Deriving the Integration by Parts Formula (6.11.1) | AP Calculus AB Notes

AP Syllabus focus:
‘Understand integration by parts as a technique based on the product rule for differentiation, leading to a formula for ∫ u dv in terms of u v − ∫ v du.’

Integration by parts arises directly from the product rule and provides a structured method for rewriting certain integrals into simpler forms. This derivation is essential for understanding why the method works.

Deriving Integration by Parts

Revisiting the Product Rule for Differentiation

The starting point for the integration by parts formula is the product rule, one of the foundational results from differential calculus.

This diagram presents the derivation of integration by parts from the product rule, ending with the formula $\int u,dv = uv - \int v,du$ . It highlights how integrating the product rule and rearranging terms produces the integration by parts identity. All elements directly reflect the algebraic relationships described in the notes. Source.

Product Rule: The derivative of a product of two differentiable functions $u(x)$ and $v(x)$ is $u'(x)v(x) + u(x)v'(x)$ .

Using this relationship allows us to connect differentiation and integration in a meaningful way.

This illustration shows how the derivative of a product can be understood geometrically, clarifying why the expression $u'v + uv'$ naturally arises. It visually supports the conceptual basis for integrating the product rule to derive integration by parts. Some geometric detail appears, but all components relate directly to the standard product rule. Source.

Because the product rule gives a derivative, integrating both sides of the equality creates a pathway to an integration formula.

Integrating the Product Rule

Beginning with the product rule expression, integrate each term with respect to the independent variable. This step uses the idea that integration can “undo” differentiation.

$\int \left(u'(x)v(x) + u(x)v'(x)\right),dx = u(x)v(x)$
$u(x)v(x)$ = Product of two differentiable functions (no units unless context supplied)

This integral identity is algebraically rearranged to isolate one of the integral expressions. Doing so transforms the formula into the standard structure used in the technique of integration by parts.

The next step is to isolate the integral involving $u(x)v'(x)$ , which is rewritten as the integral of $u,dv$ , using more compact differential notation. This notational shift supports the AP Calculus AB requirement to understand integration by parts in its familiar symbolic form.

Constructing the Integration by Parts Identity

By moving one integral to the opposite side of the equation, the relationship becomes a tool for rewriting an integral into another, often simpler, form.

This figure provides a geometric interpretation of integration by parts using labeled areas, illustrating how $\int u,dv = uv - \int v,du$ can be understood through rearranged regions. It reinforces the conceptual link between algebraic manipulation and geometric structure. The image includes some additional area details, all of which clarify the identity. Source.

$\int u(x)v'(x),dx = u(x)v(x) - \int v(x)u'(x),dx$
$u(x)$ = Chosen function to differentiate (units depend on context)
$v(x)$ = Antiderivative of $v'(x)$ , found through integration

This equation is typically expressed using differential notation to emphasize the roles of $u$ , $du$ , $v$ , and $dv$ , which are selected strategically based on the structure of the integrand. Normal integration rules require at least one sentence here to contextualize the expression before the next equation can be introduced.

$\int u,dv = uv - \int v,du$
$du$ = Differential of $u$ , obtained by differentiating $u$
$dv$ = Portion of integrand chosen to integrate to obtain $v$

This final form is the standard integration by parts formula used in AP Calculus AB.

Understanding the Structure of the Formula

Why Integration by Parts Works

Integration by parts works because it transforms an integral involving a product of functions into a new integral that may be easier to evaluate. The technique relies on selecting $u$ and $dv$ so that differentiating $u$ simplifies it and integrating $dv$ does not complicate the expression.

Key Components of the Formula

The formula connects four essential elements:

$u$ : A function chosen to differentiate
$du$ : The derivative of $u$ , identifying how $u$ changes
$dv$ : A piece of the integrand chosen for integration
$v$ : The antiderivative of $dv$

To use the formula, students should understand that the choice of $u$ and $dv$ determines whether the resulting integral $\int v,du$ becomes simpler than the original.

Strategic Selection of u and dv

When considering the structure of the integrand:

Choose $u$ so that its derivative is simpler than the original expression.
Choose $dv$ so that its antiderivative is manageable and does not introduce complexity.
The product $uv$ must be computable after finding $v$ .
The new integral $\int v,du$ should be easier to evaluate than the original.

These ideas reflect the purpose of the formula derived from the product rule: rewriting an integral into a more workable form.

Why the Derivation Matters

Deriving the formula clarifies that integration by parts is not a memorized trick but a direct consequence of a fundamental idea in calculus. By grounding the method in the product rule, students can better recognize when the technique applies and understand the relationships among $u$ , $du$ , $v$ , and $dv$ .

This derivation also reinforces the connection between differentiation and integration, highlighting the conceptual symmetry that underlies much of AP Calculus AB.

FAQ

Integration by parts is advantageous when the integrand contains a product where differentiating one factor simplifies it while integrating the other remains manageable.

It is especially helpful when:
• One part becomes simpler upon differentiation, such as logarithmic or inverse trigonometric functions.
• Substitution does not meaningfully simplify the expression because no clear inner function is present.

It therefore fills a gap where substitution is ineffective but algebraic restructuring using the product rule becomes powerful.

Multiple valid choices for u and dv may exist because the original integrand can often be partitioned in more than one way.

Different choices may:
• Produce integrals that are equally workable.
• Lead to a more complicated remaining integral, requiring additional steps.
• Sometimes result in circular calculations that bring the expression back to the original form.

Strategic choice aims to minimise complexity and reduce the number of required steps.

Both u and v must be differentiable where the integral is defined. This ensures that:
• u has a well-defined derivative forming du.
• v can be obtained by integrating dv and is valid on the interval considered.

If either function is not differentiable at a point, the integration by parts formula may still hold on subintervals, provided the functions behave suitably away from isolated discontinuities.

The formula arises from integrating a derivative, which inherently introduces an arbitrary constant.

In indefinite integrals:
• Both uv and the remaining integral contain implicit constants.
• These merge into a single constant in the final expression.

Although the constant does not affect the structure of the method, recognising its presence avoids errors when checking equality between two antiderivatives.

The method effectively undoes the product rule by expressing the integral of a product in terms of the original product minus another product-based integral.

Conceptually:
• Differentiation expands a product into two terms.
• Integration by parts condenses a product and redistributes the remaining integral.

This reverse process highlights the deep connection between differentiation and integration, reinforcing why the method emerges naturally from the product rule.

Practice Questions

Question 1 (1–3 marks)
The function u and v are differentiable. Starting from the product rule for differentiation, write down the integration by parts formula used to evaluate integrals of the form ∫u dv.

Question 1 (1–3 marks)
• 1 mark for stating the correct formula uv − ∫v du.
• 1 mark for showing it originates from ∫u dv = uv − ∫v du.
• 1 mark for correct notation linking u, du, v and dv.

Question 2 (4–6 marks)
A differentiable function f is written as the product of two functions f(x) = u(x)v(x), where u and v are differentiable.
(a) State the product rule for differentiation.
(b) Integrate both sides of the product rule with respect to x.
(c) Rearrange your result to obtain the integration by parts identity.
Your answer must clearly show each step in the derivation.

Question 2 (4–6 marks)
(a)
• 1 mark for stating the product rule: f'(x) = u'(x)v(x) + u(x)v'(x).

(b)
• 1 mark for integrating both sides correctly.
• 1 mark for writing the integral of the derivative as uv.

(c)
• 1 mark for isolating the integral of u v' to obtain uv − ∫v u'.
• 1 mark for presenting the final identity ∫u dv = uv − ∫v du in correct notation.

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Written by:

Dr Rahil Sachak-Patwa

Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.