JEE Main 2025 Apr 3 Shift 1, Mathematics Q15: Methods of Integration

Substitute $3-x^2=t^2$, so $xdx=-tdt$ and $x^2=3-t^2.$

$f(x)=\int x^2\sqrt{3-x^2}(xdx)=\int (3-t^2)t(-tdt)=\int (t^4-3t^2)dt=\frac{t^5}{5}-t^3+C.$

Back-substitute $t=\sqrt{3-x^2}$:

$f(x)=\frac{(3-x^2{)}^{5/2}}{5}-(3-x^2{)}^{3/2}+C.$

At $x=\sqrt{2}$: $3-x^2=1$, so $f(\sqrt{2})=\frac{1}{5}-1+C=C-\frac{4}{5}.$ Given $5f(\sqrt{2})=-4\Rightarrow f(\sqrt{2})=-\frac{4}{5}$, hence $C=0.$

At $x=1$: $3-x^2=2$, so

$f(1)=\frac{2^{5/2}}{5}-2^{3/2}=\sqrt{2}\left(\frac{4}{5}-2\right)=\sqrt{2}\cdot \left(-\frac{6}{5}\right)=-\frac{6\sqrt{2}}{5}.$