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

Transcript text: $\int_{0}^{1} \frac{x^{2}}{\sqrt{6-x^{3}}} d x$

Solution

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

$\boxed{\frac{2}{3} \sqrt{6} - \frac{2}{3} \sqrt{5}}$

Was this solution helpful?

Unhelpful

Helpful

Questions asked by other users

< Previous Next >