Why Boundary Terms Matter
A definite integral asks for an accumulated value across a specific interval. Integration by parts can transform that integral into another one, but the transformation includes a contribution from both endpoints. Keeping track of that contribution is essential: it may be zero, nonzero, or negative.
The formula resembles the indefinite integration-by-parts rule, but in a definite integral the boundary term is evaluated immediately. There is no arbitrary constant of integration. The remaining definite integral must then be evaluated using the same original limits unless a change of variable is made.
The Definite Integration-by-Parts Formula
Suppose \(u\) and \(v\) are differentiable on the closed interval \([a,b]\), and their derivatives are continuous there. The product rule gives \((uv)'=u'v+uv'\). Rearranging and integrating from \(a\) to \(b\) gives the definite integration-by-parts formula.
In the usual shorthand, if \(du=u'(x)\,dx\) and \(dv=v'(x)\,dx\), the same rule is \(\int_a^b u\,dv=[uv]_a^b-\int_a^b v\,du\). The notation \([uv]_a^b\) always means the value at the upper limit minus the value at the lower limit.
Why the Formula Works
The formula follows directly from the product rule and the Fundamental Theorem of Calculus. The endpoint term is not an extra convention; it is exactly what results when the derivative of a product is integrated over an interval.
\((uv)'=u'v+uv'\), so \(uv'= (uv)'-u'v\).
\(\int_a^b uv'\,dx=\int_a^b (uv)'\,dx-\int_a^b u'v\,dx\).
\(\int_a^b (uv)'\,dx=u(b)v(b)-u(a)v(a)\). Substitution gives the stated formula.
This derivation also explains why the lower endpoint must be subtracted. The Fundamental Theorem evaluates an antiderivative at the upper limit and subtracts its value at the lower limit.
A Reliable Calculation Order
For a definite integral, it helps to keep the endpoint evaluation visible instead of writing only an unevaluated bracket. Once \(u\) and \(dv\) have been selected, integrate \(dv\) to obtain \(v\), differentiate \(u\) to obtain \(du\), evaluate \(uv\) at both endpoints, and then calculate the remaining definite integral.
- Identify \(u\), \(du\), \(dv\), and \(v\).
- Write the boundary difference explicitly as \(u(b)v(b)-u(a)v(a)\).
- Subtract the remaining integral \(\int_a^b v\,du\).
- Simplify and evaluate that integral, preserving the limits.
- Check the result using an antiderivative when practical.
Integration by parts can be especially effective when differentiating one factor makes it simpler, or when integrating the other factor is straightforward. The definite limits do not change the choice of factors, but they do make the endpoint calculation an explicit part of the work.
Worked Example: A Logarithm on a Finite Interval
Evaluate \(\int_1^e \ln x\,dx\). Choose \(u=\ln x\) and \(dv=dx\). Then \(du=\frac{1}{x}\,dx\) and \(v=x\). Apply the formula:
Here \(\ln e=1\) and \(\ln 1=0\). The residual integral is \([x]_1^e=e-1\), so the result is \(1\). To check, an antiderivative is \(F(x)=x\ln x-x\), since \(F'(x)=\ln x+1-1=\ln x\). Its endpoint difference is \(F(e)-F(1)=(e-e)-(0-1)=1\), as required.
The Endpoint Term Can Vanish
Sometimes the boundary term is zero because one or both endpoint values make the product \(uv\) vanish. That should be concluded only after evaluating the product at both limits. For example, a factor \(x\) vanishes at \(x=0\), but that does not imply that the boundary term vanishes at a nonzero upper endpoint.
Worked Example: An Exponential Integral with a Nonzero Boundary Difference
Evaluate \(\int_0^1 xe^x\,dx\). Let \(u=x\), so \(du=dx\), and let \(dv=e^x\,dx\), so \(v=e^x\). Then
The boundary difference is \(e-0=e\); the lower endpoint contribution is zero, but the upper one is not. For a check, \(F(x)=e^x(x-1)\) satisfies \(F'(x)=e^x(x-1)+e^x=xe^x\). Also, \(F(1)=0\) and \(F(0)=-1\), so \(F(1)-F(0)=1\).
When the Remaining Integral Needs Simplifying
After integration by parts, the remaining integral may need algebraic simplification before it is easy to evaluate. The limits remain attached to the whole integral during that simplification. In particular, splitting an integrand into simpler terms is valid here because the definite integral is linear.
Worked Example: Simplifying the Remaining Definite Integral
Evaluate \(\int_0^1 x\ln(1+x)\,dx\). Choose \(u=\ln(1+x)\), so \(du=\frac{1}{1+x}\,dx\), and \(dv=x\,dx\), so \(v=\frac{x^2}{2}\). Then
To simplify the remaining integrand, divide \(x^2\) by \(x+1\): since \(x^2=(x+1)(x-1)+1\),
Therefore, the remaining integral is
Substituting this value gives \(\frac12\ln 2-\frac12(\ln 2-\frac12)=\frac14\). For a check, an antiderivative of the original integrand is
Differentiating, \(F'(x)=x\ln(1+x)+\frac{x^2-1}{2(1+x)}-\frac{x}{2}+\frac12\). Since \(\frac{x^2-1}{2(1+x)}=\frac{x-1}{2}\), the non-logarithmic terms cancel, leaving \(F'(x)=x\ln(1+x)\). Finally, \(F(1)=\frac14\) and \(F(0)=0\), confirming the definite integral is \(\frac14\).
A Trigonometric Example
The boundary difference can also contribute a familiar constant such as an endpoint angle. Keep the exact limits in the bracket until both endpoint values have been substituted.
Worked Example: A Product with a Trigonometric Factor
Evaluate \(\int_0^{\pi/2} x\cos x\,dx\). Take \(u=x\), \(du=dx\), \(dv=\cos x\,dx\), and \(v=\sin x\). Then
For a check, \(F(x)=x\sin x+\cos x\) has derivative \(F'(x)=\sin x+x\cos x-\sin x=x\cos x\). The endpoint difference is \(F(\frac{\pi}{2})-F(0)=\frac{\pi}{2}-1\), in agreement.
Common Pitfall: Dropping the Remaining Integral
A frequent error is to calculate the boundary term and treat it as the entire answer. Integration by parts does not say that \(\int_a^b u\,dv=[uv]_a^b\); it says to subtract the remaining integral as well.
A related sign error is to reverse the boundary difference. By definition, \([uv]_a^b=u(b)v(b)-u(a)v(a)\), not the lower value minus the upper value. Writing the two endpoint values explicitly, as in the examples, makes this mistake easier to catch.
When the Boundary Term Is Zero
The endpoint product may vanish at both limits, but that is a conclusion about the specific \(u\), \(v\), and interval—not a general feature of definite integration by parts. If \([uv]_a^b=0\), the formula reduces to \(\int_a^b u\,dv=-\int_a^b v\,du\). If only one endpoint product is zero, the other still contributes.
It is useful to write down \(u(a)v(a)\) and \(u(b)v(b)\) separately before simplifying. This is particularly helpful when an endpoint involves a logarithm, a trigonometric value, or a factor that vanishes. Only after both values have been evaluated should the boundary term be combined with the remaining integral.
Summary
Definite integration by parts is the product rule integrated over an interval. It transforms an integral into an endpoint difference minus another definite integral. To use it accurately:
- State \(u\), \(du\), \(dv\), and \(v\) clearly.
- Evaluate the boundary term at the upper endpoint and subtract its value at the lower endpoint.
- Keep the original limits on the remaining integral.
- Simplify and evaluate that integral before combining terms.
- Check the result by differentiating an antiderivative and evaluating its endpoint difference when practical.
Check Your Understanding
For each question, keep the endpoint evaluation explicit and check any computed integral where requested.
- Use integration by parts to evaluate \(\int_0^1 x\,dx\), choosing \(u=x\) and \(dv=dx\). What is the boundary difference?
- For \(\int_1^2 x\ln x\,dx\), identify a useful choice of \(u\) and \(dv\), then write the boundary term with both endpoint values shown.
- Explain why the formula includes \(u(b)v(b)-u(a)v(a)\), rather than just the upper-endpoint product.
- Evaluate \(\int_0^{\pi/2}x\sin x\,dx\) by parts and verify the result using an antiderivative.
- A student writes \(\int_a^b u\,dv=[uv]_a^b\). Identify the missing term and explain when the boundary term alone would happen to give the correct value.
