What is the area between the curves y = x^2 and y = 4 from x = -2 to x = 2?
Practice Questions
1 question
Q1
What is the area between the curves y = x^2 and y = 4 from x = -2 to x = 2?
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The area between the curves is given by ∫(from -2 to 2) (4 - x^2) dx = [4x - x^3/3] from -2 to 2 = (8 - (8/3)) - (-8 + (8/3)) = 16 - (16/3) = 32/3.
Questions & Step-by-step Solutions
1 item
Q
Q: What is the area between the curves y = x^2 and y = 4 from x = -2 to x = 2?
Solution: The area between the curves is given by ∫(from -2 to 2) (4 - x^2) dx = [4x - x^3/3] from -2 to 2 = (8 - (8/3)) - (-8 + (8/3)) = 16 - (16/3) = 32/3.
Steps: 9
Step 1: Identify the curves. We have y = x^2 (a parabola) and y = 4 (a horizontal line).
Step 2: Determine the points of intersection of the curves. Set x^2 = 4, which gives x = -2 and x = 2.
Step 3: Set up the integral to find the area between the curves from x = -2 to x = 2. The area is given by the integral of the top curve minus the bottom curve: ∫(from -2 to 2) (4 - x^2) dx.
Step 4: Calculate the integral. The integral of (4 - x^2) is 4x - (x^3)/3.
Step 5: Evaluate the integral from -2 to 2. Substitute x = 2 into the integral result: 4(2) - (2^3)/3 = 8 - (8/3).
Step 6: Substitute x = -2 into the integral result: 4(-2) - ((-2)^3)/3 = -8 + (8/3).
Step 7: Calculate the area by subtracting the two results: (8 - (8/3)) - (-8 + (8/3)).
Step 8: Simplify the expression: 8 + 8 = 16 and - (8/3) - (8/3) = - (16/3). So, 16 - (16/3) = 32/3.
Step 9: The final area between the curves from x = -2 to x = 2 is 32/3.
