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

We have $$f(x)=\int x^{3}\sqrt{3-x^{2}}\;dx$$. To integrate, put $$u=3-x^{2}\;$$ so that $$du=-2x\,dx$$, hence $$x\,dx=-\dfrac{du}{2}$$.

Rewrite $$x^{3}\,dx=x^{2}(x\,dx)=(3-u)\!\left(-\dfrac{du}{2}\right)=-\dfrac{1}{2}(3-u)\,du$$.

Therefore $$f(x)=\int -\dfrac{1}{2}(3-u)\,u^{1/2}\,du$$ $$\quad =-\dfrac{1}{2}\!\left[3\!\int u^{1/2}du-\!\int u^{3/2}du\right].$$

Using $$\int u^{n}\,du=\dfrac{u^{n+1}}{n+1}$$, we get $$\int u^{1/2}du=\dfrac{2}{3}u^{3/2},\qquad \int u^{3/2}du=\dfrac{2}{5}u^{5/2}.$$

Hence $$f(x)=-\dfrac{1}{2}\!\left[3\!\left(\dfrac{2}{3}u^{3/2}\right)-\dfrac{2}{5}u^{5/2}\right]+C =-u^{3/2}+\dfrac{1}{5}u^{5/2}+C.$$

Re-substituting $$u=3-x^{2}$$ gives $$f(x)=-(3-x^{2})^{3/2}+\dfrac{1}{5}(3-x^{2})^{5/2}+C.$$

The condition $$5f(\sqrt{2})=-4$$ fixes the constant. At $$x=\sqrt{2}$$, $$3-x^{2}=1$$, so $$f(\sqrt{2})=-1+\dfrac{1}{5}(1)+C=-\dfrac{4}{5}+C.$$ Thus $$5f(\sqrt{2})=5\!\left(-\dfrac{4}{5}+C\right)=-4+5C=-4,$$ which gives $$C=0.$$

Therefore $$f(x)=-(3-x^{2})^{3/2}+\dfrac{1}{5}(3-x^{2})^{5/2}.$$

Now evaluate at $$x=1$$: here $$3-x^{2}=2$$.

Compute the powers: $$(3-1)^{3/2}=2^{3/2}=2\sqrt{2},\qquad (3-1)^{5/2}=2^{5/2}=4\sqrt{2}.$$

Hence $$f(1)=-(2\sqrt{2})+\dfrac{1}{5}(4\sqrt{2}) =-2\sqrt{2}+\dfrac{4}{5}\sqrt{2} =\left(-\dfrac{10}{5}+\dfrac{4}{5}\right)\sqrt{2} =-\dfrac{6}{5}\sqrt{2}.$$

Thus $$f(1)=-\dfrac{6\sqrt{2}}{5}$$, which matches Option D.