Find the value of the definite integral ∫(0 to 2) (x^2 + 1) dx. (2020)

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Question: Find the value of the definite integral ∫(0 to 2) (x^2 + 1) dx. (2020)

Options:

Correct Answer: 6

Exam Year: 2020

Solution:

∫(0 to 2) (x^2 + 1) dx = [x^3/3 + x] from 0 to 2 = (8/3 + 2) - (0) = 4.

Find the value of the definite integral ∫(0 to 2) (x^2 + 1) dx. (2020)

Find the value of the definite integral ∫(0 to 2) (x^2 + 1) dx. (2020)

Step 1: Identify the integral you need to solve: ∫(0 to 2) (x^2 + 1) dx.
Step 2: Find the antiderivative of the function (x^2 + 1). The antiderivative is (x^3/3 + x).
Step 3: Evaluate the antiderivative at the upper limit (2): (2^3/3 + 2) = (8/3 + 2).
Step 4: Convert 2 into a fraction with a denominator of 3: 2 = 6/3. So, (8/3 + 6/3) = 14/3.
Step 5: Evaluate the antiderivative at the lower limit (0): (0^3/3 + 0) = 0.
Step 6: Subtract the value at the lower limit from the value at the upper limit: (14/3 - 0) = 14/3.
Step 7: Simplify the result if necessary. The final answer is 14/3.

Definite Integral – The process of calculating the area under a curve defined by a function over a specific interval.
Fundamental Theorem of Calculus – This theorem connects differentiation and integration, allowing the evaluation of definite integrals using antiderivatives.
Polynomial Integration – The technique of integrating polynomial functions, which involves applying the power rule for integration.