Please show me how to get to the answer with detailed work. Using a substitution I changeddx intoke" duWhat was my u ?What was my du ?HINT: be sure to include drWhat was my new lower limit a =HINT: turn the lower limit into a u valueWhat was my new upper limit b =What constant shows up after you do the substitution kEvaluatedr =

We have, ∫14e1xx3dx Let us consider, u=1x ⋯⋯1du=-12x-12-1dxdu=-12x3dx…

Answered: ev - dx into Using a substitution I…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Please show me how to get to the answer with detailed work.

Using a substitution I changed
dx into
ke" du
What was my u ?
What was my du ?
HINT: be sure to include dr
What was my new lower limit a =
HINT: turn the lower limit into a u value
What was my new upper limit b =
What constant shows up after you do the substitution k
Evaluate
dr =

Expert Solution

Step 1

We have,

$\int_{1}^{4} \frac{e^{\frac{1}{\sqrt{x}}}}{\sqrt{x^{3}}} d x$

Let us consider,

$\begin{array}{rcl} u & = & \frac{1}{\sqrt{x}} \dots \dots (1) \\ d u & = & \frac{- 1}{2} {(x)}^{\frac{- 1}{2} - 1} d x \\ d u & = & \frac{- 1}{2 \sqrt{x^{3}}} d x \dots \dots (2) \\ - 2 d u & = & \frac{1}{\sqrt{x^{3}}} d x \end{array}$

Now for upper and lower limit,

$\begin{array}{rcl} A t x & = & 1 \Rightarrow u = 1 \\ a n d a t x & = & 4 \Rightarrow u = \frac{1}{2} \end{array}$

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