May 16th, 2017, 09:46 AM  #1 
Member Joined: May 2017 From: Slovenia Posts: 87 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: Southern California, USA Posts: 1,410 Thanks: 715 
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 
Member Joined: May 2017 From: Slovenia Posts: 87 Thanks: 0  
May 17th, 2017, 03:15 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,649 Thanks: 681 
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: 2,624 Thanks: 1305 
$x^3 \cdot \cos(x^2) =0$ 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. 

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