My Math Forum Integrals help!

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 May 16th, 2017, 09:46 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?
 May 16th, 2017, 10:04 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 1,930 Thanks: 998 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
May 17th, 2017, 02:57 AM   #3
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 Originally Posted by romsek 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)?

 May 17th, 2017, 03:15 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,159 Thanks: 866 The given function, $f(x)= x^3cos(x^2)$ 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. $x^3 cos(x)= 0$ when x= 0, of course, and next, when $cos(x^2)= 0$ which happens are $x^2= \pi/2$. (By the way, "consists", here, is the wrong word. You probably mean "contains (1, 1/10)".) Thanks from sarajoveska
 May 17th, 2017, 03:19 AM #5 Math Team   Joined: Jul 2011 From: Texas Posts: 2,750 Thanks: 1400 $x^3 \cdot \cos(x^2) =0$ 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

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