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Find, correct to one decimal place, the value of

$$\int_{0}^{60} 2\sin(x/2) \, dx.$$

Can someone please show me how this question is done. It would be very helpful thanks!

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To re-confirm, the ranges are in radian, right? –  lab bhattacharjee Aug 6 '13 at 13:40
    
What do you know about integration? –  Daniel Littlewood Aug 6 '13 at 13:42
    
Yes the original bound were in radians. But i converted it –  Red Queen10101 Aug 6 '13 at 23:58
    
Sorry for the late reply, totally forgot about this. –  Red Queen10101 Aug 7 '13 at 0:02
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1 Answer

up vote 1 down vote accepted

To solve this integral one must use a substitution. If we set $u=x/2$, then $du=(1/2)dx$, which gives $2\,du=dx$. Thus, the integral then becomes, $$2\int_0^{30} 2\sin(u)\,du.$$ To get the limits on the integral, we used our substitution again, that is, at $x=0$, $u=0/2=0$ and at $x=60$, $u=60/2=30$. With all the pieces determined, the integral evaluates to,

$$2\int_0^{30}2\sin(u)du=2(-2\cos(u))\bigg|_0^{30}=-4(\cos(30)-1)$$

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