Questions: ∫ from 0 to 1 x²/√(6-x³) dx

∫ from 0 to 1 x²/√(6-x³) dx
Transcript text: $\int_{0}^{1} \frac{x^{2}}{\sqrt{6-x^{3}}} d x$
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Solution

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Solution Steps

Step 1: Perform Substitution

To solve the integral \(\int_{0}^{1} \frac{x^{2}}{\sqrt{6-x^{3}}} \, dx\), we perform a substitution. Let \( u = 6 - x^3 \). Then, the derivative is \( du = -3x^2 \, dx \), which implies \( dx = \frac{du}{-3x^2} \).

Step 2: Substitute and Simplify

Substitute \( x^2 \, dx = -\frac{1}{3} \, du \) into the integrand: \[ \int \frac{x^2}{\sqrt{6-x^3}} \, dx = \int -\frac{1}{3} \frac{1}{\sqrt{u}} \, du \]

Step 3: Integrate with Respect to \( u \)

The integral becomes: \[ -\frac{1}{3} \int \frac{1}{\sqrt{u}} \, du = -\frac{1}{3} \cdot 2\sqrt{u} = -\frac{2}{3} \sqrt{u} \]

Step 4: Substitute Back and Evaluate the Definite Integral

Substitute back \( u = 6 - x^3 \) to get: \[ -\frac{2}{3} \sqrt{6 - x^3} \]

Evaluate the definite integral from 0 to 1: \[ \left[-\frac{2}{3} \sqrt{6 - x^3}\right]_{0}^{1} = -\frac{2}{3} \sqrt{6 - 1^3} + \frac{2}{3} \sqrt{6 - 0^3} \] \[ = -\frac{2}{3} \sqrt{5} + \frac{2}{3} \sqrt{6} \]

Final Answer

\(\boxed{\frac{2}{3} \sqrt{6} - \frac{2}{3} \sqrt{5}}\)

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