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Vector Limits and Calculus · Tutorial 150 of 999

Integrating Piecewise Vector-Valued Functions

You will learn to find the combined breakpoint partition and evaluate each vector integral piece by piece.

Beginner 10 min read

What You'll Learn

  • State when a piecewise vector-valued function has a definite integral.
  • Find the union of the breakpoints in all components.
  • Split an interval into subintervals where one vector formula applies.
  • Integrate each vector piece and add the results.
  • Check component calculations and avoid using a formula across its breakpoint.

Why Breakpoints Matter

A vector-valued function may use different formulas on different parts of an interval. Its components may also change formulas at different parameter values. To find the definite integral correctly, divide the interval wherever any component changes formula, integrate the vector formula on each resulting subinterval, and add those integrals.

This process uses the componentwise definition of a vector integral and the additivity of definite integrals. It does not require the vector function to be continuous at each breakpoint. What matters for the usual definite integral is that its components are integrable on the interval.

Key idea: Collect all component breakpoints inside the integration interval, sort them, and use that common partition for the whole vector.

Definition and Conditions

Let \(\mathbf r:[a,b]\to\mathbb R^n\) be a vector-valued function with components \(r_1,\ldots,r_n\), so that \(\mathbf r(t)=(r_1(t),\ldots,r_n(t))\). If each component is integrable on \([a,b]\), the definite vector integral is defined componentwise:

$$ \int_a^b \mathbf r(t)\,dt = \left( \int_a^b r_1(t)\,dt,\ldots, \int_a^b r_n(t)\,dt \right). $$

A useful sufficient condition is that each component is piecewise continuous: there is a finite partition of \([a,b]\) such that the component is continuous on the open subintervals and has finite one-sided limits at the partition points. Values assigned at finitely many breakpoints do not affect the definite integral. Thus, a jump at a breakpoint does not by itself prevent integration.

To integrate a piecewise vector function, make a single partition that includes every breakpoint of every component. On each open subinterval of this partition, every component has a fixed formula, and therefore the vector has a fixed formula as well. The formula used at a single endpoint may differ from the formula on the adjacent open interval; that endpoint value does not change the integral.

A Procedure for Piecewise Vector Integrals

1
Collect the breakpoints.
List every parameter value inside the integration interval where any component changes formula.
2
Make the common partition.
Sort the breakpoints and include the integration endpoints. These values divide the interval into subintervals.
3
Write the vector formula on each piece.
On every subinterval, use the correct formula for each component.
4
Integrate and add.
Evaluate each vector integral componentwise, then add the resulting vectors.

In symbols, if the partition is \(a=c_0<c_1<\cdots<c_k=b\), additivity gives:

$$ \int_a^b \mathbf r(t)\,dt = \sum_{j=1}^{k}\int_{c_{j-1}}^{c_j}\mathbf r(t)\,dt. $$

The formula for \(\mathbf r(t)\) on \((c_{j-1},c_j)\) is used in the corresponding integral. The endpoints are included in the limits, but the value of the function at any one endpoint does not affect that integral.

Why Splitting Works

The justification comes from two facts: a vector integral is defined componentwise, and each scalar definite integral is additive over adjacent intervals. To see how they combine, let the common partition be \(a=c_0<c_1<\cdots<c_k=b\). For each component \(r_i\), scalar additivity gives:

$$ \int_a^b r_i(t)\,dt = \sum_{j=1}^{k}\int_{c_{j-1}}^{c_j}r_i(t)\,dt. $$

Writing this equation for every component and grouping the components into vectors yields:

$$ \begin{aligned} \int_a^b\mathbf r(t)\,dt &= \left(\int_a^b r_1(t)\,dt,\ldots,\int_a^b r_n(t)\,dt\right)\\ &= \sum_{j=1}^{k} \left(\int_{c_{j-1}}^{c_j}r_1(t)\,dt,\ldots, \int_{c_{j-1}}^{c_j}r_n(t)\,dt\right)\\ &= \sum_{j=1}^{k}\int_{c_{j-1}}^{c_j}\mathbf r(t)\,dt. \end{aligned} $$

Each subinterval integral can use the formula valid there. If the function is defined differently at a partition point, that change affects at most finitely many values. For piecewise continuous components, changing finitely many point values does not change their definite integrals. This is why a jump at a breakpoint needs to be accounted for in the partition, but does not require a separate integral at the breakpoint.

Worked Example: One Vector Breakpoint

Worked Example: One Vector Breakpoint

Evaluate \(\int_0^2\mathbf r(t)\,dt\), where:

$$ \mathbf r(t)= \begin{cases} (t,t^2,1),&0\le t\le 1,\\ (2,t,t-1),&1<t\le 2. \end{cases} $$

The only interior breakpoint is \(t=1\), so split the integral there:

$$ \int_0^2\mathbf r(t)\,dt = \int_0^1(t,t^2,1)\,dt + \int_1^2(2,t,t-1)\,dt. $$

For the first piece, integrate each component and evaluate at the bounds:

$$ \begin{aligned} \int_0^1(t,t^2,1)\,dt &= \left( \left[\frac{t^2}{2}\right]_0^1, \left[\frac{t^3}{3}\right]_0^1, [t]_0^1 \right)\\ &= \left(\frac12,\frac13,1\right). \end{aligned} $$

For the second piece:

$$ \begin{aligned} \int_1^2(2,t,t-1)\,dt &= \left( [2t]_1^2, \left[\frac{t^2}{2}\right]_1^2, \left[\frac{t^2}{2}-t\right]_1^2 \right)\\ &= \left(4-2,\ 2-\frac12,\ (2-2)-\left(\frac12-1\right)\right)\\ &= \left(2,\frac32,\frac12\right). \end{aligned} $$

Add the two vectors component by component:

$$ \int_0^2\mathbf r(t)\,dt = \left(\frac12,\frac13,1\right) + \left(2,\frac32,\frac12\right) = \left(\frac52,\frac{11}{6},\frac32\right). $$

As a check, the antiderivatives used on \([0,1]\) are \(t^2/2\), \(t^3/3\), and \(t\); their derivatives are \(t\), \(t^2\), and \(1\). On \([1,2]\), the antiderivatives are \(2t\), \(t^2/2\), and \(t^2/2-t\); their derivatives are \(2\), \(t\), and \(t-1\). Each matches its component formula on the interval where it was used.

Worked Example: Different Breakpoints in Different Components

Worked Example: Different Breakpoints in Different Components

Let \(\mathbf r(t)=(f(t),g(t),h(t))\) on \([0,4]\), where:

$$ f(t)= \begin{cases} t,&0\le t\le 1,\\ 2,&1<t\le 4, \end{cases} \qquad g(t)= \begin{cases} 1,&0\le t\le 2,\\ t,&2<t\le 4, \end{cases} \qquad h(t)= \begin{cases} t,&0\le t\le 3,\\ 3,&3<t\le 4. \end{cases} $$

The components change formula at \(1\), \(2\), and \(3\). The common partition is therefore \(0,1,2,3,4\). On the interiors of its subintervals, the vector formulas are:

SubintervalVector formulaIntegral on the subinterval
\([0,1]\)\((t,1,t)\)\((\frac12,1,\frac12)\)
\([1,2]\)\((2,1,t)\)\((2,1,\frac32)\)
\([2,3]\)\((2,t,t)\)\((2,\frac52,\frac52)\)
\([3,4]\)\((2,t,3)\)\((2,\frac72,3)\)

For example, on \([2,3]\), the integral is \((\int_2^3 2\,dt,\int_2^3 t\,dt,\int_2^3 t\,dt)=(2,5/2,5/2)\). The other entries follow by evaluating the antiderivatives \(t^2/2\), \(t\), or \(t^2/2\), as appropriate, at the displayed bounds.

Now add all four subinterval results:

$$ \begin{aligned} \int_0^4\mathbf r(t)\,dt &= \left(\frac12,1,\frac12\right) +\left(2,1,\frac32\right) +\left(2,\frac52,\frac52\right) +\left(2,\frac72,3\right)\\ &= \left(\frac{13}{2},8,\frac{15}{2}\right). \end{aligned} $$

Check each component over its own formula intervals. For \(f\), the integral is \(\int_0^1t\,dt+\int_1^4 2\,dt=\frac12+6=\frac{13}{2}\). For \(g\), it is \(\int_0^2 1\,dt+\int_2^4t\,dt=2+\frac{16-4}{2}=8\). For \(h\), it is \(\int_0^3t\,dt+\int_3^4 3\,dt=\frac92+3=\frac{15}{2}\). These componentwise checks agree with the sum of the four vector integrals.

Worked Example: Three Pieces on an Interval with Negative Values

Worked Example: Three Pieces on an Interval with Negative Values

Evaluate \(\int_{-1}^2\mathbf r(t)\,dt\) for:

$$ \mathbf r(t)= \begin{cases} (t^2,1-t,t+1),&-1\le t\le 0,\\ (2t,1-t,1),&0<t\le 1,\\ (2t,2,1),&1<t\le 2. \end{cases} $$

The interior breakpoints are \(0\) and \(1\). Thus:

$$ \int_{-1}^2\mathbf r(t)\,dt = \int_{-1}^0(t^2,1-t,t+1)\,dt + \int_0^1(2t,1-t,1)\,dt + \int_1^2(2t,2,1)\,dt. $$

Evaluate each piece componentwise:

$$ \begin{aligned} \int_{-1}^0(t^2,1-t,t+1)\,dt &= \left( \left[\frac{t^3}{3}\right]_{-1}^0, \left[t-\frac{t^2}{2}\right]_{-1}^0, \left[\frac{t^2}{2}+t\right]_{-1}^0 \right)\\ &= \left(\frac13,\frac32,\frac12\right),\\[4pt] \int_0^1(2t,1-t,1)\,dt &= \left([t^2]_0^1, \left[t-\frac{t^2}{2}\right]_0^1, [t]_0^1\right) = \left(1,\frac12,1\right),\\[4pt] \int_1^2(2t,2,1)\,dt &= \left([t^2]_1^2,[2t]_1^2,[t]_1^2\right) = (3,2,1). \end{aligned} $$

Adding the results gives:

$$ \int_{-1}^2\mathbf r(t)\,dt = \left(\frac13,\frac32,\frac12\right) +\left(1,\frac12,1\right) +(3,2,1) = \left(\frac{13}{3},4,\frac52\right). $$

For a check, differentiate the antiderivatives on each piece: \((t^3/3,\ t-t^2/2,\ t^2/2+t)\) differentiates to \((t^2,1-t,t+1)\); \((t^2,\ t-t^2/2,\ t)\) differentiates to \((2t,1-t,1)\); and \((t^2,\ 2t,\ t)\) differentiates to \((2t,2,1)\). The antiderivatives therefore match the three formulas used in the calculation.

A Common Mistake: Ignoring One Component's Breakpoint

In the second example, the first component changes formula at \(t=1\). If you integrate it as though \(f(t)=t\) throughout \([0,2]\), you obtain \(\int_0^2 t\,dt=2\). That is incorrect because the formula \(f(t)=t\) applies only up to \(1\); on \((1,2]\), \(f(t)=2\). The correct calculation is:

$$ \int_0^2 f(t)\,dt = \int_0^1t\,dt+\int_1^2 2\,dt = \frac12+2 = \frac52. $$

The error comes from missing a breakpoint in one component. Using a single vector formula across an interval is valid only if every component uses that formula throughout the interval. When in doubt, list the breakpoints component by component, then use their union.

What the Integral Records

The result of a definite vector integral is a vector whose components are the accumulated integrals of the corresponding scalar components. Different components can have different formulas and breakpoints; the common partition is simply a reliable way to ensure that every component is integrated over intervals where its formula is known.

Keep the roles of the partition and the endpoint values distinct. Breakpoints determine where the formulas change, so they must divide the interval. But the value assigned at an individual breakpoint does not contribute a separate amount to the integral. This lets you handle jumps without treating a single parameter value as an interval of positive length.

Takeaway: For a piecewise vector-valued function, split at every component breakpoint, integrate each resulting vector piece, and add the vectors. The calculation is correct because vector integration is componentwise and definite integrals are additive.

Check Your Understanding

Use the definition of the vector integral and the common-partition method to answer these questions.

  1. If the first component changes formula at \(t=1\) and the second at \(t=2\), which interior values belong in the common partition?
  2. Why can a piecewise continuous vector function have a definite integral even if it has a jump at a breakpoint?
  3. For \(\mathbf r(t)=(t,1)\) on \([0,1]\) and \(\mathbf r(t)=(2,t)\) on \((1,2]\), write the split-integral expression for \(\int_0^2\mathbf r(t)\,dt\).
  4. What is the specific error in using one component's formula on both sides of its breakpoint?
  5. Explain why adding the subinterval vector integrals gives the integral over the full interval.

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