May 16th, 2017, 09:46 AM  #1 
Newbie Joined: May 2017 From: Slovenia Posts: 2 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? 
May 16th, 2017, 10:04 AM  #2 
Senior Member Joined: Sep 2015 From: CA Posts: 1,202 Thanks: 613 
I'm going to guess that point just means they want you to integrate the bit of that graph above the xaxis, i.e. $x \in \left[0, \sqrt{\dfrac \pi 2}\right]$

May 17th, 2017, 02:57 AM  #3 
Newbie Joined: May 2017 From: Slovenia Posts: 2 Thanks: 0  
May 17th, 2017, 03:15 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,405 Thanks: 611 
The given function, crosses the xaxis many times so has some pieces above the xaxis, others below. romsek is suggesting that you take the first of those above the xaxis. 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)".) 
May 17th, 2017, 03:19 AM  #5 
Math Team Joined: Jul 2011 From: Texas Posts: 2,508 Thanks: 1234 
$x^3 \cdot \cos(x^2) =0$ at $x=0$ and whenever $x^2$ equals an oddinteger 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. 

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 04:35 AM 
Integrals  joshbeldon  Calculus  5  January 16th, 2015 07:41 AM 
Two integrals  Dacu  Calculus  3  June 28th, 2014 04:56 AM 
Some help with Integrals  Calii  Calculus  0  November 6th, 2010 02:26 PM 
integrals  hector manuel  Real Analysis  0  May 4th, 2009 11:16 PM 