Question

$\begin{gathered}{\boxed{\boxed{\begin{array}{cc}\bf \: if \: \: f(x) = \:\begin{cases} &\sf{1 - |x| } \: \: \: \: \: when \: \: \: |x| \leqslant 0\ \\ \\ &\sf{ |x| - 1 } \: \: \: \: \: when \: \: \: \: |x| > 0 \end{cases} \\ \\ \\ \bf \: and \\ \\ \bf \: g(x) = f(x - 1) + f(x + 1) \\ \\ \\ \sf \: then \: find \: the \: value \: of \: \: \: \displaystyle \: \int_ { - 3}^{5} \rm \: \: g(x) \: dx\end{array}}}}\end{gathered}$



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Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{f(x)=\begin{cases}\tt{1-|x|\,,\,\,\,|x|\le0}\\\\\tt{|x|-1\,,\,\,\,|x|>0}\end{cases}}$

This can be rewritten as

$\tt{f(x)=\begin{cases}\tt{-x-1\,,\,\,\,x<0}\\\tt{1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,x=0}\\\tt{x-1\,\,\,\,\,\,,\,\,\,x>0}\end{cases}}$

Now,

$\tt{f(x-1)=\begin{cases}\tt{-x\,\,\,\,\,\,\,\,\,\,\,,\,\,\,x<1}\\\tt{1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,x=1}\\\tt{x-2\,\,\,\,\,\,,\,\,\,x>1}\end{cases}}$

And,

$\tt{f(x+1)=\begin{cases}\tt{-x-2\,,\,\,\,x<-1}\\\tt{1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,x=-1}\\\tt{x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,x>-1}\end{cases}}$

Now, we have,

$\displaystyle\tt{\int^{5}_{-3}\,g(x)\,dx}$

$\displaystyle\tt{=\int^{5}_{-3}\,\{f(x-1)+f(x+1)\}\,dx}$

$\displaystyle\tt{=\int^{5}_{-3}\,f(x-1)\,dx+\int^{5}_{-3}\,f(x+1)\,dx}$

$\displaystyle\tt{=\int^{1}_{-3}\,(-x)\,dx+\int^{5}_{1}(x-2)\,dx+\int^{-1}_{-3}\,(-x-2)\,dx+\int^{5}_{-1}\,x\,dx}$

$\displaystyle\tt{=-\left[\dfrac{x^2}{2}\right]^{1}_{-3}+\left[\dfrac{x^2}{2}-2x\right]^{5}_{1}-\left[\dfrac{x^2}{2}+2x\right]^{-1}_{-3}+\left[\dfrac{x^2}{2}\right]^{5}_{-1}}$

$\displaystyle\tt{=-\left\{\dfrac{1}{2}-\dfrac{9}{2}\right\}+\left\{\dfrac{25}{2}-10-\dfrac{1}{2}+2\right\}-\left\{\dfrac{1}{2}-2-\dfrac{9}{2}+6\right\}+\left\{\dfrac{25}{2}-\dfrac{1}{2}\right\}}$

$\displaystyle\tt{=-\left\{\dfrac{1-9}{2}\right\}+\left\{\dfrac{25-1}{2}-8\right\}-\left\{\dfrac{1-9}{2}+4\right\}+\left\{\dfrac{25-1}{2}\right\}}$

$\displaystyle\tt{=-\left\{\dfrac{-8}{2}\right\}+\left\{\dfrac{24}{2}-8\right\}-\left\{\dfrac{-8}{2}+4\right\}+\left\{\dfrac{24}{2}\right\}}$

$\displaystyle\tt{=4+4-0+12}$

$\displaystyle\tt{=20}$

Answer 2

Answer:

We have,

$\tt{f(x)=\begin{cases}\tt{1-|x|\,,\,\,\,|x|\le0}\\\\\tt{|x|-1\,,\,\,\,|x|>0}\end{cases}}$

This can be rewritten as

$\tt{f(x)=\begin{cases}\tt{-x-1\,,\,\,\,x<0}\\\tt{1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,x=0}\\\tt{x-1\,\,\,\,\,\,,\,\,\,x>0}\end{cases}}$

Now,

$\tt{f(x-1)=\begin{cases}\tt{-x\,\,\,\,\,\,\,\,\,\,\,,\,\,\,x<1}\\\tt{1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,x=1}\\\tt{x-2\,\,\,\,\,\,,\,\,\,x>1}\end{cases}}$

And,

$\tt{f(x+1)=\begin{cases}\tt{-x-2\,,\,\,\,x<-1}\\\tt{1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,x=-1}\\\tt{x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,x>-1}\end{cases}}$

Now, we have,

$\displaystyle\tt{\int^{5}_{-3}\,g(x)\,dx}$

$\displaystyle\tt{=\int^{5}_{-3}\,\{f(x-1)+f(x+1)\}\,dx}$

$\displaystyle\tt{=\int^{5}_{-3}\,f(x-1)\,dx+\int^{5}_{-3}\,f(x+1)\,dx}$

$\displaystyle\tt{=\int^{1}_{-3}\,(-x)\,dx+\int^{5}_{1}(x-2)\,dx+\int^{-1}_{-3}\,(-x-2)\,dx+\int^{5}_{-1}\,x\,dx}$

$\displaystyle\tt{=-\left[\dfrac{x^2}{2}\right]^{1}_{-3}+\left[\dfrac{x^2}{2}-2x\right]^{5}_{1}-\left[\dfrac{x^2}{2}+2x\right]^{-1}_{-3}+\left[\dfrac{x^2}{2}\right]^{5}_{-1}}$

$\displaystyle\tt{=-\left\{\dfrac{1}{2}-\dfrac{9}{2}\right\}+\left\{\dfrac{25}{2}-10-\dfrac{1}{2}+2\right\}-\left\{\dfrac{1}{2}-2-\dfrac{9}{2}+6\right\}+\left\{\dfrac{25}{2}-\dfrac{1}{2}\right\}}$

$\displaystyle\tt{=-\left\{\dfrac{1-9}{2}\right\}+\left\{\dfrac{25-1}{2}-8\right\}-\left\{\dfrac{1-9}{2}+4\right\}+\left\{\dfrac{25-1}{2}\right\}}$

$\displaystyle\tt{=-\left\{\dfrac{-8}{2}\right\}+\left\{\dfrac{24}{2}-8\right\}-\left\{\dfrac{-8}{2}+4\right\}+\left\{\dfrac{24}{2}\right\}}$

$\displaystyle\tt{=4+4-0+12}$

$\displaystyle\tt{=20}$

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