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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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Verified Answer
The correct answer is:
$\frac{5-x^3}{\sqrt{12-x}}$
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)
$$
2 I=\int_{-5}^5\left(\frac{x^3+5}{\sqrt{12+x}}+\frac{-x^3+5}{\sqrt{12-x}}\right) d x
$$
Since $f(-x)=f(x)$
$\begin{aligned} & \text { Hence } 2 I=2 \int_0^5\left(\frac{x^3+5}{\sqrt{12+x}}+\frac{5-x^3}{\sqrt{12-x}}\right) d x \\ & \Rightarrow \int_{-5}^5 \frac{x^3+5}{\sqrt{12+x}} d x=\int_0^5\left(\frac{x^3+5}{\sqrt{12+x}}+\frac{5-x^3}{\sqrt{12-x}}\right) d x \\ & \text { By comparision, } g(x)=\frac{5-x^3}{\sqrt{12-x}} .\end{aligned}$
$$
\Rightarrow I=\int_{-5}^5 \frac{(5-5-x)^3+5}{\sqrt{12+(5-5-x)}} d x...(By Property)...(2)
$$
Adding equation (1) \& (2)
$$
2 I=\int_{-5}^5\left(\frac{x^3+5}{\sqrt{12+x}}+\frac{-x^3+5}{\sqrt{12-x}}\right) d x
$$
Since $f(-x)=f(x)$
$\begin{aligned} & \text { Hence } 2 I=2 \int_0^5\left(\frac{x^3+5}{\sqrt{12+x}}+\frac{5-x^3}{\sqrt{12-x}}\right) d x \\ & \Rightarrow \int_{-5}^5 \frac{x^3+5}{\sqrt{12+x}} d x=\int_0^5\left(\frac{x^3+5}{\sqrt{12+x}}+\frac{5-x^3}{\sqrt{12-x}}\right) d x \\ & \text { By comparision, } g(x)=\frac{5-x^3}{\sqrt{12-x}} .\end{aligned}$
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