The integral provided in the image is:

$\int_0^1 \left( 1 - 4 \sqrt[3]{x} \right) dx.$

Let's solve it step by step.

Step 1: Break the integral into two parts

[ \int_0^1 \left( 1 - 4 \sqrt[3]{x} \right) dx = \int_0^1 1 , dx - \int_0^1 4 \sqrt[3]{x} , dx. ]

Step 2: Solve $\int_0^1 1 \, dx$

The integral of a constant $1$ is: $\int_0^1 1 \, dx = x \Big|_0^1 = 1 - 0 = 1.$

Step 3: Solve $\int_0^1 4 \sqrt[3]{x} \, dx$

Rewrite $\sqrt[3]{x}$ as $x^{1/3}$, so: $\int_0^1 4 \sqrt[3]{x} \, dx = 4 \int_0^1 x^{1/3} \, dx.$

Use the power rule for integration: $\int x^n dx = \frac{x^{n+1}}{n+1} + C, \quad \text{for } n \neq -1.$

Here, $n = \frac{1}{3}$, so: $\int x^{1/3} dx = \frac{x^{1/3 + 1}}{1/3 + 1} = \frac{x^{4/3}}{4/3} = \frac{3}{4} x^{4/3}.$

Substitute the limits: $4 \int_0^1 x^{1/3} dx = 4 \cdot \frac{3}{4} x^{4/3} \Big|_0^1 = 3 \cdot \left[ 1^{4/3} - 0^{4/3} \right] = 3.$

Step 4: Combine the results

[ \int_0^1 \left( 1 - 4 \sqrt[3]{x} \right) dx = \int_0^1 1 , dx - \int_0^1 4 \sqrt[3]{x} , dx = 1 - 3 = -2. ]

Final Answer:

$\int_0^1 \left( 1 - 4 \sqrt[3]{x} \right) dx = -2.$

Would you like a detailed breakdown of any step?

Related Questions:

1. How do we handle improper integrals when the limits are infinite?
2. What changes in the solution if the power of $x$ is negative, e.g., $x^{-1/3}$?
3. What are common applications of definite integrals in physics or engineering?
4. How do you calculate integrals with irrational exponents like $x^{\sqrt{2}}$?
5. Can definite integrals be calculated geometrically, and how?

Tip:

Always rewrite roots or radicals in terms of fractional exponents to simplify the integration process!