Let $f(x)=\int x^{3}\sqrt{3-x^{2}} dx$. If $5f(\sqrt{2})=-4$, then $f(1)$ is equal to
- (1) $-\frac{2\sqrt{2}}{5}$
- (2) $-\frac{6\sqrt{2}}{5}$
- (3) $-\frac{4\sqrt{2}}{5}$
- (4) $-\frac{8\sqrt{2}}{5}$
Solution:
$$f(x) = \int x^{3}\sqrt{3-x^{2}} dx = \int x^{2}\sqrt{3-x^{2}} \cdot x dx$$
Let $3-x^{2} = t^{2}$
$$-2x dx = 2t dt \Rightarrow x dx = -t dt$$
$$x^{2} = 3-t^{2}$$
$$f(x) = \int (3-t^{2}) \cdot t \cdot (-t dt)$$
$$f(x) = \int (t^{4} – 3t^{2}) dt$$
$$f(x) = \frac{t^{5}}{5} – t^{3} + c$$
$$f(x) = \frac{(3-x^{2})^{5/2}}{5} – (3-x^{2})^{3/2} + c$$
$$f(\sqrt{2}) = \frac{(3-(\sqrt{2})^{2})^{5/2}}{5} – (3-(\sqrt{2})^{2})^{3/2} + c$$
$$f(\sqrt{2}) = \frac{(1)^{5/2}}{5} – (1)^{3/2} + c$$
$$f(\sqrt{2}) = \frac{1}{5} – 1 + c = -\frac{4}{5} + c$$
$$5f(\sqrt{2}) = -4$$
$$5\left(-\frac{4}{5} + c\right) = -4$$
$$-4 + 5c = -4 \Rightarrow 5c = 0 \Rightarrow c = 0$$
$$f(1) = \frac{(3-1^{2})^{5/2}}{5} – (3-1^{2})^{3/2} + 0$$
$$f(1) = \frac{2^{5/2}}{5} – 2^{3/2}$$
$$f(1) = \frac{4\sqrt{2}}{5} – 2\sqrt{2}$$
$$f(1) = \sqrt{2}\left(\frac{4}{5} – 2\right)$$
$$f(1) = \sqrt{2}\left(\frac{4 – 10}{5}\right)$$
$$f(1) = -\frac{6\sqrt{2}}{5}$$
Ans. (2)