Let f(x) = int x^3 sqrt{3-x^2} dx. If 5f(sqrt | vedclass

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(D) Let $3-x^2 = t^2$. Then $-2x dx = 2t dt$,which implies $x dx = -t dt$.Also,$x^2 = 3-t^2$.Substituting these into the integral:$f(x) = \int (3-t^2)

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$-\frac{6\sqrt{2}}{5}$

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(D) Let $3-x^2 = t^2$. Then $-2x dx = 2t dt$,which implies $x dx = -t dt$.Also,$x^2 = 3-t^2$.Substituting these into the integral:$f(x) = \int (3-t^2) \cdot t \cdot (-t dt) + C$$f(x) = \int (t^4 - 3t^2) dt + C$$f(x) = \frac{t^5}{5} - t^3 + C$Substituting back $t = \sqrt{3-x^2}$:$f(x) = \frac{(3-x^2)^{5/2}}{5} - (3-x^2)^{3/2} + C$Given $5f(\sqrt{2}) = -4$,we calculate $f(\sqrt{2})$:$f(\sqrt{2}) = \frac{(3-2)^{5/2}}{5} - (3-2)^{3/2} + C = \frac{1}{5} - 1 + C = -\frac{4}{5} + C$.Since $5f(\sqrt{2}) = -4$,we have $5(-\frac{4}{5} + C) = -4$,which gives $-4 + 5C = -4$,so $C = 0$.Thus,$f(x) = \frac{(3-x^2)^{5/2}}{5} - (3-x^2)^{3/2}$.Now,calculate $f(1)$:$f(1) = \frac{(3-1)^{5/2}}{5} - (3-1)^{3/2} = \frac{2^{5/2}}{5} - 2^{3/2} = \frac{4\sqrt{2}}{5} - 2\sqrt{2} = \sqrt{2}(\frac{4}{5} - 2) = \sqrt{2}(\frac{4-10}{5}) = -\frac{6\sqrt{2}}{5}$.

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

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