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

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(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 $x = 1$

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$\sqrt{3}$

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: $I = \int_3^0 \sqrt{u} \left(-\frac{1}{2}\right) \, du = \frac{1}{2} \int_0^3 u^{1/2} \, du$. $I = \frac{1}{2} \left[ \frac{u^{3/2}}{3/2} \right]_0^3 = \frac{1}{2} \cdot \frac{2}{3} \left[ u^{3/2} \right]_0^3$. $I = \frac{1}{3} (3^{3/2} - 0) = \frac{1}{3} (3 \sqrt{3}) = \sqrt{3}$.

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

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