Let f(x)=\int x^3 \sqrt3-x^2 d x . If 5 f(\sqrt2)=-4 , then f(1) is equal to

Indefinite Integrals PYQ — JEE Main 2025 · 3 Apr · Shift 1 · Q32 with Solution

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Correct answer: A

We need an antiderivative of $f(x)=\int x^3\sqrt{3-x^2}\,dx.$ So let $I=\int x^3\sqrt{3-x^2}\,dx.$
Rewrite: $x^3=x^2\cdot x=(3-(3-x^2))x.$ A simpler substitution is $u=3-x^2 \implies du=-2x\,dx, \quad x\,dx=-\frac{1}{2}du.$ Also, $x^2=3-u.$ Therefore,

=\int (3-u)u^{1/2}\left(-\frac12 du\right).$$ So, $$I=-\frac12\int (3u^{1/2}-u^{3/2})\,du.$$ 3. Integrate: $$I=-\frac12\left(3\cdot \frac{2}{3}u^{3/2}-\frac{2}{5}u^{5/2}\right)+C = -u^{3/2}+\frac{1}{5}u^{5/2}+C.$$ Now substitute back $u=3-x^2$: $$f(x)=-(3-x^2)^{3/2}+\frac15(3-x^2)^{5/2}+C.$$ Factor: $$f(x)=(3-x^2)^{3/2}\left(\frac{3-x^2}{5}-1\right)+C = -\frac{x^2+2}{5}(3-x^2)^{3/2}+C.$$ 4. Use the condition $5f(\sqrt2)=-4$. At $x=\sqrt2$, $$3-x^2=3-2=1,$$ so $$f(\sqrt2)=-\frac{2+2}{5}(1)^{3/2}+C=-\frac45+C.$$ Given $$5f(\sqrt2)=-4 \implies f(\sqrt2)=-\frac45.$$ Hence, $$-\frac45+C=-\frac45 \implies C=0.$$ 5. Now compute $f(1)$: $$f(1)=-\frac{1+2}{5}(3-1)^{3/2} =-\frac35(2)^{3/2}.$$ Since $$(2)^{3/2}=2\sqrt2,$$ we get $$f(1)=-\frac35\cdot 2\sqrt2=-\frac{6\sqrt2}{5}.$$ 6. Check options: - A: $-\frac{6\sqrt2}{5}$ ✓ - B: $-\frac{8\sqrt2}{5}$ ✗ - C: $-\frac{2\sqrt2}{5}$ ✗ - D: $-\frac{4\sqrt2}{5}$ ✗ Therefore, the correct answer is **Option A**.

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View written solutionFree

Correct answer: A

We need an antiderivative of $f(x)=\int x^3\sqrt{3-x^2}\,dx.$ So let $I=\int x^3\sqrt{3-x^2}\,dx.$
Rewrite: $x^3=x^2\cdot x=(3-(3-x^2))x.$ A simpler substitution is $u=3-x^2 \implies du=-2x\,dx, \quad x\,dx=-\frac{1}{2}du.$ Also, $x^2=3-u.$ Therefore,

=\int (3-u)u^{1/2}\left(-\frac12 du\right).$$ So, $$I=-\frac12\int (3u^{1/2}-u^{3/2})\,du.$$ 3. Integrate: $$I=-\frac12\left(3\cdot \frac{2}{3}u^{3/2}-\frac{2}{5}u^{5/2}\right)+C = -u^{3/2}+\frac{1}{5}u^{5/2}+C.$$ Now substitute back $u=3-x^2$: $$f(x)=-(3-x^2)^{3/2}+\frac15(3-x^2)^{5/2}+C.$$ Factor: $$f(x)=(3-x^2)^{3/2}\left(\frac{3-x^2}{5}-1\right)+C = -\frac{x^2+2}{5}(3-x^2)^{3/2}+C.$$ 4. Use the condition $5f(\sqrt2)=-4$. At $x=\sqrt2$, $$3-x^2=3-2=1,$$ so $$f(\sqrt2)=-\frac{2+2}{5}(1)^{3/2}+C=-\frac45+C.$$ Given $$5f(\sqrt2)=-4 \implies f(\sqrt2)=-\frac45.$$ Hence, $$-\frac45+C=-\frac45 \implies C=0.$$ 5. Now compute $f(1)$: $$f(1)=-\frac{1+2}{5}(3-1)^{3/2} =-\frac35(2)^{3/2}.$$ Since $$(2)^{3/2}=2\sqrt2,$$ we get $$f(1)=-\frac35\cdot 2\sqrt2=-\frac{6\sqrt2}{5}.$$ 6. Check options: - A: $-\frac{6\sqrt2}{5}$ ✓ - B: $-\frac{8\sqrt2}{5}$ ✗ - C: $-\frac{2\sqrt2}{5}$ ✗ - D: $-\frac{4\sqrt2}{5}$ ✗ Therefore, the correct answer is **Option A**.