Integration by Parts

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 7.1 – Integration by Parts

Expected Educational Results

• Objective 03–01: I can use integration by parts to evaluate an indefinite integral using FTC-I.

• Objective 03–02: I can use integration by parts to evaluate a definite integral using FTC-II.

• Objective 03–03: I can use substitution and integration by parts to evaluate an indefinite integral using FTC-I.

• Objective 03–04: I can use substitution and integration by parts to evaluate a definite integral using FTC-II.

Bloom’s Taxonomy

A modern version of Bloom’s Taxonomy is included here to recognize various different levels of understanding and to encourage you to work towards higher-order understanding (those at the top of the pyramid). All Objectives, Investigations, Activities, etc. are color-coded with the level of understanding.

Integration by Parts

Definition: Integration by Parts

If $U=f(x)$$U=f(x)$ is a differentiable function and $dV=g^{\prime }(x)$$dV=g^{\,\prime}(x)$ is a integrable function, then $dU=f^{\prime }(x)$$dU=f^{\,\prime}(x)$ and $V=g(x)$$V=g(x)$. Thus, $\int UdV=UV-\int VdU$$\int{U\,dV}=UV-\int{V\,dU}$

Derivation of Integration by Parts

NOTE: The derivation of the Method of Integration by Parts will not be assessed. It is provided here to help you conceptually understand the formula for the method.

Suppose $f(x)$$f(x)$ and $g(x)$$g(x)$ are continuous and differentiable functions.

1. Use the product rule to differentiate the product of $f(x)$$f(x)$ and $g(x)$$g(x)$: ${\displaystyle \frac{d}{dx}\left[f(x)g(x)\right]=f^{\prime }(x)g(x)+f(x)g^{\prime }(x)}$$\displaystyle\frac{d}{dx} \left[{\color{green}{f(x)g(x)}}\right]={\color{orange}f^{\,\prime}(x)g(x)}+{\color{red}f(x)g^{\,\prime}(x)}$

2. Integrate both sides: ${\displaystyle \int \frac{d}{dx}\left[f(x)g(x)\right]dx=\int f^{\prime }(x)g(x)dx+\int f(x)g^{\prime }(x)dx}$$\displaystyle\int{\frac{d}{dx} \left[{\color{green}{f(x)g(x)}}\right]\,dx}=\int{{\color{orange}f^{\,\prime}(x)g(x)}\,dx}+\int{{\color{red}f(x)g^{\,\prime}(x)}\,dx}$

3. Use FTC-$1$$1$ to simplify the left side of the above equation: ${\displaystyle f(x)g(x)=\int f^{\prime }(x)g(x)dx+\int f(x)g^{\prime }(x)dx}$$\displaystyle {\color{green}{f(x)g(x)}}=\int{{\color{orange}f^{\,\prime}(x)g(x)}\,dx}+\int{{\color{red}f(x)g^{\,\prime}(x)}\,dx}$

4. Rewriting: ${\displaystyle \int f(x)g^{\prime }(x)dx=f(x)g(x)-\int f^{\prime }(x)g(x)dx}$$\displaystyle\int{{\color{red}f(x)g^{\,\prime}(x)}\,dx}={\color{green}{f(x)g(x)}}-\int{{\color{orange}f^{\,\prime}(x)g(x)}\,dx}$

5. Let $U=f(x)$$U=f(x)$ and $dV=g^{\prime }(x)dx$$dV=g^{\,\prime}(x)dx$, then $dU=f^{\prime }(x)dx$$dU=f^{\,\prime}(x)dx$ and $V=g(x)$$V=g(x)$, then $\int UdV=UV-\int VdU$$\int{{\color{red}U\,dV}}={\color{green}UV}-\int{{\color{orange}V\,dU}}$

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Last Modified: Thursday, 26 August 2021 - 11:53 (EDT)