My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum


Thanks Tree4Thanks
  • 1 Post By romsek
  • 1 Post By Country Boy
  • 2 Post By skeeter
Reply
 
LinkBack Thread Tools Display Modes
May 16th, 2017, 10:46 AM   #1
Member
 
Joined: May 2017
From: Slovenia

Posts: 89
Thanks: 0

Integrals help!

Ok, so I got stuck here.

Find the area bounded by the x axis and the graph of the function f(x) = x^3cos(x^2) and also consists the point (1, 1/10). The last part with the point confused me, any help?
sarajoveska is offline  
 
May 16th, 2017, 11:04 AM   #2
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: Southern California, USA

Posts: 1,601
Thanks: 816

I'm going to guess that point just means they want you to integrate the bit of that graph above the x-axis, i.e. $x \in \left[0, \sqrt{\dfrac \pi 2}\right]$
Thanks from greg1313
romsek is online now  
May 17th, 2017, 03:57 AM   #3
Member
 
Joined: May 2017
From: Slovenia

Posts: 89
Thanks: 0

Quote:
Originally Posted by romsek View Post
I'm going to guess that point just means they want you to integrate the bit of that graph above the x-axis, i.e. $x \in \left[0, \sqrt{\dfrac \pi 2}\right]$
I still don't get it. How did you get 0, sqrt(pi/2)?
sarajoveska is offline  
May 17th, 2017, 04:15 AM   #4
Math Team
 
Joined: Jan 2015
From: Alabama

Posts: 2,820
Thanks: 750

The given function, crosses the x-axis many times so has some pieces above the x-axis, others below. romsek is suggesting that you take the first of those above the x-axis. when x= 0, of course, and next, when which happens are .

(By the way, "consists", here, is the wrong word. You probably mean "contains (1, 1/10)".)
Thanks from sarajoveska
Country Boy is offline  
May 17th, 2017, 04:19 AM   #5
Math Team
 
Joined: Jul 2011
From: Texas

Posts: 2,656
Thanks: 1327

$x^3 \cdot \cos(x^2) =0$ at $x=0$ and whenever $x^2$ equals an odd-integer multiple of $\dfrac{\pi}{2}$

FYI, $0 < \dfrac{1}{10} < \sqrt{\dfrac{\pi}{2}}$, and $x^3 \cdot \cos(x^2) > 0$ on that same interval.

$\displaystyle A = \int_0^{\sqrt{\pi/2}} x^3 \cdot \cos(x^2) \, dx$

Let $t=x^2 \implies dt = 2x \, dx$ ...

$\displaystyle A = \dfrac{1}{2} \int_0^{\pi/2} t \cdot \cos(t) \, dt$

Looks like integration by parts from this point ... I'll let you finish.
Thanks from greg1313 and sarajoveska
skeeter is offline  
Reply

  My Math Forum > College Math Forum > Calculus

Tags
integrals



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
integrals sin and cos The_Ys_Guy Calculus 4 July 8th, 2016 05:35 AM
Integrals joshbeldon Calculus 5 January 16th, 2015 07:41 AM
Two integrals Dacu Calculus 3 June 28th, 2014 05:56 AM
Some help with Integrals Calii Calculus 0 November 6th, 2010 03:26 PM
integrals hector manuel Real Analysis 0 May 5th, 2009 12:16 AM





Copyright © 2017 My Math Forum. All rights reserved.