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Question Number 205269 by cherokeesay last updated on 14/Mar/24

Answered by som(math1967) last updated on 14/Mar/24

$${\int }_{-2}^{2}2f\left(x\right)dx[\because f\left(x\right)=f(-x)]$$$$=2{\int }_{-2}^{2}f\left(x\right)dx$$$$=4{\int }_0^{2}f\left(x\right)dx$$$$[\because f\left(x\right)=f(-x)\therefore {\int }_{-2}^{2}f\left(x\right)dx=2{\int }_0^{2}f\left(x\right)dx]$$$$=4\times -4=-16$$

Commented by cherokeesay last updated on 14/Mar/24

$$thankyou.$$

Answered by Sutrisno last updated on 14/Mar/24

$${\int }_{-2}^2(f\left(x\right)+f(-x))dx$$$$=2{\int }_0^2(f\left(x\right)+f(-x))dx$$$$=2({\int }_0^{2}f\left(x\right)dx+{\int }_0^{2}f(-x)dx)$$$$=2({\int }_0^{2}f\left(x\right)dx+{\int }_0^{2}f\left(x\right)dx)$$$$=2(-4-4)$$$$=-16$$

Commented by cherokeesay last updated on 14/Mar/24

$$thanks!$$

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