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Chapter 5: Q 92. (page 431)

Show that choosing a different anti-derivative $v + C$ of $d v$ will yield an equivalent formula for integration by parts,
as follows:
(a) Explain why what we need to show is that $u v - 鈭� v d u = u (v + C) - 鈭� (v + C) d u$ .
(b) Rewrite the equation from part (a), substituting $u = u (x)$ , $v = v (x)$ , and $d u = u' (x) d x$ .
(c) Prove the equality you wrote down in part (b).

Short Answer

Expert verified

Part (a) We have added the constant term, which was eliminated during the derivative.

Part (b) $u (x) v (x) - 鈭� v (x) u' (x) d x = u (x) (v (x) + C) - 鈭� (v (x) + C) u' (x) d x$

Part (c) Proved in the step.

Step by step solution

01

Part (a) Step 1. Explain the result.

Consider the formula of by part,

$鈭� u d v = u v - 鈭� v d u$

Substitute $v = v + C$ in the above integral.

$u v - 鈭� v d u = u (v + C) - 鈭� (v + C) d u$

In the above by part method, by use v+C in place of v because we are doing integration and take the first function in terms of v. As the constant term C is added because the derivative of constant is zero.

02

Part 2. Step 1. Explanation.

Substitute the given values into the given equation.

$u (x) v (x) - 鈭� v (x) u' (x) d x = u (x) (v (x) + C) - 鈭� (v (x) + C) u' (x) d x$

03

Part (c) Step 1. Explanation.

Prove the equality you wrote down in part (b).

let y be a function,

$y = u v$

Do the derivative of the functions.

$\begin{array}{l} \frac{d y}{d x} = \frac{d (u v)}{d x} \\ = u \frac{d v}{d x} + v \frac{d u}{d x} \end{array}$

Rearrange the rule to solve it.

$u \frac{d v}{d x} = \frac{d (u v)}{d x} - v \frac{d u}{d x}$

Integrate both sides.

$\begin{array}{l} 鈭� u \frac{d v}{d x} d x = 鈭� \frac{d (u v)}{d x} d x - 鈭� v \frac{d u}{d x} d x \\ 鈭� u d v = u v - v 鈭� d u \end{array}$

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