QuestionMarch 3, 2026

(b) int _(0)^3(2x)/(sqrt (9-x^2))dx

(b) int _(0)^3(2x)/(sqrt (9-x^2))dx
(b) int _(0)^3(2x)/(sqrt (9-x^2))dx

Solution
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Answer

6 Explanation 1. Use substitution Let u = 9 - x^2 \Rightarrow du = -2x\,dx. When x = 0, u = 9; when x = 3, u = 0. 2. Rewrite the integral \int_{0}^{3} \frac{2x}{\sqrt{9 - x^2}} \, dx = \int_{u=9}^{0} \frac{-du}{\sqrt{u}} 3. Adjust limits and integrate \int_{0}^{9} u^{-1/2} \, du = \left[ 2 u^{1/2} \right]_{0}^{9} = 2\sqrt{9} - 2\sqrt{0} 4. Simplify 2 \times 3 - 0 = 6

Explanation

1. Use substitution <br /> Let $u = 9 - x^2 \Rightarrow du = -2x\,dx$. <br />When $x = 0$, $u = 9$; when $x = 3$, $u = 0$. <br /><br />2. Rewrite the integral <br /> $\int_{0}^{3} \frac{2x}{\sqrt{9 - x^2}} \, dx = \int_{u=9}^{0} \frac{-du}{\sqrt{u}}$ <br /><br />3. Adjust limits and integrate <br /> $\int_{0}^{9} u^{-1/2} \, du = \left[ 2 u^{1/2} \right]_{0}^{9} = 2\sqrt{9} - 2\sqrt{0}$ <br /><br />4. Simplify <br /> $2 \times 3 - 0 = 6$
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