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.