Integral Calculus

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)

Step 1: Understand the problem
We need to find the area of the region bounded by the curve y = x, the x-axis, and the vertical lines x = 0 and x = 2.

Step 2: Set up the integral for the area
The area under the curve y = x from x = 0 to x = 2 is given by the definite integral:
Area = ∫ from 0 to 2 of x dx

Step 3: Calculate the integral
The integral of x with respect to x is (x²)/2.
So, Area = [ (x²)/2 ] evaluated from 0 to 2.

Step 4: Evaluate the definite integral
Substitute the upper limit x = 2:
(2)² / 2 = 4 / 2 = 2
Substitute the lower limit x = 0:
(0)² / 2 = 0
So, Area = 2 - 0 = 2 square units.

Step 5: Conclusion
The area of the region bounded by the curve y = x, the x-axis, x = 0 and x = 2 is 2 square units.

Final Answer: (C) 2