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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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)

Vedclass · Accepted Answer

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

Vedclass · Answer

(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

Similar Questions

$\int \frac{x}{\sqrt{4 - x^4}} \, dx = $

$\int \frac{1}{x^2 \sqrt{1-x^2}} \cdot d x = \dots + C$. Where,$(0 < |x| < 1)$.

If $\int \frac{\sqrt{1-x^2}}{x^4} \,dx = A(x)\left(\sqrt{1-x^2}\right)^{m} + c$ for a suitably chosen integer $m$ and a function $A(x)$,where $c$ is a constant of integration,then $(A(x))^{m}$ equals

Integrate the function: $\frac{3x^{2}}{x^{6}+1}$

$\int \sin^3 x \cos^2 x \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module