Find area under curve x^2=y from x=3 to x=5

To find the area under the curve \( x^2 = y \) from \( x = 3 \) to \( x = 5 \), we can proceed as follows:

### Step 1: Rewrite the equation
The equation \( x^2 = y \) implies that \( y = x^2 \).

### Step 2: Set up the integral
We need to integrate \( y = x^2 \) with respect to \( x \) from \( x = 3 \) to \( x = 5 \):
\[
\text{Area} = \int_{3}^{5} x^2 \, dx
\]

### Step 3: Integrate
The integral of \( x^2 \) is:
\[
\int x^2 \, dx = \frac{x^3}{3}
\]

### Step 4: Evaluate the integral
Now, we substitute the limits:
\[
\text{Area} = \left[ \frac{x^3}{3} \right]_{3}^{5} = \frac{5^3}{3} - \frac{3^3}{3}
\]

Calculating each part:
\[
= \frac{125}{3} - \frac{27}{3} = \frac{125 - 27}{3} = \frac{98}{3}
\]

### Final Answer
The area under the curve \( x^2 = y \) from \( x = 3 \) to \( x = 5 \) is:
\[
\frac{98}{3}
\]

Would you like more details or have any questions about this solution?

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Here are five related questions:
1. How would the solution change if the bounds were from \( x = 0 \) to \( x = 2 \)?
2. What if the function were \( y = x^3 \) instead of \( y = x^2 \)?
3. Can we calculate the area under \( y = x^2 \) if the region is bounded by both \( x \)- and \( y \)-axes?
4. What if we were asked to find the volume when \( y = x^2 \) is rotated around the x-axis?
5. How would we find the area between \( y = x^2 \) and \( y = 2x \)?

**Tip:** When setting up integrals for areas, always ensure your function represents the correct curve and that the limits are carefully chosen for the area desired.

Find the area under the curve x^2 = y from x = 3 to x = 5 using definite integration. This solution walks through setting up and evaluating the integral, applying key calculus concepts and the Fundamental Theorem of Calculus. Ideal for Grades 11-12 students exploring integration and area calculations.

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