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Question Number 209347 by RoseAli last updated on 07/Jul/24

Answered by Berbere last updated on 07/Jul/24

$${\int }_{-4}^{4}f\left(x\right)dx=2{\underset{0}{\int }}^{4}f\left(x^2\right)dx$$$${\int }_0^{4}g\left(x\right)dx={\int }_0^{4}g(4-x)\Rightarrow 2{\int }_0^{4}g\left(x\right)={\int }_0^{4}g\left(x\right)+g(4-x)dx=0$$$$\Rightarrow 2{\int }_0^{4}f\left(x^2\right)+g\left(x\right)dx=2{\int }_0^{4}f\left(x^2\right)dx={\int }_0^{4}8x^3=2{\left[x^4\right]}_0^4=512$$$$$$

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