What is the area under the curve y = 2x^2 + 3 from x = 0 to x = 2?

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Question: What is the area under the curve y = 2x^2 + 3 from x = 0 to x = 2?

Options:

Correct Answer: 12

Solution:

The area under the curve is given by ∫(from 0 to 2) (2x^2 + 3) dx = [(2/3)x^3 + 3x] from 0 to 2 = (16/3 + 6) = 10.

What is the area under the curve y = 2x^2 + 3 from x = 0 to x = 2?

What is the area under the curve y = 2x^2 + 3 from x = 0 to x = 2?

Step 1: Identify the function for which we want to find the area under the curve. In this case, the function is y = 2x^2 + 3.
Step 2: Set up the integral to find the area under the curve from x = 0 to x = 2. This is written as ∫(from 0 to 2) (2x^2 + 3) dx.
Step 3: Calculate the integral of the function. The integral of 2x^2 is (2/3)x^3 and the integral of 3 is 3x. So, the integral becomes (2/3)x^3 + 3x.
Step 4: Evaluate the integral from the lower limit (0) to the upper limit (2). This means we will calculate [(2/3)(2^3) + 3(2)] - [(2/3)(0^3) + 3(0)].
Step 5: Calculate (2/3)(2^3) which is (2/3)(8) = 16/3 and 3(2) which is 6. So, we have (16/3 + 6).
Step 6: Convert 6 into a fraction with a denominator of 3. This is 6 = 18/3. Now add (16/3 + 18/3) = 34/3.
Step 7: The final area under the curve from x = 0 to x = 2 is 34/3.

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