Definite Integrals: Part 2
Signed area, units, applications and velocity-time graphs.
On this page
- Interpreting a definite integral as signed area
- Positive signed area
- When the curve crosses the axis
- Worked examples: negative area and cancellation
- Applications and units
- Hotel net profit example
- Velocity-time graph example
- Calculator/GDC workflow
- Optional extension: rectangles to the integral
- Practice questions
1Interpreting a definite integral as signed area
In Part 1, the definite integral was connected to accumulation over an interval. Graphically, this accumulation is best understood as signed area.
For a continuous function \(f\) on \([a,b]\),
\[ \int_a^b f(x)\,dx \]is the sum of the signed areas between the graph of \(y=f(x)\) and the \(x\)-axis from \(x=a\) to \(x=b\).
- If the graph is above the \(x\)-axis, the contribution is positive.
- If the graph is below the \(x\)-axis, the contribution is negative.
When the graph stays above the \(x\)-axis, the definite integral matches the ordinary area under the curve.
2Worked example: positive signed area
Worked example 1
Evaluate and interpret \(\displaystyle \int_0^2 (x^2+1)\,dx\).
The graph lies entirely above the \(x\)-axis on \([0,2]\), so every signed contribution is positive.
Answer \(\displaystyle \int_0^2 (x^2+1)\,dx=\frac{14}{3}\). The signed area is positive because the graph is above the \(x\)-axis throughout the interval.
3What changes when the curve crosses the axis?
If a graph crosses the \(x\)-axis, some signed contributions are positive and some are negative. The definite integral gives their signed sum:
\[ \int_a^b f(x)\,dx=\text{positive contributions}+\text{negative contributions}. \]Here, the outer regions contribute positively, while the middle region contributes negatively.
This page focuses on definite integrals and signed areas. A definite integral may be positive, zero or negative because signed contributions can cancel.
Total geometrical area is a different question: every region is counted positively. That is treated separately on area-under-curve and area-between-curves pages.
4More worked examples: negative area and cancellation
Example 2: a negative signed area
Evaluate \(\displaystyle \int_0^4 (x-3)\,dx\) and interpret the answer.
The integral is negative because the negative contribution from \(0\) to \(3\) is larger than the positive contribution from \(3\) to \(4\).
\[ \text{Signed area}=\frac12(1)(1)-\frac12(3)(3)=\frac12-\frac92=-4. \]Answer The signed area is \(-4\).
Example 3: signed area when the graph crosses the axis
Evaluate and interpret \(\displaystyle \int_{-2}^2 (x^2-1)\,dx\).
The graph crosses the \(x\)-axis when
\[ x^2-1=0\quad\Rightarrow\quad x=-1,\,1. \] \[ \begin{aligned} \int_{-2}^2(x^2-1)\,dx &=\left[\frac{x^3}{3}-x\right]_{-2}^2\ &=\left(\frac83-2\right)-\left(-\frac83+2\right)\ &=\frac43. \end{aligned} \]Answer The signed area is \(\frac43\). The negative contribution between \(-1\) and \(1\) cancels part of the positive contribution from the two outer regions.
Example 4: cancellation can give a zero integral
Explain why \(\displaystyle \int_{-2}^2 x^3\,dx=0\) does not mean that the function has no positive or negative contributions.
The function \(x^3\) is below the \(x\)-axis on \([-2,0]\) and above the \(x\)-axis on \([0,2]\).
\[ \int_{-2}^2 x^3\,dx=\left[\frac{x^4}{4}\right]_{-2}^2=4-4=0. \]The negative contribution on \([-2,0]\) exactly cancels the positive contribution on \([0,2]\).
Conclusion The integral is zero because signed contributions cancel, not because nothing is happening on the graph.
5Applications of definite integrals and signed area
If the vertical axis has units \(U_y\) and the horizontal axis has units \(U_x\), then the units of area on the graph are
\[ U_y\times U_x. \]So, in an applications question, the definite integral is not automatically interpreted as “square units”; it is interpreted as the product of the axis units.
| Vertical-axis units | Horizontal-axis units | Units of signed area | Contextual interpretation |
|---|---|---|---|
| \(\$ / \text{day}\) | days | \(\$\) | net profit / net loss |
| Euros / hour | hours | Euros | net money earned / lost |
| metres / second | seconds | metres | displacement |
| litres / minute | minutes | litres | net volume added / removed |
| people / hour | hours | people | net change in number of people |
Cashflow rate
Let \(r(t)\) be measured in dollars per day, and let \(t\) be measured in days.
\(\left(\frac{\$}{\text{day}}\right)(\text{day})=\$\). The signed area represents total money in minus total money out.
Earnings rate
Let \(e(t)\) be measured in Euros per hour, and let \(t\) be measured in hours.
\(\left(\frac{\text{Euros}}{\text{hour}}\right)(\text{hour})=\text{Euros}\). The signed area represents earnings gained minus deductions or losses.
Velocity-time graph
Let \(v(t)\) be measured in metres per second, and let \(t\) be measured in seconds.
\(\left(\frac{m}{s}\right)(s)=m\). The signed area represents displacement: forward movement minus backward movement.
In every context above, regions above the horizontal axis represent positive contributions, while regions below it represent negative contributions. The signed area tells us the net change in the quantity being accumulated.
6Example 5: a hotel’s net profit over 12 months
Application example
Suppose \(p(t)\) is a hotel’s profit rate, measured in thousand Euros per month, where \(t\) is time in months and \(p(t)=3\sin\left(\frac{\pi t}{6}\right)\), \(0\leq t\leq 12\). Interpret \(\displaystyle \int_0^{12}p(t)\,dt\).
The units of the signed area are
\[ \left(\frac{\text{thousand Euros}}{\text{month}}\right)(\text{month})=\text{thousand Euros}. \]So the integral represents the hotel’s net profit over the year.
\[ \begin{aligned} \int_0^{12} p(t)\,dt &=\int_0^{12} 3\sin\left(\frac{\pi t}{6}\right)\,dt\ &=\left[-\frac{18}{\pi}\cos\left(\frac{\pi t}{6}\right)\right]_0^{12}\ &=0. \end{aligned} \]Interpretation The positive signed area from profitable months exactly cancels the negative signed area from loss-making months. The net profit is \(0\) thousand Euros over the 12 months.
This does not mean “nothing happened”; it means gains and losses balance overall.
7Example 6: velocity-time graph — displacement as signed area
Kinematics example
A particle has velocity \(v(t)=t^2-4t+3\), \(0\leq t\leq 4\). Find its displacement from \(t=0\) to \(t=4\).
Displacement is the signed area under the velocity-time graph:
\[ \begin{aligned} \int_0^4(t^2-4t+3)\,dt &=\left[\frac{t^3}{3}-2t^2+3t\right]_0^4\ &=\frac{64}{3}-32+12\ &=\frac43. \end{aligned} \]The velocity changes sign at
\[ t^2-4t+3=0\quad\Rightarrow\quad t=1,\,3. \]Some parts of the journey contribute positively to displacement, and some contribute negatively.
Answer The displacement is \(\frac43\) metres.
Distance travelled is a total geometrical-area idea: it counts all movement positively and would require \(\int |v(t)|\,dt\).
8Using technology appropriately
- Graph the function and identify where it crosses the horizontal axis inside the interval.
- If asked for a definite integral, compute \(\displaystyle \int_a^b f(x)\,dx\).
- Decide which parts of the graph contribute positively and which parts contribute negatively.
- Interpret the result in context. A signed-area answer can be positive, zero or negative.
If a question asks for total geometrical area enclosed by the curve and the \(x\)-axis, that is a different task. You split at intercepts and count every region positively, often using absolute values.
9Optional extension: from rectangles to the definite integral
This section connects the visual “area under a curve” idea to the formal notation of a definite integral.
Divide \([a,b]\) into \(n\) equal parts. Each rectangle has width
\[\Delta x=\frac{b-a}{n}.\]If \(x_i^*\) is a sample point in the \(i\)-th strip, then the approximate signed area of that rectangle is \(f(x_i^*)\Delta x\).
The total approximate signed area is
\[S_n=\sum_{i=1}^n f(x_i^*)\Delta x.\]This is called a Riemann sum: a sum of the signed areas of rectangles.
As the number of rectangles increases,
\[n\to\infty,\qquad \Delta x=\frac{b-a}{n}\to 0.\]In this limiting process, the small width \(\Delta x\) is replaced by the differential \(dx\), which represents an infinitely small change in \(x\). At the same time, the finite sum symbol \(\sum\) is replaced by the integral sign \(\int\).
finite sum
capital S
for sum
vertically →
integral sign
So the approximation becomes exact:
\[ \int_a^b f(x)\,dx=\lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)\Delta x. \]You can think of the definite integral as what happens when \(\sum f(x_i^*)\Delta x\) is pushed to the limit:
- the number of rectangles tends to infinity, \(n\to\infty\);
- the rectangle width tends to zero, \(\Delta x\to 0\);
- the width \(\Delta x\) is replaced by \(dx\);
- the sigma sign \(\sum\) becomes the integral sign \(\int\).
Read the integral as a sum of infinitely many infinitely thin rectangular strips, each with height \(f(x)\) and base width \(dx\).
10Practice questions
Try these without looking at the answers first. Sketch the sign of the graph whenever possible.
- Evaluate \(\displaystyle \int_0^4 (5-x)\,dx\) and interpret the result as a signed area.
Show solution
\[\int_0^4(5-x)\,dx=\left[5x-\frac{x^2}{2}\right]_0^4=20-8=12.\]The graph is above the \(x\)-axis on \([0,4]\), so the signed area is positive.
- Evaluate \(\displaystyle \int_0^5 (x-3)\,dx\). Explain why the answer is negative.
Show solution
\[\int_0^5(x-3)\,dx=\left[\frac{x^2}{2}-3x\right]_0^5=\frac{25}{2}-15=-\frac52.\]The answer is negative because the negative contribution from \(0\) to \(3\) is larger than the positive contribution from \(3\) to \(5\).
- For \(f(x)=x^2-4\) on \([-3,3]\), find \(\displaystyle \int_{-3}^3(x^2-4)\,dx\) and interpret the sign.
Show solution
\[\int_{-3}^3(x^2-4)\,dx=\left[\frac{x^3}{3}-4x\right]_{-3}^3=(9-12)-(-9+12)=-6.\]The negative signed area between \(-2\) and \(2\) is larger than the two positive outer contributions.
- Find \(k\) if \(\displaystyle \int_0^2(kx+1)\,dx=8\).
Show solution
\[\int_0^2(kx+1)\,dx=\left[\frac{kx^2}{2}+x\right]_0^2=2k+2.\]So \(2k+2=8\), hence \(k=3\).
- Explain why \(\displaystyle \int_0^{2\pi}\sin x\,dx=0\).
Show solution
\[\int_0^{2\pi}\sin x\,dx=\left[-\cos x\right]_0^{2\pi}=-1-(-1)=0.\]The positive contribution on \([0,\pi]\) cancels the negative contribution on \([\pi,2\pi]\).
AI SL note: this exact trigonometric integral is extension for AI SL.
- A velocity is given by \(v(t)=t-2\), \(0\leq t\leq5\). Find the displacement.
Show solution
\[\int_0^5(t-2)\,dt=\left[\frac{t^2}{2}-2t\right]_0^5=\frac{25}{2}-10=\frac52.\]The displacement is \(\frac52\) units of distance.
- Given \(\displaystyle \int_{-2}^2 f(x)\,dx=-3\) and \(\displaystyle \int_{-2}^0 f(x)\,dx=5\), find \(\displaystyle \int_0^2 f(x)\,dx\).
Show solution
Use additivity of definite integrals:
\[\int_{-2}^2 f(x)\,dx=\int_{-2}^0 f(x)\,dx+\int_0^2 f(x)\,dx.\]So \(-3=5+\int_0^2 f(x)\,dx\), hence \(\displaystyle \int_0^2 f(x)\,dx=-8\).
- Find the signed area \(\displaystyle \int_0^4(x-1)(x-3)\,dx\).
Show solution
Expand first:
\[ (x-1)(x-3)=x^2-4x+3. \]Now integrate:
\[ \begin{aligned} \int_0^4 (x-1)(x-3)\,dx &=\int_0^4 (x^2-4x+3)\,dx\\ &=\left[\frac{x^3}{3}-2x^2+3x\right]_0^4\\ &=\left(\frac{64}{3}-32+12\right)-0\\ &=\frac{64}{3}-20\\ &=\frac{4}{3}. \end{aligned} \]Answer The signed area is \(\displaystyle \frac{4}{3}\).