Best solutions to - "Integrating a product, one factor a derivative ∫dx(t)dtx(t)2dt"

Name: Integrating a product, one factor a derivative ∫dx(t)dtx(t)2dt
Uploaded: 2022-02-23T17:22:57+00:00
Description: Integrating a product, one factor a derivative ∫dx(t)dtx(t)2dt

Betsy Rhone

Answered question

2022-01-21

Integrating a product, one factor a derivative

\int \frac{d x (t)}{d t} x {(t)}^{2} d t

Answer & Explanation

Elois Puryear

Beginner2022-01-21Added 30 answers

If integration by parts gave you 0, its

Dabanka4v

Beginner2022-01-22Added 36 answers

I’m going to assume that

\frac{x (t)}{d t}

is a typo for

\frac{d (x (t))}{d t}

.
Look at a specific example, say with

x (t) = \sin t

. Then

\int \frac{d (x (t))}{d t} x {(t)}^{2} d t = \int \cos t \sin^{2} t d t

,
a problem that you would most likely solve by making the substitution

u = \sin t, d u = \cos t d t

, and integrating

\int u^{2} d u

.
But you don’t have to know what x(t) is to make this substitution. If you let

u = x (t)

, then

d u = \frac{d (x (t))}{d t} d t

,
and

\int \frac{d (x (t))}{d t} x {(t)}^{2} d t = \int u^{2} d u = \frac{u^{3}}{3} + C = \frac{x {(t)}^{3}}{3} + C

RizerMix

Expert2022-01-27Added 656 answers

Also, since

\int \frac{d x (t)}{d t} x (t)^{2} d t = \int \frac{d f (t)}{d t} d t

, where

f (t) = \frac{1}{3} x (t)^{3}

, we can use the Second Fundamental Theorem of Calculus to conclude that

\int \frac{d x (t)}{d t} x (t)^{2} d t = f (t) + C = \frac{1}{3} x (t)^{3} + C

. No substitution needed.

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