Integration review Evaluate the following integrals. dpJo4 -l04 - V

Given Data The integral function is ∫01dp4-p. Let, p=addpp=ddpa12p=dadpdp=2pda=2adap=a0a0=0a1=1=1

Answered: dp Jo4 - l04 - V

Elementary Algebra

17th Edition

ISBN:9780998625713

Author:Lynn Marecek, MaryAnne Anthony-Smith

Publisher:Lynn Marecek, MaryAnne Anthony-Smith

Chapter1: Foundations

Section1.7: Decimals

Problem 1.200TI: Divide: 4.11(0.12) .

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Integration review Evaluate the following integrals.

Expert Solution

Step 1

Given Data

The integral function is $\int_{0}^{1} \frac{d p}{4 - \sqrt{p}}$ .

Let,

$\begin{array}{rcl} \sqrt{p} & = & a \\ \frac{d}{d p} (\sqrt{p}) & = & \frac{d}{d p} (a) \\ \frac{1}{2 \sqrt{p}} & = & \frac{d a}{d p} \\ d p & = & 2 \sqrt{p} d a \\ = & 2 a d a \\ \sqrt{p} & = & a_{0} \\ a_{0} & = & 0 \\ a_{1} & = & \sqrt{1} \\ = & 1 \end{array}$

Step 2

Put the values in the expression $\int_{0}^{1} \frac{d p}{4 - \sqrt{p}}$ ,

$\begin{array}{rcl} \int_{0}^{1} \frac{d p}{4 - \sqrt{p}} & = & \int_{0}^{1} \frac{2 a d a}{4 - a} \\ = & 2 \int_{0}^{1} \frac{d a}{4 - a} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (I) \end{array}$

Let,

$\begin{array}{rcl} 4 - a & = & w \\ \frac{d}{d a} (4 - a) & = & \frac{d w}{d a} \\ - 1 & = & \frac{d w}{d a} \\ d a & = & - d w \\ w_{a = 0} & = & 4 - a \\ = & 4 - 0 \\ = & 4 \\ w_{a = 1} & = & 4 - 1 \\ = & 3 \end{array}$

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