Find the area of the curve where y= 4x² and the intersection of (1/2, 1) and (1, 4)

To find the area under the curve \( y = 4x^2 \) between the points \( \left( \frac{1}{2}, 1 \right) \) and \( (1, 4) \), we need to integrate the function from \( x = \frac{1}{2} \) to \( x = 1 \).

### Step-by-step solution:

1. The equation of the curve is given by \( y = 4x^2 \).
2. We want to compute the area under the curve between the limits \( x = \frac{1}{2} \) and \( x = 1 \).
3. The formula for the area under the curve is given by the definite integral:
   \[
   A = \int_{\frac{1}{2}}^1 4x^2 \, dx
   \]
4. Now, integrate:
   \[
   \int 4x^2 \, dx = \frac{4x^3}{3}
   \]
5. Apply the limits of integration:
   \[
   A = \left[ \frac{4x^3}{3} \right]_{\frac{1}{2}}^1
   \]

6. First, substitute the upper limit \( x = 1 \):
   \[
   A = \frac{4(1)^3}{3} = \frac{4}{3}
   \]
7. Now, substitute the lower limit \( x = \frac{1}{2} \):
   \[
   A = \frac{4\left(\frac{1}{2}\right)^3}{3} = \frac{4 \times \frac{1}{8}}{3} = \frac{4}{24} = \frac{1}{6}
   \]

8. Subtract the values to find the total area:
   \[
   A = \frac{4}{3} - \frac{1}{6}
   \]
   Find a common denominator:
   \[
   A = \frac{8}{6} - \frac{1}{6} = \frac{7}{6}
   \]

Thus, the area under the curve between \( \left( \frac{1}{2}, 1 \right) \) and \( (1, 4) \) is \( \frac{7}{6} \) square units.

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Do you want more details or have any questions?

Here are some related questions:
1. How do you find the area under any quadratic curve?
2. What are the steps to compute a definite integral?
3. Can we use numerical integration methods to find the area?
4. How does the Fundamental Theorem of Calculus apply to this problem?
5. What happens if the curve's limits are extended beyond these points?

**Tip:** Always check the curve's shape and the boundaries when calculating area using definite integrals, especially if the curve dips below the x-axis!

Discover how to calculate the area under the curve y = 4x² between the points (1/2, 1) and (1, 4) using definite integration. This detailed solution includes step-by-step calculations, key formulas, and relevant theorems. Perfect for students in Grades 11-12 or early college studying calculus.

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