If f(x) = frac{x^3+5}{sqrt{12+x}} and int_{-5 | vedclass

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(A) Let $I = \int_{-5}^5 \frac{x^3+5}{\sqrt{12+x}} dx$. Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we have: $I = \int_{-5}^5 \frac{(

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$\frac{5-x^3}{\sqrt{12-x}}$

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(A) Let $I = \int_{-5}^5 \frac{x^3+5}{\sqrt{12+x}} dx$. Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we have: $I = \int_{-5}^5 \frac{(-5+5-x)^3+5}{\sqrt{12+(-5+5-x)}} dx = \int_{-5}^5 \frac{-x^3+5}{\sqrt{12-x}} dx$. Adding the two expressions for $I$: $2I = \int_{-5}^5 \left( \frac{x^3+5}{\sqrt{12+x}} + \frac{5-x^3}{\sqrt{12-x}} \right) dx$. Since the integrand is even,we can write: $2I = 2 \int_0^5 \left( \frac{x^3+5}{\sqrt{12+x}} + \frac{5-x^3}{\sqrt{12-x}} \right) dx$. Thus,$I = \int_0^5 \left( \frac{x^3+5}{\sqrt{12+x}} + \frac{5-x^3}{\sqrt{12-x}} \right) dx$. Given $\int_{-5}^5 f(x) dx = \int_0^5 (f(x) + g(x)) dx$,we compare the integrands: $\int_0^5 (f(x) + g(x)) dx = \int_0^5 \left( \frac{x^3+5}{\sqrt{12+x}} + \frac{5-x^3}{\sqrt{12-x}} \right) dx$. Therefore,$g(x) = \frac{5-x^3}{\sqrt{12-x}}$.

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

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