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Question Number 32380 by mondodotto@gmail.com last updated on 24/Mar/18

Commented by prof Abdo imad last updated on 24/Mar/18

$$letproveiffisoddandimtegrablein[-a,a]$$$${\int }_{-a}^{a}f\left(x\right)dx=0wehave$$$${\int }_{-a}^{a}f\left(x\right)dx={\int }_{-a}^{0}f\left(x\right)dx+{\int }_0^{a}f\left(x\right)dxch.x=-t$$$$give{\int }_{-a}^{0}f\left(x\right)dx={\int }_a^{0}f(-t)(-dt)$$$$={\int }_a^{0}f\left(t\right)dt=-{\int }_0^{a}f\left(t\right)dt\Rightarrow$$$${\int }_{-a}^{a}f\left(x\right)dx=-{\int }_0^{a}f\left(x\right)dx+{\int }_0^{a}f\left(x\right)dx=0$$

Commented by prof Abdo imad last updated on 24/Mar/18

$$thefunctionx\to x^{99}{cos}^{4}xisoddso$$$${\int }_{-1}^1{x}^{99}{cos}^{4}xdx=0.$$

Answered by Joel578 last updated on 24/Mar/18

$$f\left(x\right)=x^{99}{\cos }^{4}x$$$$f(-x)={(-x)}^{99}{\cos }^4(-x)=-x^{99}{\cos }^{4}x=-f\left(x\right)$$$$\mathrm{So},f\left(x\right)\mathrm{is}\mathrm{an}\mathrm{odd}\mathrm{function}$$$${\int }_{-1}^1{x}^{99}{\cos }^{4}xdx=0$$

Commented by mondodotto@gmail.com last updated on 24/Mar/18

$$\mathrm{how}?$$

Commented by Joel578 last updated on 24/Mar/18

$$\mathrm{Sorry},{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{corrected}\mathrm{now}.\mathrm{Thank}\mathrm{you}$$

Commented by MJS last updated on 24/Mar/18

$$\mathrm{but}{\mathrm{it}}^{\prime }\mathrm{s}uneven,\mathrm{so}{\stackrel{1}{\int }}_{-1}f\left(x\right)=0$$

Commented by MJS last updated on 24/Mar/18

$$...\mathrm{I}\mathrm{thought}\mathrm{it}\mathrm{must}\mathrm{have}\mathrm{been}\mathrm{a}\mathrm{typo}$$

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