This note covers definite integrals and their main use: finding areas. You will see how the limits turn an anti-derivative into a number, how areas below the axis are treated, and how to find the area between two curves. These skills also give distance, total change and many modelling answers.
Key ideas
A definite integral has limits. If \(F'(x)=f(x)\), then \[\int_a^b f(x)\,dx=F(b)-F(a).\] The constant \(C\) cancels, so it is not needed. Write the result in square brackets first, \([F(x)]_a^b\), then substitute the upper limit and subtract the value at the lower limit.
Meaning. For a curve with \(f(x)\ge 0\) on \(a\le x\le b\), the definite integral is the area between the curve, the \(x\)-axis and the lines \(x=a\) and \(x=b\). Where the curve is below the \(x\)-axis, the integral counts that part as negative. So a definite integral is a signed area.
Properties.
- Swapping the limits changes the sign: \(\int_b^a f\,dx=-\int_a^b f\,dx\).
- Equal limits give zero: \(\int_a^a f\,dx=0\).
- Intervals can be joined: \(\int_a^b f\,dx+\int_b^c f\,dx=\int_a^c f\,dx\).
- Constants and sums behave as usual: \(\int_a^b kf\,dx=k\int_a^b f\,dx\).
Area between a curve and the \(x\)-axis. Find where the curve crosses the axis. Split the interval at each crossing. Integrate each part, take the size of each answer, and add. Equivalently, the area is \(\int_a^b|f(x)|\,dx\). Your GDC can evaluate this with the definite integral feature if you enter the function with its absolute value.
Area between two curves. If \(f(x)\ge g(x)\) on \(a\le x\le b\), then \[\text{Area}=\int_a^b\big(f(x)-g(x)\big)\,dx.\] This works whether the curves are above or below the axis, because only the vertical gap matters. If the curves cross inside the interval, split at each crossing so that the "upper" curve is always the same one in each piece. The limits are often the \(x\)-coordinates of the intersection points. Find these by solving \(f(x)=g(x)\), or with the intersection feature on your GDC.
Substitution in a definite integral. When you substitute \(u=g(x)\), you can change the limits too. The new limits are \(g(a)\) and \(g(b)\). Then you never need to return to \(x\). This is cleaner and avoids errors.
Unknown limits or constants. Some questions give the area and ask for a limit or a constant. Evaluate the definite integral in terms of the unknown, set it equal to the given value, then solve the resulting equation. For instance, if \(\int_1^k\dfrac1x\,dx=3\), then \(\ln k-\ln 1=3\), so \(k=e^3\approx 20.1\). When the equation is hard to solve by hand, plot the left side as a graph on your GDC and use the intersection feature.
Areas with a line or the \(y\)-axis. Boundaries such as \(x=a\) are vertical lines and simply provide limits. A boundary on the \(y\)-axis means the lower limit is \(x=0\). Always say in words which curve is on top and which limits you use before integrating.
Total change. Because \(\int_a^b f'(x)\,dx=f(b)-f(a)\), integrating a rate of change over an interval gives the total change in the quantity over that interval.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(\displaystyle\int_a^b f(x)\,dx=F(b)-F(a)\) | Definite integral. \(F\) is any anti-derivative of \(f\). |
| \(\displaystyle\int_a^b |f(x)|\,dx\) | Area between the curve and the \(x\)-axis, all counted as positive. |
| \(\displaystyle\int_a^b \big(f(x)-g(x)\big)\,dx\) | Area between two curves, with \(f\ge g\) on the interval. |
| \(\displaystyle\int_a^b f\,dx=-\int_b^a f\,dx\) | Reversing the limits reverses the sign. |
| \(\displaystyle\int_a^c f\,dx=\int_a^b f\,dx+\int_b^c f\,dx\) | Splitting an interval at \(b\). |
| \(u=g(x)\): limits become \(g(a),g(b)\) | Definite integral by substitution. |
Worked examples
Example 1 A landscape designer models the edge of a flower bed by the curves \(y=6x-x^2\) and \(y=x^2\), with \(x\) and \(y\) in metres. Find the area enclosed between the curves.
Find the intersections: \(6x-x^2=x^2\), so \(6x-2x^2=0\), so \(2x(3-x)=0\). Hence \(x=0\) or \(x=3\).
Between these, test \(x=1\): \(6(1)-1=5\) and \(1^2=1\). So \(y=6x-x^2\) is the upper curve.
Area \(=\displaystyle\int_0^3\big((6x-x^2)-x^2\big)\,dx=\int_0^3(6x-2x^2)\,dx\).
\(=\left[3x^2-\dfrac{2x^3}{3}\right]_0^3=\left(27-18\right)-0=9\).
Answer: 9 m\(^2\).
Example 2 Find (a) \(\displaystyle\int_{-2}^{2}(x^3-4x)\,dx\) and (b) the total area between the curve \(y=x^3-4x\) and the \(x\)-axis for \(-2\le x\le 2\).
(a) An anti-derivative is \(\dfrac{x^4}{4}-2x^2\). At \(x=2\): \(4-8=-4\). At \(x=-2\): \(4-8=-4\). The integral is \(-4-(-4)=0\).
(b) The positive and negative parts cancel in part (a), so we must treat them separately. Factorise: \(x^3-4x=x(x-2)(x+2)\), so the curve crosses the axis at \(-2\), \(0\), \(2\).
On \(0\le x\le 2\): \(\displaystyle\int_0^2(x^3-4x)\,dx=(4-8)-0=-4\). The curve is below the axis, so this area is 4.
On \(-2\le x\le 0\): \(\displaystyle\int_{-2}^0(x^3-4x)\,dx=0-(-4)=4\). The curve is above the axis, so this area is 4.
Total area \(=4+4=8\). You can confirm this on your GDC by integrating \(|x^3-4x|\) from \(-2\) to \(2\).
Answer: (a) 0; (b) 8 square units.
Example 3 Evaluate \(\displaystyle\int_0^2\frac{x}{x^2+1}\,dx\), giving your answer to 3 s.f.
Let \(u=x^2+1\). Then \(du=2x\,dx\), so \(x\,dx=\tfrac12\,du\).
Change the limits: when \(x=0\), \(u=1\). When \(x=2\), \(u=5\).
The integral becomes \(\displaystyle\int_1^5\frac{1}{u}\cdot\frac12\,du=\frac12\big[\ln u\big]_1^5=\frac12(\ln5-\ln1)=\frac12\ln5\).
\(\tfrac12\ln5=0.8047\ldots\)
Answer: \(\tfrac12\ln5\approx 0.805\).
Common errors
- Common error: treating a definite integral as an area when the curve goes below the axis. The signed integral can be zero or negative.
- Common error: subtracting in the wrong order, \(F(a)-F(b)\), which gives the wrong sign.
- Common error: in "area between curves", subtracting the lower from the upper curve incorrectly, or using \(\int f-\int g\) with different limits. Subtract the functions first, then integrate once.
- Common error: changing to \(u\) but keeping the original \(x\)-limits.
- Common error: adding \(+C\) to a definite integral. It cancels, so leave it out.
Exam tip: Sketch the curves on your GDC first. It shows you the limits, which curve is on top, and where a curve crosses the axis. Then use the graph to check that your final area is sensible in size and positive.
Check yourself
- Given \(\int_1^5 f(x)\,dx=7\) and \(\int_1^3 f(x)\,dx=2\), find \(\int_3^5 f(x)\,dx\) and \(\int_5^1 f(x)\,dx\).
Answer
\(\int_3^5 f\,dx=7-2=5\). Reversing the limits, \(\int_5^1 f\,dx=-7\). - Evaluate \(\displaystyle\int_0^1 e^{2x}\,dx\) to 3 s.f.
Answer
\(\left[\tfrac12e^{2x}\right]_0^1=\tfrac12(e^2-1)=\tfrac12(7.389-1)=3.19\). - Find the area enclosed by \(y=x^2\) and \(y=2x+3\).
Answer
\(x^2=2x+3\) gives \(x^2-2x-3=0\), so \(x=-1\) or \(x=3\). The line is above the parabola between them. Area \(=\int_{-1}^3(2x+3-x^2)\,dx=\left[x^2+3x-\tfrac{x^3}{3}\right]_{-1}^3=9-\left(-\tfrac53\right)=\tfrac{32}{3}\approx 10.7\). - Use the substitution \(u=\cos x\) to evaluate \(\displaystyle\int_0^{\pi/2}\sin x\cos^2x\,dx\).
Answer
\(du=-\sin x\,dx\). Limits: \(x=0\to u=1\), \(x=\tfrac\pi2\to u=0\). The integral is \(\int_1^0-u^2\,du=\int_0^1u^2\,du=\tfrac13\).