ext { Let } f(x)=\int x^3 \sqrt{3-x^2} d x ext {. If } 5 f(\sqrt{2})=-4 ext {, then } f(1) ext { is equal to }

Question

ext { Let } f(x)=\int x^3 \sqrt{3-x^2} d x 	ext {. If } 5 f(\sqrt{2})=-4 	ext {, then } f(1) 	ext { is equal to }

Accepted Answer

Answer: A. -\frac{6 \sqrt{2}}{5} \begin{aligned} & \int x^3 \sqrt{3-x^2} d x \ & 3-x^2=t^2 \ & -2 x d x=2 t d t \ & =-\int t^2\left(3-t^2ight) d t=\int t^4-3 t^2 d t \ & =\frac{t^5}{5}-t^3+c \end{aligned} \begin{aligned} & f(x)=\frac{\left(3-x^2ight)^{\frac{5}{2}}}{5}-\left(3-x^2ight)^{\frac{3}{2}}+c \ & \because f(\sqrt{2})=-\frac{4}{5} \ & \Rightarrow \quad-\frac{4}{5}=\frac{1}{5}-1+c \Rightarrow c=0 \ & \begin{aligned} 	herefore \quad f(x) & =\frac{\left(3-x^2ight)^{\frac{5}{2}}}{5}-\left(3-x^2ight)^{\frac{3}{2}} \ 	ext { Now } f(1) & =\frac{2^{\frac{5}{2}}}{5}-2^{\frac{3}{2}}=2^{\frac{3}{2}}\left[\frac{2}{5}-1ight] \ & =2 \sqrt{2}\left(-\frac{3}{5}ight)=-\frac{6}{5} \sqrt{2} \end{aligned} \end{aligned}

$$ \text { Let } f(x)=\int x^3 \sqrt{3-x^2} d x \text {. If } 5 f(\sqrt{2})=-4 \text {, then } f(1) \text { is equal to } $$

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