Pergunta
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5. [-/1 Points]
Find the indefinite integral and check the result by differentiation.
int sqrt [3](3-2x^2)(-4x)dx
square +c
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Your last submission is used for your score. 5. [-/1 Points] Find the indefinite integral and check the result by differentiation. int sqrt [3](3-2x^2)(-4x)dx square +c DETAILS MY NOTES
Solução

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MarcosProfissional · Tutor por 6 anos
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To find the indefinite integral \(\int \sqrt[3]{3-2x^2}(-4x) \, dx\), we can use the substitution method.<br /><br />Let \( u = 3 - 2x^2 \). Then, \( du = -4x \, dx \).<br /><br />Now, rewrite the integral in terms of \( u \):<br /><br />\[<br />\int \sqrt[3]{3-2x^2}(-4x) \, dx = \int \sqrt[3]{u} \, du<br />\]<br /><br />Next, integrate \(\sqrt[3]{u}\):<br /><br />\[<br />\int \sqrt[3]{u} \, du = \frac{u^{4/3}}{\frac{4}{3}} + C = \frac{3}{4}u^{4/3} + C<br />\]<br /><br />Finally, substitute back \( u = 3 - 2x^2 \):<br /><br />\[<br />\int \sqrt[3]{3-2x^2}(-4x) \, dx = \frac{3}{4}(3 - 2x^2)^{4/3} + C<br />\]<br /><br />To check the result by differentiation, we differentiate \(\frac{3}{4}(3 - 2x^2)^{4/3} + C\):<br /><br />\[<br />\frac{d}{dx} \left( \frac{3}{4}(3 - 2x^2)^{4/3} + C \right) = \frac{3}{4} \cdot \frac{4}{3}(3 - 2x^2)^{-1/3} \cdot (-4x) = \sqrt[3]{3-2x^2}(-4x)<br />\]<br /><br />Since the derivative matches the original integrand, our solution is correct.<br /><br />Thus, the indefinite integral is:<br /><br />\[<br />\boxed{\frac{3}{4}(3 - 2x^2)^{4/3} + C}<br />\]
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