Chapter 6, Problem 1RCC

To determine

To state: The rule for integration by parts.

Explanation of Solution

Explanation to state the rule for integration by parts:

The rule that corresponds to the Product Rule for differentiation is called as the integration by parts.

The product Rule states that if f and g are differentiable functions, then

$\frac{d}{dx}\left[f\left(x\right)g\left(x\right)\right]=f\left(x\right)g^{\prime }\left(x\right)+g\left(x\right)f^{\prime }\left(x\right)$

Integrate both sides of the Equation for indefinite integrals.

$\begin{array}{l}f\left(x\right)g\left(x\right)={\displaystyle \int \left[f\left(x\right)g^{\prime }\left(x\right)+g\left(x\right)f^{\prime }\left(x\right)\right]dx}\\ f\left(x\right)g\left(x\right)={\displaystyle \int f\left(x\right)g^{\prime }\left(x\right)dx}+{\displaystyle \int g\left(x\right)f^{\prime }\left(x\right)dx}\\ {\displaystyle \int f\left(x\right)g^{\prime }\left(x\right)dx}=f\left(x\right)g\left(x\right)-{\displaystyle \int g\left(x\right)f^{\prime }\left(x\right)dx}\end{array}$ (1)

Consider

$\begin{array}{l}u=f\left(x\right)\\ v=g\left(x\right)\end{array}$

Differentiate both sides of the above Equations.

$\begin{array}{l}du=f^{\prime }\left(x\right)dx\\ dv=g^{\prime }\left(x\right)dx\end{array}$

Substitute u for $f\left(x\right)$, v for $g\left(x\right)$, du for $f^{\prime }\left(x\right)dx$, and dv for $g^{\prime }\left(x\right)dx$ in Equation (1).

${\displaystyle \int udv}=uv-{\displaystyle \int vdu}$ (2)

Therefore, the rule for integration by parts is $\underset{\_}{{\displaystyle \int udv}=uv-{\displaystyle \int vdu}}$.

Example to use the rule for integration by parts:

Consider the integral function $y={\displaystyle \int x{f}^{\prime }\left(x\right)dx}$.

Consider $u=x$

Differentiate both sides of the Equation.

$du=dx$

Consider $dv=f^{\prime }\left(x\right)dx$

Integrate both sides of the Equation.

$v=f\left(x\right)$

Substitute x for u, dx for du, $f\left(x\right)$ for v, and $f^{\prime }\left(x\right)dx$ for dv in Equation (2).

${\displaystyle \int x{f}^{\prime }\left(x\right)dx}=xf\left(x\right)-{\displaystyle \int f\left(x\right)dx}$

Therefore, the value of the integral function ${\displaystyle \int x{f}^{\prime }\left(x\right)dx}$ is $xf\left(x\right)-{\displaystyle \int f\left(x\right)dx}$.