[ Let f(x)= x^3 √3-x^2 d x .

Question

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

Accepted Answer

$$\begin{aligned}
& \int x^3 \sqrt{3-x^2} d x \
& 3-x^2=t^2 \
& -2 x d x=2 t d t \
& =-\int t^2\left(3-t^2\right) d t=\int t^4-3 t^2 d t \
& =\frac{t^5}{5}-t^3+c
\end{aligned}$$

$$\begin{aligned}
& f(x)=\frac{\left(3-x^2\right)^{\frac{5}{2}}}{5}-\left(3-x^2\right)^{\frac{3}{2}}+c \
& \because f(\sqrt{2})=-\frac{4}{5} \
& \Rightarrow \quad-\frac{4}{5}=\frac{1}{5}-1+c \Rightarrow c=0 \
& \begin{aligned}
\therefore \quad f(x) & =\frac{\left(3-x^2\right)^{\frac{5}{2}}}{5}-\left(3-x^2\right)^{\frac{3}{2}} \
\text { Now } f(1) & =\frac{2^{\frac{5}{2}}}{5}-2^{\frac{3}{2}}=2^{\frac{3}{2}}\left[\frac{2}{5}-1\right] \
& =2 \sqrt{2}\left(-\frac{3}{5}\right)=-\frac{6}{5} \sqrt{2}
\end{aligned}
\end{aligned}$$

Solutions · 1