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

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

Options:

Correct Answer: 4

Exam Year: 2020

Solution:

The integral ∫(2x + 1)dx = x^2 + x. Evaluating from 0 to 2 gives (2^2 + 2) - (0 + 0) = 4.

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

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

Step 1: Identify the integral you need to solve, which is ∫(2x + 1)dx.
Step 2: Find the antiderivative of the function (2x + 1). The antiderivative is x^2 + x.
Step 3: Set up the definite integral by evaluating the antiderivative from the lower limit (0) to the upper limit (2).
Step 4: Substitute the upper limit (2) into the antiderivative: (2^2 + 2) = 4 + 2 = 6.
Step 5: Substitute the lower limit (0) into the antiderivative: (0^2 + 0) = 0.
Step 6: Subtract the value from the lower limit from the value from the upper limit: 6 - 0 = 6.
Step 7: The final answer for the integral ∫(2x + 1)dx from 0 to 2 is 6.

Definite Integral – The process of calculating the area under a curve defined by a function over a specific interval.
Fundamental Theorem of Calculus – Relates differentiation and integration, allowing the evaluation of definite integrals using antiderivatives.