Question

Using integration, evaluate the area of the region bounded by the curve \( y = x^2 \), the lines \( y = 1 \) and \( y = 3 \), and the y-axis.

Answer 1

Step 1: Understand the problem
We need to find the area of the region bounded by the curve y = x², the horizontal lines y = 1 and y = 3, and the y-axis (x = 0).

Step 2: Express x in terms of y
Since y = x², solving for x gives:
x = √y

Step 3: Set up the integral with respect to y
The area bounded between y = 1 and y = 3 and the curve is given by:
Area = ∫ from y = 1 to y = 3 of x dy = ∫ from 1 to 3 of √y dy

Step 4: Evaluate the integral
Integral of √y dy is:
∫ y^(1/2) dy = (2/3) y^(3/2) + C

Step 5: Calculate definite integral values
Area = [(2/3) y^(3/2)] evaluated from 1 to 3 = (2/3) [3^(3/2) - 1^(3/2)]

Step 6: Simplify the expression
3^(3/2) = 3^(1) * 3^(1/2) = 3 * √3
So, Area = (2/3) (3√3 - 1)

Step 7: Conclusion
The area of the region bounded by the curve y = x², the lines y = 1 and y = 3, and the y-axis is (2/3)(3√3 - 1) square units.

Final Answer: (2/3)(3√3 - 1)