If $f(x)=\frac{x^3+5}{\sqrt{12+x}}$ and $\int_{-5}^5 f(x) d x=\int_0^5(f(x)+g(x)) d x$, then $g(x)=$

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Aicharya · Accepted Answer

Let $I=\int_{-5}^5 \frac{x^3+5}{\sqrt{12+x}} d x...(1)$ $ \Rightarrow I=\int_{-5}^5 \frac{(5-5-x)^3+5}{\sqrt{12+(5-5-x)}} d x...(By Property)...(2) $ Adding equation (1) \& (2) $ 2 I=\int_{-5}^5\left(\frac{x^3+5}{\sqrt{12+x}}+\frac{-x^3+5}{\sqrt{12-x}}ight) d x $ Since $f(-x)=f(x)$ $\begin{aligned} & 	ext { Hence } 2 I=2 \int_0^5\left(\frac{x^3+5}{\sqrt{12+x}}+\frac{5-x^3}{\sqrt{12-x}}ight) d x \ & \Rightarrow \int_{-5}^5 \frac{x^3+5}{\sqrt{12+x}} d x=\int_0^5\left(\frac{x^3+5}{\sqrt{12+x}}+\frac{5-x^3}{\sqrt{12-x}}ight) d x \ & 	ext { By comparision, } g(x)=\frac{5-x^3}{\sqrt{12-x}} .\end{aligned}$

If $f(x)=\frac{x^3+5}{\sqrt{12+x}}$ and $\int_{-5}^5 f(x) d x=\int_0^5(f(x)+g(x)) d x$, then $g(x)=$

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