Let f(x) = x^3√3-x^2,dx.

Question

Let $f(x) = \displaystyle\int x^3\sqrt{3-x^2}\,dx$. If $5f(\sqrt2)=-4$, then $f(1)$ is equal to:

Accepted Answer

Let $3-x^2=t^2$, $-2x\,dx=2t\,dt$:

$$f(x) = -\int t^2(3-t^2)\,dt = \int(t^4-3t^2)\,dt = \frac{t^5}5-t^3+c = \frac{(3-x^2)^{5/2}}5-(3-x^2)^{3/2}+c$$

From $f(\sqrt2)=-4/5$: $-\dfrac45 = \dfrac15-1+c \implies c=0$.

$$f(1) = \frac{2^{5/2}}5-2^{3/2} = 2^{3/2}\left(\frac25-1\right) = 2\sqrt2\left(-\frac35\right) = -\frac{6\sqrt2}5$$

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