Integration by Parts: Definite Integrals

10 min read

Textbook Reference

OpenStax Calculus Volume 2 is free and openly licensed. OpenStax labels the definite-integral version of the formula Theorem 3.2.

Key idea

A definite integral by parts is the same trade you already know, with the bounds carried along.

You already know two things separately. From the previous lesson, you know the indefinite formula $\int u\, dv = uv - \int v\, du$. From the Fundamental Theorem of Calculus, Part 2, you know that a definite integral is evaluated by plugging the limits into an antiderivative and subtracting. The definite version of integration by parts simply fuses these: the product term $uv$ gets evaluated at the two limits, and the leftover integral keeps its limits too.

Here is the picture. When you do parts on a definite integral, the answer has two pieces. The first piece, $uv$, is already an antiderivative-style expression, so you evaluate it from the lower limit to the upper limit right away, the way the Fundamental Theorem tells you to. The second piece is still an integral, $\int v\, du$, and it keeps the same limits, to be finished separately. Nothing about the choice of $u$ and $dv$ changes. Only the bookkeeping at the end changes.

This means: if you can already do the indefinite integral, you can do the definite one. The only new habit is remembering to evaluate the $uv$ term at the bounds, and not letting the bracket notation trip you up.

Prerequisite Check

Before this lesson, make sure you can do all of the following:

• Apply the indefinite formula $\int u\, dv = uv - \int v\, du$ and choose $u$ by LIATE
• Evaluate a definite integral with the Fundamental Theorem of Calculus, Part 2: $\int_a^b f(x)\, dx = F(b) - F(a)$
• Read and apply bracket notation: $\big[F(x)\big]_a^b = F(b) - F(a)$
• Recall basic values: $\ln 1 = 0$, $e^0 = 1$, $\arctan 1 = \dfrac{\pi}{4}$, $\arctan 0 = 0$

If the indefinite version is shaky, review the Integration by Parts Formula lesson first. If evaluating at bounds feels uncertain, review FTC Part 2.

Quick Reference

The formula (definite integrals). \[ \int_a^b u\, dv = \Big[uv\Big]_a^b - \int_a^b v\, du \]

The bracket $\big[uv\big]_a^b$ means: evaluate the product $uv$ at the upper limit $b$, then subtract its value at the lower limit $a$. The remaining integral $\int_a^b v\, du$ keeps the same limits.

Two equivalent strategies.

Both give the same number. The bounds-carried strategy is the one shown below.

Key Concepts

1. Where the Bracket Comes From

Goal. See why the definite formula is just the indefinite one with FTC Part 2 applied.

Step 1. Start from the indefinite identity, valid as functions of $x$: \[ \int u\, dv = uv - \int v\, du. \]

Step 2. Take the definite integral of both sides from $a$ to $b$. The left side becomes $\int_a^b u\, dv$. On the right, the term $uv$ is an explicit function, so by FTC Part 2 its definite integral contribution is $\big[uv\big]_a^b$, and the integral term keeps its limits: \[ \int_a^b u\, dv = \Big[uv\Big]_a^b - \int_a^b v\, du. \]

Recap. The bracket on $uv$ is exactly the “evaluate at the limits and subtract” step from the Fundamental Theorem. No new idea is needed; the definite formula is the indefinite formula wearing limits.

2. The Procedure

To evaluate $\int_a^b (\text{product})\, dx$ by parts:

1. Choose $u$ and $dv$ by LIATE, exactly as in the indefinite case.
2. Compute $du$ and $v$ (no $+C$).
3. Write $\big[uv\big]_a^b - \int_a^b v\, du$.
4. Evaluate the bracket $\big[uv\big]_a^b = (uv)\big|_{x=b} - (uv)\big|_{x=a}$ as a number.
5. Finish the remaining integral $\int_a^b v\, du$, also as a number.
6. Combine the two numbers. There is no $+C$ on a definite integral.

(A frequent slip is to forget that the bracket needs both limits substituted and subtracted. Another is to leave a $+C$ on a definite answer, where it does not belong.)

3. Worked Example: A Definite Integral with a Logarithm

Example 1. Evaluate $\displaystyle\int_1^e \ln x\, dx$.

Goal and thought process. The integrand is a lone logarithm, so by LIATE let $u = \ln x$ and $dv = dx$, the same setup as the indefinite case. The limits $1$ and $e$ are chosen because $\ln 1 = 0$ and $\ln e = 1$, which keeps the arithmetic clean.

Step 1. Choose. $u = \ln x$, $dv = dx$, so $du = \dfrac{1}{x}\, dx$ and $v = x$.

Step 2. Apply the formula with limits. \[ \int_1^e \ln x\, dx = \Big[x\ln x\Big]_1^e - \int_1^e x\cdot\frac{1}{x}\, dx = \Big[x\ln x\Big]_1^e - \int_1^e 1\, dx. \]

Step 3. Evaluate the bracket. \[ \Big[x\ln x\Big]_1^e = e\ln e - 1\cdot\ln 1 = e(1) - 1(0) = e. \]

Step 4. Finish the remaining integral. \[ \int_1^e 1\, dx = \big[x\big]_1^e = e - 1. \]

Step 5. Combine. \[ \int_1^e \ln x\, dx = e - (e - 1) = 1. \]

Boxed answer. \[ \boxed{\int_1^e \ln x\, dx = 1} \]

Sanity check. The graph of $\ln x$ from $1$ to $e$ rises from $0$ to $1$ over a width of $e - 1 \approx 1.72$. The region sits under a curve that ends at height $1$, so an area of exactly $1$ is the right order of magnitude. The exact value $1$ is a clean, reassuring result.

4. Worked Example: An Inverse Trig Definite Integral

Example 2. Evaluate $\displaystyle\int_0^1 \arctan x\, dx$.

Goal and thought process. A lone inverse trig function, so $u = \arctan x$, $dv = dx$ (Inverse trig ranks high on LIATE). Then a substitution finishes the leftover integral, a two-technique pattern that is common.

Step 1. Choose. $u = \arctan x$, $dv = dx$, so $du = \dfrac{1}{1+x^2}\, dx$ and $v = x$.

Step 2. Apply the formula with limits. \[ \int_0^1 \arctan x\, dx = \Big[x\arctan x\Big]_0^1 - \int_0^1 \frac{x}{1+x^2}\, dx. \]

Step 3. Evaluate the bracket. \[ \Big[x\arctan x\Big]_0^1 = 1\cdot\arctan 1 - 0\cdot\arctan 0 = 1\cdot\frac{\pi}{4} - 0 = \frac{\pi}{4}. \]

Step 4. Finish the remaining integral by substitution. Let $w = 1 + x^2$, $dw = 2x\, dx$, so $x\, dx = \tfrac{1}{2}\, dw$. When $x = 0$, $w = 1$; when $x = 1$, $w = 2$: \[ \int_0^1 \frac{x}{1+x^2}\, dx = \frac{1}{2}\int_1^2 \frac{1}{w}\, dw = \frac{1}{2}\big[\ln w\big]_1^2 = \frac{1}{2}(\ln 2 - \ln 1) = \frac{1}{2}\ln 2. \]

Step 5. Combine. \[ \boxed{\int_0^1 \arctan x\, dx = \frac{\pi}{4} - \frac{1}{2}\ln 2} \]

Recap. Two habits to carry away: when you change variables inside a definite integral, change the limits to match the new variable (here $1$ and $2$, not $0$ and $1$); and a single parts step often leaves an integral that a quick substitution finishes.

Quick self-check. In Example 2, why did the limits on the substituted integral become $1$ and $2$ instead of staying $0$ and $1$?

Show answer

Because the variable changed from $x$ to $w = 1 + x^2$. Definite-integral limits are values of the variable of integration. When $x = 0$, $w = 1 + 0 = 1$; when $x = 1$, $w = 1 + 1 = 2$. Substituting the new limits lets you finish without converting back to $x$.

5. Antiderivative-First, the Alternative Order

For some problems it is tidier to find the whole indefinite antiderivative first, then evaluate once at the end. For $\int_0^1 x e^{-x}\, dx$, first do the indefinite integral by parts ($u = x$, $dv = e^{-x}\, dx$, so $v = -e^{-x}$): \[ \int x e^{-x}\, dx = -x e^{-x} - \int (-e^{-x})\, dx = -x e^{-x} - e^{-x} + C = -(x+1)e^{-x} + C. \] Then apply FTC Part 2 once: \[ \int_0^1 x e^{-x}\, dx = \Big[-(x+1)e^{-x}\Big]_0^1 = -(2)e^{-1} - \big(-(1)e^{0}\big) = -\frac{2}{e} + 1 = 1 - \frac{2}{e}. \] Both orders are correct. Use whichever keeps your work cleaner.

Common Errors Summary

Common Misconceptions

Leveled Practice

Level 1 -- Direct Application

Problem 1. Evaluate $\displaystyle\int_0^1 x e^{x}\, dx$.

Problem 2. Evaluate $\displaystyle\int_0^{\pi} x \sin x\, dx$.

Level 2 -- Two Techniques Together

Problem 3. Evaluate $\displaystyle\int_1^2 x\ln x\, dx$.

Problem 4. Evaluate $\displaystyle\int_0^1 \arctan x\, dx$ on your own, then confirm it equals $\dfrac{\pi}{4} - \dfrac{1}{2}\ln 2$ from Example 2.

Level 3 -- Reasoning

Problem 5. A student finds the indefinite antiderivative $\int x\sin x\, dx = -x\cos x + \sin x + C$ and then writes $\int_0^{\pi} x\sin x\, dx = -x\cos x + \sin x + C$ as the final answer. Identify both errors.

Mastery Checklist

You have mastered this skill when you can do all of the following without referring to notes:

• State the definite formula $\int_a^b u\, dv = \big[uv\big]_a^b - \int_a^b v\, du$
• Explain why the $uv$ term is evaluated at the limits (it comes from FTC Part 2)
• Evaluate $\int_1^e \ln x\, dx$ and $\int_0^{\pi} x\sin x\, dx$ from scratch
• Change the limits correctly when a substitution finishes the leftover integral
• Avoid writing $+C$ on a definite answer
• Choose between bounds-carried and antiderivative-first and get the same number either way

Mental Model

A definite integral by parts has two accounts to settle, not one.

The indefinite method already taught you the trade: hand over $\int u\, dv$, receive a finished product $uv$ plus a new integral $\int v\, du$. For a definite integral, you settle each account at the two endpoints. The product $uv$ is a completed expression, so you read its value at the top limit and subtract its value at the bottom limit right away, the way the Fundamental Theorem of Calculus says to settle any antiderivative. The leftover integral is still an integral, so it stays on the books with its limits until you finish it. The final answer is the difference of the first account and the second.

The only mistakes that are special to the definite case live in that settling step: forgetting to subtract the lower-limit value, forgetting to update the limits after a substitution, or carrying a $+C$ that has no business on a number.

Connections

Within Techniques of Integration (MATH162)

• Built directly on: the indefinite Integration by Parts Formula and FTC Part 2. The definite case adds only the evaluation step.
• Leads to: repeated integration by parts, where a definite integral may need two or more parts steps before the bracket and the leftover are both settled.
• Used in applications: definite integrals by parts appear when computing average values, moments, work, and the coefficients of series expansions, all later in the course.

Toward Later Mathematics

• Probability: the expected value of a continuous random variable is a definite integral $\int x f(x)\, dx$ that is frequently evaluated by parts.
• Transforms: the Laplace transform $\int_0^{\infty} f(t)e^{-st}\, dt$ is an improper definite integral routinely handled by parts, then a limit.

Audience Notes

For students who find math intimidating: The new step is small. You already know how to do the integral; now you just evaluate the $uv$ piece at the two numbers and subtract, the same as any definite integral. Watch two things: subtract the bottom value (even when it is zero), and never write $+C$ on a definite answer.

For students interested in proof: The definite formula assumes $u$ and $v$ are continuously differentiable on $[a, b]$, so that the bracket term is well defined at both endpoints and FTC Part 2 applies. When an endpoint is a point of discontinuity or infinity, the integral is improper and you evaluate the bracket as a limit, which is the bridge to the improper-integral material later in the chapter.

For students interested in careers: Definite integrals computed by parts are everywhere in applied work: expected values in statistics, signal energy in engineering, and transform coefficients in control theory all rely on exactly this evaluation pattern.

Back to Techniques of Integration | Previous: Integration by Parts Formula | Next: Repeated Integration by Parts