Let _-2^2(| x|+[x x])dx=2(3- 2)+ , where [ ] is the greatest integer function.

Question

Let $\int _{-2}^{2}(|\sin x|+[x\sin x])dx=2(3-\cos 2)+\beta $, where [ \cdot  ] is the greatest integer function. Then $\beta \sin (\frac{\beta }{2})$ equals:

Accepted Answer

Let the given integral be denoted by $I$. We can split $I$ into two separate integrals:

$I=\int _{-2}^{2}(|\sin x|+[x\sin x])dx=\int _{-2}^{2}|\sin x|dx+\int _{-2}^{2}[x\sin x]dx$

Let ${I}_{1}=\int _{-2}^{2}|\sin x|dx$ and ${I}_{2}=\int _{-2}^{2}[x\sin x]dx$.

Evaluate ${I}_{1}: 1: Evaluate ${I}\_{1}$

Based on the property of definite integrals, if $f(x)$ is an even function (i.e., $f(-x)=f(x)$), then ${\int }\_{a}\^{-a}f(x)dx=2{\int }\_{a}\^{0}f(x)dx$.

${I}\_{1}=2\int \_{0}\^{2}|\sin x|dx$

In the interval $[0,2]$, $2$ radians is less than $\pi (\approx 3.14)$, so $\sin x>0$. Hence, $|\sin x|=\sin x$.

${I}\_{1}=2\int \_{0}\^{2}\sin xdx$

${I}\_{1}=2[-\cos x{]}\_{2}\^{0}=2(-\cos 2-(-\cos 0))$

${I}\_{1}=2(1-\cos 2)=2-2\cos 2$

Evaluate ${I}\_{2}: 2: Evaluate ${I}_{2}$

Let $f(x)=x\sin x$. We check if it is even or odd:

$f(-x)=(-x)\sin (-x)=(-x)(-\sin x)=x\sin x=f(x)$.

Since $f(x)$ is an even function, $[x\sin x]$ is also an even function.

${I}_{2}=2\int _{0}^{2}[x\sin x]dx$

Now, let us analyze the behavior of $f(x)=x\sin x$ in the interval $x\in [0,2]$:

When $x=0$, $f(0)=0$.

When $x=2$, $f(2)=2\sin 2$.

Notice that $\pi /2So, $1.732Let's verify if $f(x)$ is strictly increasing in $[0,2]$. We differentiate $f(x)$ with respect to $x$:

${f}^{'}(x)=\sin x+x\cos x$

For $x\in (0,\pi /2]$, both $\sin x>0$ and $\cos x\ge 0$, so ${f}^{'}(x)>0$.

For $x\in (\pi /2,2]$, $\cos x${f}\^{'}(x)=\cos x(	an x+x)$

Since $	an x$ is negative in the second quadrant and $2$ radians is smaller than ${	an }\^{-1}(-2)$, the expression $(	an x+x)$ evaluates to a negative value. Therefore, $\cos x$ (negative) $	imes$ $(	an x+x)$ (negative) will yield ${f}\^{'}(x)>0$.

Since $f(x)$ strictly increases from $0$ to a value between $1$ and $2$, there is exactly one real number $\alpha \in (0,2)$ at which $f(\alpha )=1$. So,

$\alpha \sin \alpha =1$

For $x\in [0,\alpha )$, the function values are $0\le x\sin xFor $x\in [\alpha ,2]$, the function values are $1\le x\sin xNow, split the integral ${I}\_{2}$ at $x=\alpha $:

${I}\_{2}=2(\int \_{0}\^{\alpha }0dx+\int \_{\alpha }\^{2}1dx)$

${I}\_{2}=2[x{]}\_{2}\^{\alpha }=2(2-\alpha )=4-2\alpha $

Finding the value of $\beta \sin (\frac{\beta }{2}): 3: Finding the value of $\beta \sin (\frac{\beta }{2})$

Combining ${I}_{1}$ and ${I}_{2}$, we get the total integral:

$I={I}_{1}+{I}_{2}=(2-2\cos 2)+(4-2\alpha )=6-2\cos 2-2\alpha $

$I=2(3-\cos 2)-2\alpha $

According to the question, the total integral evaluates to:

$I=2(3-\cos 2)+\beta $

Comparing our result with the given equation, we find:

$\beta =-2\alpha $

Finally, substituting $\beta =-2\alpha $ into the expression $\beta \sin (\frac{\beta }{2})$:

$\beta \sin (\frac{\beta }{2})=-2\alpha \cdot \sin (\frac{-2\alpha }{2})$

$\beta \sin (\frac{\beta }{2})=-2\alpha \cdot \sin (-\alpha )=-2\alpha (-\sin \alpha )$

$\beta \sin (\frac{\beta }{2})=2(\alpha \sin \alpha )$

From our earlier deduction in Step 2, we established that $\alpha \sin \alpha =1$. Therefore:

$\beta \sin (\frac{\beta }{2})=2(1)=2$

Option B is the correct answer.

Solutions · 1