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

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

Options:

Correct Answer: 2

Exam Year: 2020

Solution:

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

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

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

Step 1: Identify the integral you need to solve: ∫(0 to 1) (x^2 + 2x) dx.
Step 2: Break down the integral into two parts: ∫(0 to 1) x^2 dx + ∫(0 to 1) 2x dx.
Step 3: Find the antiderivative of x^2, which is (x^3)/3.
Step 4: Find the antiderivative of 2x, which is x^2.
Step 5: Combine the antiderivatives: (x^3)/3 + x^2.
Step 6: Evaluate the combined antiderivative from 0 to 1: [(1^3)/3 + (1^2)] - [(0^3)/3 + (0^2)].
Step 7: Calculate the values: (1/3 + 1) - (0) = 1/3 + 1 = 4/3.

Definite Integrals – The question tests the ability to evaluate a definite integral, which involves finding the area under the curve of a function between specified limits.
Polynomial Integration – The integral involves a polynomial function, requiring knowledge of basic integration rules for polynomials.
Fundamental Theorem of Calculus – The solution requires applying the Fundamental Theorem of Calculus to evaluate the integral at the given limits.