Math Problem Statement

Solution

To determine the area of the figure bounded by the lines given:

  1. y=x23y = \sqrt[3]{x^2}
  2. y=0y = 0
  3. x=1x = 1

we need to set up the definite integral of y=x23y = \sqrt[3]{x^2} from x=0x = 0 to x=1x = 1.

The integral to calculate the area is:

01x23dx\int_{0}^{1} \sqrt[3]{x^2} \, dx

To solve this, let's use the substitution method:

Let u=x2u = x^2, then du=2xdxdu = 2x \, dx or dx=du2xdx = \frac{du}{2x}.

Thus, the integral becomes:

01(u1/3)du2u=1201u1/3u1/2du=1201u1/6du\int_{0}^{1} (u^{1/3}) \frac{du}{2\sqrt{u}} = \frac{1}{2} \int_{0}^{1} u^{1/3} u^{-1/2} \, du = \frac{1}{2} \int_{0}^{1} u^{-1/6} \, du

This can be integrated using the power rule:

[ \frac{1}{2} \int_{0}^{1} u^{-1/6} , du = \frac{1}{2} \left[ \frac{u^{5/6}}{5/6} \right]{0}^{1} = \frac{1}{2} \left( \frac{6}{5} \left[ u^{5/6} \right]{0}^{1} \right) = \frac{1}{2} \cdot \frac{6}{5} \left( 1^{5/6} - 0^{5/6} \right) = \frac{1}{2} \cdot \frac{6}{5} = \frac{3}{5} = 0.6 ]

Therefore, the area of the figure is 0.6.

Would you like any further details or explanations?

Here are some questions you might want to ask:

  1. How do you find the area under a curve?
  2. What is the power rule for integration?
  3. How does substitution work in integration?
  4. What are definite integrals?
  5. Can you show another example of finding an area using integrals?

Tip: When integrating, always double-check your substitutions and limits to ensure accuracy.

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Math Problem Analysis

Mathematical Concepts

Calculus
Integral calculus
Definite integrals

Formulas

Integral of u^(1/3)

Theorems

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Suitable Grade Level

Advanced High School