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International Baccalaureate · DP 2: Analysis and Approaches SL

Definite Integrals

⌚ ~106 min ▢ 11 blocks ✓ 50 exercises ➣ Prerequisites : Introduction to Integration, Equations
I Techniques of Integration
I.1 Integration by Inspection
Method — Integration by Inspection
A common strategy is to make an educated guess for the antiderivative, differentiate it, and adjust any constant factors. This is often called “integration by inspection” or “guess and check”.
  1. Guess: Look at the function and guess a possible antiderivative.
  2. Differentiate: Differentiate the guess.
  3. Adjust: Compare the result with the integrand. If the result differs from the integrand by a constant factor, divide the guess by this factor; otherwise, change the guess.
Example
Find the integral of \(\sin(3x)\).

  1. Guess: The integral of \(\sin\) is \(-\cos\). A good guess for the antiderivative is \(F(x) = -\cos(3x)\).
  2. Differentiate: Using the chain rule, the derivative of our guess is \(F'(x) = -(-\sin(3x)) \cdot 3 = 3\sin(3x)\).
  3. Adjust: Our result, \(3\sin(3x)\), is 3 times larger than the original integrand, \(\sin(3x)\). We must therefore divide our initial guess by 3.
The correct antiderivative is \(-\dfrac{1}{3}\cos(3x)\). So, $$ \int \sin(3x) \, \mathrm{d}x = -\frac{1}{3}\cos(3x) + C $$

Example
Find \(\displaystyle\int x\cos(x^2)\,\mathrm{d}x\).

The integrand is not of the form \(f(ax + b)\): the inner function \(x^2\) is not linear. Its derivative, \(2x\), appears in the integrand, up to a constant factor.
  1. Guess: An antiderivative of \(\cos\) is \(\sin\), so a first guess is \(F(x) = \sin(x^2)\).
  2. Differentiate: By the chain rule, \(F'(x) = \cos(x^2) \times 2x = 2x\cos(x^2)\).
  3. Adjust: The result, \(2x\cos(x^2)\), is twice the integrand \(x\cos(x^2)\): the guess is divided by \(2\).
Check: $$\begin{aligned}[t] \left(\frac{1}{2}\sin(x^2)\right)' &= \frac{1}{2} \times \cos(x^2) \times 2x &&\text{(chain rule)}\\ &= x\cos(x^2). \end{aligned}$$ So $$ \int x\cos(x^2) \, \mathrm{d}x = \frac{1}{2}\sin(x^2) + C. $$ For \(\cos(x^2)\) alone, the method fails: the derivative of \(\sin(x^2)\) is \(2x\cos(x^2)\), which differs from \(\cos(x^2)\) by the factor \(2x\), and this factor is not a constant.

Skills to practice
  • Finding Antiderivatives by Inspection
I.2 Integration by Substitution
Integration by substitution is a powerful technique that reverses the chain rule for differentiation. It is used when an integrand contains both a function and its derivative.
Consider the integral \(\int (2x^3+1)^7 (6x^2) \, \mathrm{d}x\). We can see that the expression contains an “inner function”, \(u = 2x^3+1\), and its derivative, \(u'(x) = 6x^2\). This structure is a clear indicator for substitution.
We define a new variable, \(u = 2x^3+1\). Differentiating this with respect to \(x\) gives \(u'(x)=6x^2\). In differential form, we can write \(\mathrm{d}u = 6x^2\,\mathrm{d}x\). Now we can substitute both \(u\) and \(\mathrm{d}u\) into the original integral: $$\begin{aligned} \int \underbrace{(2x^3+1)^{7}}_{u^{7}} \underbrace{(6x^2)\,\mathrm{d}x}_{\mathrm{d}u} &= \int u^{7}\,\mathrm{d}u \\ &= \frac{1}{8}u^{8}+C\end{aligned}$$ Finally, we substitute back to express the result in terms of \(x\): $$ \int (2x^3+1)^7 (6x^2) \, \mathrm{d}x =\frac{1}{8}(2x^3+1)^{8}+C $$
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Proposition — Integration by Substitution
Let \(u\) be differentiable on an interval, with \(u'\) continuous, and let \(f\) be continuous on an interval that contains the values of \(u\).
  • Indefinite Integral: $$\int f(u(x))u'(x) \, \mathrm{d}x = \int f(u) \, \mathrm{d}u $$
  • Definite Integral: $$\int_a^b f(u(x))u'(x) \, \mathrm{d}x = \int_{u(a)}^{u(b)} f(u) \, \mathrm{d}u $$

Let \(F\) be an antiderivative of \(f\). By the chain rule, $$\begin{aligned}[t] \big(F(u(x))\big)' &= F'(u(x)) \times u'(x) &&\text{(chain rule)}\\ &= f(u(x))\,u'(x). &&(F' = f) \end{aligned}$$ So \(F \circ u\) is an antiderivative of \(f(u(x))\,u'(x)\), which is the first formula; by the fundamental theorem of calculus, both sides of the second formula are equal to \(F(u(b)) - F(u(a))\).

Example
Find \(\displaystyle\int_0^1 2x(x^2 + 1)^3\;\mathrm{d}x\).

  • Substitution: Let \(u = x^2+1\). Then \(\mathrm{d}u = 2x\,\mathrm{d}x\).
  • Change limits:
    • When \(x=0\), \(u = 0^2+1=1\).
    • When \(x=1\), \(u = 1^2+1=2\).
  • Integrate: We substitute to get a simpler integral in terms of \(u\): $$\begin{aligned} \int_0^1 \underbrace{(x^2 + 1)^3}_{u^3} \underbrace{2x\;\mathrm{d}x}_{\mathrm{d}u} &= \int_1^2 u^3 \, \mathrm{d}u \\ &= \left[\frac{u^4}{4}\right]_1^2 \\ &= \frac{2^4}{4} - \frac{1^4}{4} \\ &= 4 - \frac{1}{4}\\ &= \frac{15}{4} \end{aligned}$$

Skills to practice
  • Integrating by Substitution
  • Evaluating Definite Integrals by Substitution
II Areas
II.1 The Area of a Region
The definite integral \(\int_a^b f(x)\,\mathrm{d}x\) is the signed area between the graph of \(y=f(x)\) and the \(x\)-axis. This means that areas above the x-axis are positive, while areas below are negative.
However, when a problem asks for the geometric area or simply the area of a region, it refers to the physical, positive space the region occupies. In this case, we must ensure that all parts of the region, whether they are above or below the x-axis, contribute a positive value to the total. This total geometric area needs its own method.
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Method — Calculating Geometric Area Bounded by a Curve and the x-axis
To find the total geometric area \(\mathcal{A}\) bounded by a curve \(y=f(x)\) and the x-axis from \(x=a\) to \(x=b\) (\(f\) continuous on \([a, b]\), with finitely many roots in this interval):
  1. Find the \(x\)-intercepts: Determine where the curve meets the \(x\)-axis (the zeros of \(f\)) by solving \(f(x)=0\). Identify any roots \(c_1, c_2, \dots\) that lie within the interval \([a,b]\).
  2. Split the integral: Divide the main integral into smaller integrals at each intercept found in the previous step: on each sub-interval, \(f\) keeps a constant sign (it is continuous and does not vanish inside).
  3. Calculate each definite integral: Compute the integral for each sub-interval. Some of these will correspond to positive values (where \(f(x) \ge 0\)) and some to negative values (where \(f(x) \le 0\)).
  4. Sum the absolute values: The total geometric area is the sum of the absolute values of these integrals. If the integral of a sub-region is negative, take its positive value before adding it to the total.
Example
Find the total geometric area between the curve \(y=x^2-1\) and the \(x\)-axis from \(x=0\) to \(x=2\).

  1. Find intercepts: We solve \(f(x)=x^2-1=0\), which gives roots at \(x=1\) and \(x=-1\). The only root within our interval of integration \([0,2]\) is \(x=1\).
  2. Split the integral: We must split the total area calculation at \(x=1\).
    • From \(x=0\) to \(x=1\), the function is below the \(x\)-axis (Area \(\mathcal{A}_1\)).
    • From \(x=1\) to \(x=2\), the function is above the \(x\)-axis (Area \(\mathcal{A}_2\)).
    Figure
  3. Calculate each integral: $$ \int_0^1 (x^2-1)\, \mathrm{d}x = \left[\frac{x^3}{3}-x\right]_0^1 = \left(\frac{1}{3}-1\right) - 0 = -\frac{2}{3} $$ $$ \int_1^2 (x^2-1)\, \mathrm{d}x = \left[\frac{x^3}{3}-x\right]_1^2 = \left(\frac{8}{3}-2\right) - \left(\frac{1}{3}-1\right) = \frac{2}{3} - \left(-\frac{2}{3}\right) = \frac{4}{3} $$
  4. Sum the absolute values: $$ \text{Total Area} = \left|-\frac{2}{3}\right| + \left|\frac{4}{3}\right| = \frac{2}{3} + \frac{4}{3} = \frac{6}{3} = 2 $$
The total geometric area is 2 square units.

Definition — Geometric Area Formula
The total geometric area \(\mathcal{A}\) between the graph of a function \(f(x)\) and the \(x\)-axis from \(x=a\) to \(x=b\) is given by the integral of the absolute value of the function: $$ \mathcal{A} = \int_a^b |f(x)|\, \mathrm{d}x $$
Note
The method of splitting the integral and summing the absolute values is equivalent to this formal definition of the total geometric area.
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Skills to practice
  • Evaluating Areas with Geometric Formulas
  • Finding Areas with Antiderivatives
II.2 The Area Between Two Curves
We can extend the concept of finding the area under a single curve to find the area of a region enclosed between two different curves. The area under a curve \(y=f(x)\) is a special case where the second curve is simply the \(x\)-axis, \(y=0\).
The general idea is intuitive: when \(0 \le g(x) \le f(x)\) on \([a, b]\), the area of the region bounded by the upper function \(f(x)\) and the lower function \(g(x)\) is the area under \(f(x)\) minus the area under \(g(x)\). By linearity, this difference is a single integral, and the formula remains valid whenever \(g \le f\) on \([a, b]\), even when the curves go below the \(x\)-axis, since it is the area under the non-negative function \(f - g\): $$\begin{aligned} \textcolor{olive}{\mathcal{A}} &= \textcolor{colordef}{ \int_a^b f(x)\; \mathrm{d}x} - \textcolor{colorprop}{ \int_a^b g(x)\; \mathrm{d}x} \\ &= \int_a^b (f(x)-g(x))\; \mathrm{d}x \end{aligned} $$
Figure
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Proposition — Area Between Curves
Let \(f\) and \(g\) be continuous on \([a,b]\). If \(f(x) \geq g(x)\) for all \(x\) in \([a,b]\), the area \(\mathcal{A}\) of the region bounded by the curves \(y=f(x)\) and \(y=g(x)\) and the lines \(x=a\) and \(x=b\) is given by: $$ \mathcal{A} = \int_a^b (f(x)-g(x))\; \mathrm{d}x $$
This can be remembered as the integral of the upper function minus the lower function.
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Method — Calculating the Area Between Two Curves
To find the total geometric area enclosed by two curves \(y=f(x)\) and \(y=g(x)\) (\(f\) and \(g\) continuous, with finitely many intersection points on the interval considered):
  1. Find points of intersection. Solve the equation \(f(x)=g(x)\) to find where the curves meet: these points give the bounds \(a\) and \(b\) when they are not given and, when they are, the points inside \([a, b]\) where the upper and lower functions may swap. Choose the consecutive intersection points that bound the region sought.
  2. Identify the upper and lower function. On each interval between intersection points, determine which function is on top. Between two consecutive intersection points, \(f - g\) is continuous and does not vanish, so it keeps a constant sign: testing one point is enough (a quick sketch helps).
  3. Set up and evaluate the integral(s). On each interval \([c, d]\) between consecutive intersection points or bounds, calculate \(\int_c^d (\text{upper function} - \text{lower function}) \, \mathrm{d}x\), which is non-negative. The total area is the sum of these integrals.
Example
Find the area of the region enclosed by the curves \(y=x+1\) and \(y=x^2-1\).

  1. Find intersection points: We set the two functions equal to each other: $$ x+1 = x^2-1 \iff x^2-x-2=0 \iff(x-2)(x+1)=0 $$ The curves intersect at \(x=-1\) and \(x=2\). These are our limits of integration.
  2. Identify the upper function: The curves do not meet between \(x=-1\) and \(x=2\), so the difference \((x+1)-(x^2-1)\), which is continuous, keeps a constant sign on \((-1, 2)\): testing one point is enough. At \(x=0\): for \(y=x+1\), \(y(0)=1\); for \(y=x^2-1\), \(y(0)=-1\). Since \(1 > -1\), the line \(y=x+1\) is the upper function on \([-1,2]\).
    Figure
  3. Set up and evaluate: $$\begin{aligned} \mathcal{A} &= \int_{-1}^2 \left( (x+1) - (x^2-1) \right)\, \mathrm{d}x \\ &= \int_{-1}^2 (-x^2+x+2)\, \mathrm{d}x \\ &= \left[ -\frac{x^3}{3} + \frac{x^2}{2} + 2x \right]_{-1}^2 \\ &= \left(-\frac{8}{3} + \frac{4}{2} + 4\right) - \left(\frac{1}{3} + \frac{1}{2} - 2\right) \\ &= \left(-\frac{16}{6} + \frac{12}{6} + \frac{24}{6}\right) - \left(\frac{2}{6} + \frac{3}{6} - \frac{12}{6}\right) \\ &= \frac{20}{6} - \left(-\frac{7}{6}\right) \\ &= \frac{27}{6} \\ &= \frac{9}{2} \end{aligned}$$
The area of the enclosed region is \(\frac{9}{2}\) square units.

Skills to practice
  • Calculating the Area Enclosed Between Two Curves
III Integration with Technology
III.1 Definite Integrals with Technology
Method — Evaluating a Definite Integral with Technology
A graphing calculator evaluates a definite integral \(\displaystyle\int_a^b f(x)\,\mathrm{d}x\) numerically, even when no formula gives its value. It offers two tools: the integral template of the calculation screen, where the integrand and the two bounds are typed, and the integral tool of the graph screen, which asks for the lower bound, then the upper bound, and shades the region. Both return a decimal approximation of the signed area.
  1. Write the integral first, on paper: the integrand, the bounds and the differential, \(\displaystyle\int_a^b f(x)\,\mathrm{d}x\). This expression is the method; the calculator only evaluates it. For an area, a sketch first shows where the curve lies above or below the \(x\)-axis between \(a\) and \(b\).
  2. Set the angle unit to radians when \(f(x)\) contains \(\sin\), \(\cos\) or \(\tan\).
  3. Enter the integral: the integrand, written with the variable \(X\) of the calculator, then the lower bound \(a\) and the upper bound \(b\), in this order.
  4. Report the value to 3 significant figures, unless the question asks for another accuracy.
  5. Interpret the sign: the value is a signed area. It is the area of the region only when \(f(x) \ge 0\) on \([a, b]\); when \(f(x) \le 0\) on the whole of \([a, b]\), the area is \(-\displaystyle\int_a^b f(x)\,\mathrm{d}x\). When the curve crosses the \(x\)-axis between \(a\) and \(b\), the area of the region is \(\displaystyle\int_a^b |f(x)|\,\mathrm{d}x\), evaluated with the absolute-value function of the calculator; alternatively, the zeros of \(f\) in \([a, b]\) are found with the zero tool, \([a, b]\) is split at these zeros, and the absolute values of the integrals over the pieces, from \(a\) to the first zero and from the last zero to \(b\) included, are added.
The mock-up evaluates \(\displaystyle\int_0^2 \sqrt{x^3+1}\,\mathrm{d}x\) with each tool; the window of the graph screen is \(-1 \le x \le 3\), \(-1 \le y \le 5\).
Figure
Example
The region \(R\) is bounded by the curve \(y = \sqrt{x^3+1}\), the \(x\)-axis and the lines \(x = 0\) and \(x = 2\).
  1. Write down an integral that gives the area of \(R\).
  2. Find the area of \(R\) with technology.
  3. Check that this value is plausible, with a rectangle that lies under the curve and a rectangle that contains \(R\).

  1. For \(0 \le x \le 2\), \(x^3 + 1 \ge 1\), so \(\sqrt{x^3+1} \ge 1 > 0\): the curve lies above the \(x\)-axis, and the area of \(R\) is $$\mathcal{A} = \int_0^2 \sqrt{x^3+1}\,\mathrm{d}x.$$
  2. The integral template, with the integrand \(\sqrt{X^3+1}\) and the bounds \(0\) and \(2\), displays \(3.2413093\); the integral tool of the graph screen, with the lower bound \(0\) and the upper bound \(2\), displays the same value. The area of \(R\) is \(\mathcal{A} \approx 3.24\) square units.
  3. On \([0, 2]\), \(x^3 + 1\) increases, so \(f(x) = \sqrt{x^3+1}\) increases from \(f(0) = 1\) to \(f(2) = \sqrt{9} = 3\). The rectangle of width \(2\) and height \(f(0)\) lies under the curve, and the rectangle of width \(2\) and height \(f(2)\) contains \(R\): $$\begin{aligned}[t] 2 \times f(0) &\le \mathcal{A} \le 2 \times f(2) &&\text{(one rectangle below, one rectangle above)}\\ 2 &\le \mathcal{A} \le 6. \end{aligned}$$ The value \(3.24\) lies between \(2\) and \(6\).

Example
Let \(f(x) = x^3 - 4x\).
  1. Evaluate \(\displaystyle\int_{-1}^{3} f(x)\,\mathrm{d}x\) with technology.
  2. Find the area of the region bounded by the curve of \(f\), the \(x\)-axis and the lines \(x = -1\) and \(x = 3\).
  3. Explain why the two values differ.

  1. The integral template, with the integrand \(X^3 - 4X\) and the bounds \(-1\) and \(3\), displays \(4\): to 3 significant figures, $$\int_{-1}^{3} f(x)\,\mathrm{d}x \approx 4.00.$$
  2. The zero tool, in the window \(-2 \le x \le 3.5\), \(-5 \le y \le 16\), displays the zeros \(-2\), \(0\) and \(2\); the factorisation \(f(x) = x(x - 2)(x + 2)\) confirms them. In \([-1, 3]\) the curve crosses the \(x\)-axis at \(0\) and at \(2\): it lies above the axis on \([-1, 0]\) and on \([2, 3]\), below it on \([0, 2]\).
    Figure
    For the three pieces, the template displays \(1.75\), \(-4\) and \(6.25\): \(\displaystyle\int_{-1}^{0} f(x)\,\mathrm{d}x \approx 1.75\), \(\displaystyle\int_{0}^{2} f(x)\,\mathrm{d}x \approx -4\) and \(\displaystyle\int_{2}^{3} f(x)\,\mathrm{d}x \approx 6.25\). Each part counts positively in the area: $$\begin{aligned}[t] \mathcal{A} &= \left|\int_{-1}^{0} f(x)\,\mathrm{d}x\right| + \left|\int_{0}^{2} f(x)\,\mathrm{d}x\right| + \left|\int_{2}^{3} f(x)\,\mathrm{d}x\right| &&\text{(split at the zeros)}\\ &\approx 1.75 + 4 + 6.25 &&\text{(values displayed)}\\ &\approx 12.0. \end{aligned}$$ The template with the integrand \(|X^3 - 4X|\) displays \(12\) directly. The area of the region is about \(12.0\) square units.
  3. The integral is a signed area: the part below the axis, on \([0, 2]\), counts negatively, so \(1.75 - 4 + 6.25 = 4\). The area counts every part positively: \(1.75 + 4 + 6.25 = 12\).

Skills to practice
  • Setting Up an Integral for the Calculator
  • Evaluating Integrals and Areas with Technology
III.2 The Area Between Two Curves with Technology
Method — Finding the Area Between Two Curves with Technology
When the points where two curves meet cannot be found by algebra, or when no antiderivative of \(f - g\) is known, a graphing calculator finds the area of the region enclosed by the curves \(y = f(x)\) and \(y = g(x)\).
  1. Graph both functions, \(Y_1 = f(x)\) and \(Y_2 = g(x)\), in a window that shows the whole region.
  2. Find every point of intersection with the intersection tool, as in the Method “Solving an Equation Graphically with Technology”, and store each \(x\)-coordinate in a memory of the calculator. The bounds keep all their digits: the area is rounded once, at the end. A window alone does not show that no point of intersection is missing: check their number with the degree of \(f(x) - g(x)\) when it is a polynomial, or with the behaviour of the curves outside the window.
  3. Write the area before evaluating it. Between two consecutive points of intersection \(c\) and \(d\), the area of the part is \(\displaystyle\int_c^d \big(\text{upper function} - \text{lower function}\big)\,\mathrm{d}x\), the upper function being read on the graph; the area of the region is the sum of these parts. Equivalently, the area is \(\displaystyle\int_a^b |f(x) - g(x)|\,\mathrm{d}x\), where \(a\) and \(b\) are the first and the last points of intersection.
  4. Evaluate with the stored bounds, and report the area to 3 significant figures. The integral \(\displaystyle\int_a^b \big(f(x) - g(x)\big)\,\mathrm{d}x\), taken across a crossing, is not the area: it subtracts the parts where \(g\) is above \(f\).
The mock-up shows the region enclosed by \(y = x^3 - 3x\) and \(y = \frac{1}{2}x + \frac{1}{2}\), in the window \(-2.5 \le x \le 2.5\), \(-3 \le y \le 3\): the last point of intersection found, then its three \(x\)-coordinates stored as \(A\), \(B\), \(C\) and the area evaluated in one integral.
Figure
Example
Find the area of the region enclosed by the curves \(y = x^3 - 3x\) and \(y = \dfrac{1}{2}x + \dfrac{1}{2}\).

  1. Points of intersection. In the window \(-2.5 \le x \le 2.5\), \(-3 \le y \le 3\), the line meets the cubic three times. The intersection tool displays \(X = -1.7948321\), \(X = -0.1437050\) and \(X = 1.9385372\), stored as \(A\), \(B\) and \(C\). The equation \(x^3 - 3x = \frac{1}{2}x + \frac{1}{2}\) has degree \(3\), so it has at most three solutions: none is missing.
  2. Upper function. On \((A, B)\), at \(x = -1\) the cubic gives \(2\) and the line gives \(0\): the cubic is above. On \((B, C)\), at \(x = 1\) the cubic gives \(-2\) and the line gives \(1\): the line is above.
    Figure
  3. Area. The region has two parts, so the area is written as a sum before it is evaluated: $$\begin{aligned}[t] \mathcal{A} &= \int_A^B \Big(x^3 - 3x - \big(\tfrac{1}{2}x + \tfrac{1}{2}\big)\Big)\,\mathrm{d}x + \int_B^C \Big(\tfrac{1}{2}x + \tfrac{1}{2} - (x^3 - 3x)\Big)\,\mathrm{d}x &&\text{(upper minus lower)}\\ &\approx 2.1815022 + 4.0509567 &&\text{(values displayed)}\\ &\approx 6.23. \end{aligned}$$ The single integral \(\displaystyle\int_A^C \left|x^3 - 3x - \big(\tfrac{1}{2}x + \tfrac{1}{2}\big)\right|\mathrm{d}x\) displays the same total, \(6.2324589\). The area is \(\mathcal{A} \approx 6.23\) square units.
Without the split, \(\displaystyle\int_A^C \Big(x^3 - 3x - \big(\tfrac{1}{2}x + \tfrac{1}{2}\big)\Big)\,\mathrm{d}x\) displays \(-1.8694545 = 2.1815022 - 4.0509567\): the second part is subtracted, and the value is not an area.

Skills to practice
  • Finding the Area Between Two Curves with Technology
IV Going further
Skills to practice
  • Solving Problems without Technology
  • Solving Problems with Technology
  • Quiz