5.2.1 vol of sin x^2 ; 0\le x \le \dfrac{\pi}{2}

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karush
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volume of the solid $y=\sin (x^2)\quad 0\le x \le \dfrac{\pi}{2}$
$\displaystyle \int_0^{\pi/2}\sin (x^2)\ dx$
ok think this should be area not volume but hope my int is set up ok
 
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The integral you have is indeed area. Your statement of “volume” is incomplete.
Is this a volume of rotation with respect to a given axis of rotation, or is it a volume
of similar cross sections whose base lies in a region defined by the given curve,
or … something else?
 
If you really intend $\int_0^{\pi/2} \sin(x^2) dx$, that cannot be integrated in terms of elementary function but can be approximated to any desired accuracy.