3−4k for k≤0; A(k)=4k−32/3+(8/3)(4−k)^(3/2) for 0 At k=0 and k=4, the expressions meet consistently, so no artificial jump is introduced. This parameter extension illustrates why absolute-value integration is naturally piecewise when the roots move into or out of the integration domain. Setting k=3 recovers f(x)=x²−4x+3, the opening example. The middle-case formula gives A(3)=12−32/3+(8/3)(1)^(3/2). Since 12=36/3, the result is (36−32+8)/3=12/3=4. This matches the direct three-region integral exactly. Checking a familiar parameter is a useful way to validate a derived piecewise formula before presenting it as a general result. Because |f(x)|≥0, its definite integral over a nonzero-length interval cannot be negative. For a continuous function, ∫_a^b|f(x)|dx=0 implies f(x)=0 throughout that interval. A nonzero quadratic cannot vanish identically across an interval of positive width. Therefore the absolute-value integral of a genuine nonzero quadratic over a nondegenerate interval is strictly positive, even if its signed integral happens to be zero. This provides a strong answer check. If a calculation produces a negative total area or zero for a nonzero polynomial over a substantial interval, revisit the sign chart or endpoint subtraction. If f is interpreted as a signed velocity function v(t), then ∫v gives displacement and ∫|v| gives total distance over a time interval. A velocity polynomial with two simple roots can describe forward movement, a reversal and then forward movement again. The sign-chart and piecewise integration steps are identical to those used for geometric area. Only the contextual units and interpretation change. This transfer is useful because it reveals a general mathematical principle: integrating magnitude counts all contributions positively, while integrating a signed quantity allows cancellation. Students should be able to recognise that principle across different word problems. A graphing calculator can plot the original quadratic and confirm the crossing points at x=1 and x=3. It can also numerically evaluate ∫₀⁴|x²−4x+3|dx, giving 4. But a numerical result alone does not show why three intervals are required. The exact calculation explains the positive and negative regions, retains rational fractions and supports a proof of completeness. Use technology to check arithmetic or visualise the three signed regions. For a school solution, the sign chart and separate definite integrals remain the clearest explanation of why the absolute-value area is not just |4/3|. First factor the quadratic or solve its roots. Keep only real roots that fall inside the stated integration interval, and sort them in increasing order. Second test the polynomial’s sign on every subinterval. Third integrate f where it is non-negative and integrate −f where it is negative. Fourth evaluate the antiderivative at each boundary, add the resulting non-negative contributions and state positive area units where applicable. Finally compare the total with the magnitude of the signed integral. The total absolute area must be at least as large. This check helps catch a missed negative region or a sign reversal error. Question one: evaluate ∫₀⁴|x²−4x+3|dx. Answer: 4. Question two: evaluate the ordinary signed ∫₀⁴(x²−4x+3)dx. Answer: 4/3. Question three: find the area magnitude between the roots x=1 and x=3 of the same quadratic. Answer: 4/3. Question four: evaluate ∫₀²|x²−4x+3|dx. Answer: 2. Question five: evaluate ∫₀⁴|−x²+4x−3|dx. Answer: 4. Question six: evaluate ∫₀⁴|(x−2)²|dx. Answer: 16/3. Question seven: for f_k=x²−4x+k, state when no sign reversal is needed in the absolute integral on [0,4]. Answer: k≥4, where f_k is non-negative, or k≤0, where one global sign reversal suffices. For every task, identify the signs and relevant roots before evaluating an antiderivative. One student calculates the signed integral correctly but reports it as total area. Another identifies two roots but flips the sign after every root without checking for multiplicity. A third keeps a negative area contribution in the final total. A fourth handles the graph but forgets that some roots may lie outside the integration interval. These errors require different repairs: compare displacement with distance for the first, use a repeated-root example for the second, draw positive region areas for the third and practise interval filtering for the fourth. A changed polynomial is a better test of understanding than repeating the same three integrals. The learner should independently reconstruct where absolute values must alter the integrand. Complete the square in the opening quadratic: f(x)=x²−4x+3=(x−2)²−1. Introduce u=x−2. The interval 0≤x≤4 becomes −2≤u≤2, and the absolute integrand is |u²−1|. This is an even function because replacing u with −u does not change u². Therefore the total integral is 2∫₀²|u²−1|du. Only the positive half-interval needs explicit integration. Split at u=1: the expression is 1−u² on [0,1] and u²−1 on [1,2]. Their areas are 2/3 and 4/3. Doubling the sum gives 2(2)=4, matching the original three-region computation. The substitution u=x−2 shifts the centre of the parabola to the origin without changing the horizontal scale or any area. The reflected values u and −u have the same absolute integrand, so their contributions over the symmetric interval [−2,2] are equal. This is an exact property of even functions, not an approximation based on a visually symmetric sketch. The original [0,4] interval is symmetric about the parabola’s vertex x=2, which is why the shortcut works here. For an interval not centred on the vertex, the two halves generally have different bounds and cannot simply be doubled. A symmetry method must be justified by both function and domain. The absolute-value integral over [0,4] equals 4. The interval width is also 4, so the average value of |f(x)| is 4/4=1. The signed average of f is (4/3)/4=1/3. These are different because f is negative during the middle portion. If the same f describes velocity, 1 would represent average speed in suitable units, while 1/3 would represent signed average velocity. The arithmetic is identical to the geometric-area interpretation. This comparison helps students remember that applying absolute value before integration changes the accumulated quantity, whereas dividing by the interval width only converts a total into an average. For a polynomial with several real roots, first sort only the roots lying strictly within the integration interval. These divide the domain into sign-consistent subintervals unless a repeated root leaves the sign unchanged. Choose a test point in each subinterval. Where f is positive, add ∫f; where f is negative, add −∫f. The same antiderivative can be reused for every piece, with the signs handled outside the evaluations. At the end, every contribution should be non-negative. This procedure works beyond quadratics, although higher-degree factorisation and extra sign changes may create more regions. The logic is always rooted in the sign of the original integrand over the permitted domain. The absolute integrand |x²−4x+3| is non-negative and reaches 3 at x=0 and x=4. It reaches 1 at the vertex x=2. Therefore it is bounded above by 3 throughout [0,4], so the total area cannot exceed 3·4=12. The result 4 lies within that necessary bound. A lower bound can be obtained from known subregions: each of the three exact positive contributions is 4/3, so the total must exceed the contribution of any one piece. These bounds cannot replace exact integration, but they catch impossible signs and implausibly large numerical answers before a student submits the result. Integrals of absolute-value quadratics combine factorisation, polynomial sign charts, definite integration and geometric area. The parameter-dependent piecewise formula is enrichment that can be introduced after students master ordinary curve-area questions. For Singapore, consult the relevant SEAB syllabus references for 2026 O-Level Additional Mathematics or 2027 SEC G3, as applicable. Related eduKateSG guides include How to Calculate Total Distance When Velocity Changes Direction, How to Find a Function’s Average Value from a Definite Integral, and How to Find the Area Between a Parabola and Its Chord. More guides appear in the Additional Mathematics archive. Families can enquire about suitable three-student Mathematics tutorials near Sixth Avenue MRT. The transferable skill is identifying exactly where a signed function becomes negative and converting each contribution into its correct magnitude before adding them. Properly taught kids shine a bright light into the future.Factor the quadratic before integrating
Use a sign chart to decide the integrands
Write the absolute-value integral as three pieces
Find one antiderivative for all pieces
Evaluate the antiderivative at every boundary
Calculate the first positive region
Calculate the middle region without a negative area
Calculate the final positive region
Add the three absolute contributions
Compare with the ordinary signed integral
Why taking one absolute value at the end fails
A shorter correction formula when the negative region is known
The between-roots area of any quadratic
Check the general quadratic formula against the example
A shorter interval may contain only one root
An interval entirely between two roots
A reflected quadratic has the same absolute area
A repeated root does not necessarily change sign
How a parameter changes the sign pattern
Calculate the signed integral for the parameter family
The case k≥4 needs no sign splitting
The case k≤0 lies below the axis on the interval
The crossing case 0
Obtain the complete parameter-area formula
Verify the parameter formula at k=3
An absolute-value integral can be zero only in a special case
The same method applies to total distance
Choose a calculator check without replacing the proof
A concise examination routine
Practice questions with checked answers
Common mistakes and first-error repairs
A symmetry method gives the total area in one shorter calculation
Why the symmetry shortcut is legitimate
Average magnitude is another useful quantity
A general absolute-area strategy for several roots
A quick numerical bound checks the total area
Syllabus fit and connected reading