int_1^2 x sqrt{4-x^2} , dx = | vedclass

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

Question

(A) Let $I = \int_1^2 x \sqrt{4-x^2} \, dx$. Substitute $u = 4-x^2$, then $du = -2x \, dx$, which implies $x \, dx = -\frac{1}{2} \, du$. When $x = 1$

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(A) Let $I = \int_1^2 x \sqrt{4-x^2} \, dx$. Substitute $u = 4-x^2$, then $du = -2x \, dx$, which implies $x \, dx = -\frac{1}{2} \, du$. When $x = 1$, $u = 4 - (1)^2 = 3$. When $x = 2$, $u = 4 - (2)^2 = 0$. Substituting these into the integral: $I = \int_3^0 \sqrt{u} \left(-\frac{1}{2}\right) \, du = \frac{1}{2} \int_0^3 u^{1/2} \, du$. $I = \frac{1}{2} \left[ \frac{u^{3/2}}{3/2} \right]_0^3 = \frac{1}{2} \cdot \frac{2}{3} \left[ u^{3/2} \right]_0^3$. $I = \frac{1}{3} (3^{3/2} - 0) = \frac{1}{3} (3 \sqrt{3}) = \sqrt{3}$.

$\int_1^2 x \sqrt{4-x^2} \, dx =$

Similar Questions

Evaluate the definite integral $\int_{0}^{1} x e^{x^{2}} d x$.

$\int_0^3 \frac{dx}{(x+2) \sqrt{x+1}} = $

$\int_0^{\pi / 4} \frac{\cos ^2 x \sin ^2 x}{\left(\cos ^3 x+\sin ^3 x\right)^2} d x$ is equal to

$\int\limits_0^1 {\frac{{{{\tan }^{ - 1}}x}}{x}\,dx} = $

The value of the integral $\int_0^{\log 5} {\frac{{{e^x}\sqrt {{e^x} - 1} }}{{{e^x} + 3}}} \,dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module