Evaluate the integral int_{0}^{2} x sqrt{x+2} | vedclass

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

Question

(C) We need to evaluate the integral $I = \int_{0}^{2} x \sqrt{x+2} \, dx$.Let $x+2 = t^{2}$. Then $x = t^{2}-2$ and $dx = 2t \, dt$.Change the limits

Vedclass · Accepted Answer

$32/15$

Vedclass · Answer

(C) We need to evaluate the integral $I = \int_{0}^{2} x \sqrt{x+2} \, dx$.Let $x+2 = t^{2}$. Then $x = t^{2}-2$ and $dx = 2t \, dt$.Change the limits of integration:When $x=0$,$t^{2} = 0+2 = 2$,so $t = \sqrt{2}$.When $x=2$,$t^{2} = 2+2 = 4$,so $t = 2$.Substituting these into the integral:$I = \int_{\sqrt{2}}^{2} (t^{2}-2) \sqrt{t^{2}} \cdot (2t) \, dt$$I = \int_{\sqrt{2}}^{2} (t^{2}-2) \cdot t \cdot 2t \, dt$$I = 2 \int_{\sqrt{2}}^{2} (t^{4}-2t^{2}) \, dt$Now,integrate with respect to $t$:$I = 2 \left[ \frac{t^{5}}{5} - \frac{2t^{3}}{3} \right]_{\sqrt{2}}^{2}$$I = 2 \left[ \left( \frac{32}{5} - \frac{16}{3} \right) - \left( \frac{(\sqrt{2})^{5}}{5} - \frac{2(\sqrt{2})^{3}}{3} \right) \right]$$I = 2 \left[ \left( \frac{96-80}{15} \right) - \left( \frac{4\sqrt{2}}{5} - \frac{4\sqrt{2}}{3} \right) \right]$$I = 2 \left[ \frac{16}{15} - \left( \frac{12\sqrt{2}-20\sqrt{2}}{15} \right) \right]$$I = 2 \left[ \frac{16}{15} - \left( -\frac{8\sqrt{2}}{15} \right) \right] = 2 \left( \frac{16+8\sqrt{2}}{15} \right) = \frac{32+16\sqrt{2}}{15}$

Evaluate the integral $\int_{0}^{2} x \sqrt{x+2} \, dx$ using the substitution $x+2=t^{2}$.

Similar Questions

$\int_0^{\pi /4} \frac{\sec^2 x}{(1 + \tan x)(2 + \tan x)} \,dx = $

$\int_0^{\pi / 2} e^{\sin x} \cdot \cos x \, dx =$

$\int_{\frac{1}{3}}^1 \frac{(x-x^3)^{\frac{1}{3}}}{x^4} dx = $ . . . . . . .

If $\frac{d[f(x)]}{dx} = g(x)$ for $a \le x \le b$,then $\int_a^b f(x)g(x) dx$ equals

The value of $\int_0^2 \frac{3^{\sqrt{x}}}{\sqrt{x}} \, dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module