The number of elements in the set S=x:x [0,100] and _0^xt^2 (x-t)dt=x^2 is ____ Answer: 16

Question

The number of elements in the set $S={x:x\in [0,100]$ and $\int _{0}^{x}{t}^{2}\sin (x-t)dt={x}^{2}}$ is $____$

Answer: 16

Accepted Answer

$\begin{gathered} {\int }_{x}^{0}{t}^{2}\sin (x-t)dt={x}^{2} \ {\int }_{a}^{0}f(t)dt={\int }_{a}^{0}f(a-t)dt \ {\int }_{x}^{0}(x-t{)}^{2}\sin (x-(x-t))dt={\int }_{x}^{0}(x-t{)}^{2}\sin tdt={x}^{2} \ {\int }_{x}^{0}({x}^{2}-2xt+{t}^{2})\sin tdt={x}^{2} \ {x}^{2}{\int }_{x}^{0}\sin tdt-2x{\int }_{x}^{0}t\sin tdt+{\int }_{x}^{0}{t}^{2}\sin tdt={x}^{2} \end{gathered}$

differentiating with respect to $x$ (Leibniz rule):

$\begin{gathered} 2x{\int }_{x}^{0}\sin tdt+{x}^{2}\sin x-2{\int }_{x}^{0}t\sin tdt-2x(x\sin x)+{x}^{2}\sin x=2x \ 2x{\int }_{x}^{0}\sin tdt-2{\int }_{x}^{0}t\sin tdt=2x \end{gathered}$

differentiating again:

$\begin{gathered} 2{\int }_{x}^{0}\sin tdt+2x\sin x-2x\sin x=2 \ 2[-\cos t{]}_{x}^{0}=2 \ -(\cos x-\cos 0)=1 \ -\cos x+1=1\Rightarrow \cos x=0 \end{gathered}$

finding number of elements in $S={x:x\in [0,100]$ and $\cos x=0}$

$\begin{gathered} x=(2n+1)\frac{\pi }{2} \ 0\le (2n+1)\frac{\pi }{2}\le 100\Rightarrow 0\le 2n+1\le \frac{200}{\pi }\approx 63.66 \ 2n\le 62.66\Rightarrow n\le 31.33 \end{gathered}$

for $n\in {0,1,2,\ldots ,31}$, there are 32 values.

the number of elements in set $S$ is 32 .

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