Why Boundary Terms Matter
A definite integral asks for the accumulated value of a function over an interval. When integration by parts is used, that accumulation is related to a product evaluated at the two ends of the interval. The endpoint values are part of the formula, not optional details: leaving one out can change the answer, even if the remaining integral is evaluated correctly.
The key idea comes directly from the product rule. For differentiable functions \(u\) and \(v\), the derivative of their product contains both \(u'v\) and \(uv'\). Rearranging that rule and integrating across an interval produces a formula with a boundary expression. We will derive the formula, interpret its endpoint notation, and use it in several definite integrals.
What the Boundary Notation Means
The bracket notation records the value of the expression at the upper endpoint minus its value at the lower endpoint:
Both evaluations are required. The lower-endpoint value is subtracted, including when the expression itself is negative there. If either endpoint contribution is zero, it can of course be omitted after it has been evaluated.
In differential notation, set \(du=u'(x)\,dx\) and \(dv=v'(x)\,dx\). The same formula is commonly written
This is the definite integration by parts formula. The integral on the right is still a definite integral over the same interval. Unlike an indefinite integral formula, no arbitrary constant is added.
Deriving the Formula
Assume \(u\) and \(v\) are continuously differentiable on the closed interval \([a,b]\). The product rule gives
Rearrange this identity to isolate the product \(u(x)v'(x)\):
Now integrate both sides from \(a\) to \(b\). Linearity of the integral separates the two terms, and the Fundamental Theorem of Calculus evaluates the integral of the derivative:
This proves the formula. The endpoint difference appears because the Fundamental Theorem evaluates an antiderivative at the upper endpoint and subtracts its value at the lower endpoint. That subtraction is the source of the minus sign in the boundary notation.
Write the integrand as \(u(x)v'(x)\), so that \(u\) and a convenient antiderivative \(v\) are identified.
Evaluate \(u(b)v(b)-u(a)v(a)\), keeping the upper and lower values distinct.
Subtract \(\int_a^b v(x)u'(x)\,dx\) from the complete boundary difference.
When practical, differentiate an antiderivative for the original integrand and confirm the endpoint evaluation.
Worked Examples
Worked Example: An Exponential Integral on a Finite Interval
Evaluate \(\int_0^1 xe^x\,dx\) by integration by parts.
Write the integrand as \(u\,dv\), with \(u=x\) and \(dv=e^x\,dx\). Then \(du=dx\) and \(v=e^x\). Apply the definite formula:
For a check, an antiderivative is \(xe^x-e^x\), since
Its endpoint difference is \((e-e)-(0-1)=1\), matching the result.
Worked Example: A Trigonometric Integral With a Nonzero Boundary Term
Evaluate \(\int_0^\pi x\sin x\,dx\).
Take \(u=x\), so \(du=dx\), and \(dv=\sin x\,dx\), so \(v=-\cos x\). The boundary expression and remaining integral are
The boundary term is \(\pi\), so it cannot be discarded. To check, an antiderivative is \(-x\cos x+\sin x\). Its derivative is \(-\cos x+x\sin x+\cos x=x\sin x\), and its endpoint difference is \(\pi-0=\pi\).
Worked Example: An Integral Whose Boundary Term Is Zero
Evaluate \(\int_0^1 x\cos(\pi x)\,dx\). Use \(u=x\), \(du=dx\), \(dv=\cos(\pi x)\,dx\), and \(v=\frac{\sin(\pi x)}{\pi}\). Then
The boundary expression vanishes because \(\sin(0)=\sin(\pi)=0\), but the remaining integral does not. A check is the antiderivative \(\frac{x\sin(\pi x)}{\pi}+\frac{\cos(\pi x)}{\pi^2}\). Its derivative is \(x\cos(\pi x)\), and its endpoint difference is \(-1/\pi^2-1/\pi^2=-2/\pi^2\).
Reading the Formula Carefully
The boundary term and the remaining integral play separate roles. First calculate the complete endpoint difference; then subtract the remaining definite integral. It is usually safer to write the upper and lower evaluations on separate terms rather than relying on mental arithmetic.
| Part of the formula | Meaning |
|---|---|
| \(\bigl[uv\bigr]_a^b\) | \(u(b)v(b)-u(a)v(a)\), with the lower value subtracted |
| \(\int_a^b v\,du\) | The new definite integral, over the original interval |
| The minus sign before the integral | The remaining integral is subtracted from the full boundary difference |
A boundary term may be nonzero, as in the integral of \(x\sin x\) from \(0\) to \(\pi\); it may also vanish, as in the integral of \(x\cos(\pi x)\) from \(0\) to \(1\). Neither outcome should be assumed from the appearance of the integrand. Evaluate the expression at both limits to determine it.
A Common Boundary-Term Mistake
For \(\int_0^\pi x\sin x\,dx\), one incorrect calculation is
This happens to give the correct boundary contribution in this example because the lower-endpoint product is zero. But the calculation is incomplete as a general method: the bracket means that the value at \(0\) must be subtracted, not ignored. For example, in \(\bigl[x e^x\bigr]_0^1\), keeping only the upper value would give \(e\), while the correct boundary difference is \(e-0=e\); this example also masks the omission. A nonzero lower endpoint exposes it: \(\bigl[x e^x\bigr]_1^2\) is \(2e^2-e\), not \(2e^2\).
The correction is to write both evaluations explicitly before simplifying:
Another possible error is to subtract the remaining integral before evaluating the endpoint difference correctly. Keeping the formula in the form “upper value minus lower value minus remaining integral” makes the signs visible.
Why Definite Integration by Parts Is Useful
Integration by parts lets a definite integral be replaced by another one whose factors may be easier to integrate. The endpoints also carry useful information: on some intervals, the product \(uv\) vanishes at one or both limits, simplifying the calculation. On other intervals, those endpoint values provide a substantial part of the answer.
The method is especially reliable when each piece is recorded separately: identify \(u\) and \(dv\), compute \(du\) and \(v\), evaluate \(\bigl[uv\bigr]_a^b\), and then subtract the remaining integral. Finally, a derivative check can confirm that the antiderivative used in the endpoint evaluation differentiates to the original integrand.
Check Your Understanding
For each question, pay attention to both the limits and the boundary expression.
- State the definite integration by parts formula using \(u\), \(v\), \(du\), and \(dv\).
- What does \(\bigl[uv\bigr]_a^b\) mean when written without bracket notation?
- Use integration by parts to set up, but do not necessarily simplify, \(\int_0^2 xe^x\,dx\). Identify the boundary term explicitly.
- In \(\int_0^\pi x\sin x\,dx\), which endpoint contribution is nonzero?
- Why does the definite formula contain no arbitrary constant?
