To solve the definite integral

$\int_{0}^{5} 2x (25 - x^2)^{1/3} \, dx,$

we can use the substitution $u = 25 - x^2$.

Step 1: Find the substitution

Let: $u = 25 - x^2.$ Then, we differentiate $u$ with respect to $x$: $\frac{du}{dx} = -2x \implies du = -2x \, dx \implies dx = -\frac{du}{2x}.$

Step 2: Change the limits of integration

When $x = 0$: $u = 25 - 0^2 = 25.$ When $x = 5$: $u = 25 - 5^2 = 0.$

Step 3: Substitute and rewrite the integral

Substituting into the integral gives: $\int_{0}^{5} 2x (25 - x^2)^{1/3} \, dx = \int_{25}^{0} 2x u^{1/3} \left(-\frac{du}{2x}\right).$

The $2x$ and $-\frac{1}{2x}$ cancel out, leading to: $= \int_{25}^{0} -u^{1/3} \, du = \int_{0}^{25} u^{1/3} \, du.$

Step 4: Determine $a, b, f(u)$

From the substitution, we have:

• $a = 0$
• $b = 25$
• $f(u) = u^{1/3}$

Step 5: Compute the integral

Now we can compute the integral: $\int_{0}^{25} u^{1/3} \, du.$ The antiderivative of $u^{1/3}$ is: $\frac{u^{4/3}}{4/3} = \frac{3}{4} u^{4/3}.$

Evaluating this from $0$ to $25$: $\left[ \frac{3}{4} u^{4/3} \right]_{0}^{25} = \frac{3}{4} \left( 25^{4/3} - 0^{4/3} \right).$

Calculating $25^{4/3}$: $25^{4/3} = (5^2)^{4/3} = 5^{8/3}.$

Thus, we evaluate: $\frac{3}{4} \cdot 5^{8/3}.$

Final Result

Therefore, the exact value of the integral is: $\int_{0}^{5} 2x (25 - x^2)^{1/3} \, dx = \frac{3}{4} \cdot 25^{4/3}.$

If you need further details or have any questions, let me know!

Here are some related questions you might consider:

1. How does the substitution method work in general?
2. Can you solve similar integrals using different substitutions?
3. What are the properties of definite integrals?
4. How can you check if your substitution is correct?
5. What is the significance of the limits of integration in definite integrals?

Tip: When choosing a substitution, look for expressions in the integrand that resemble derivatives of other parts of the integrand.