Question: Find the value of β« from 1 to 2 of (3x^2 - 2) dx.
Options:
1
2
3
4
Correct Answer: 1
Solution:
The integral evaluates to [x^3 - 2x] from 1 to 2 = (8 - 4) - (1 - 2) = 4 + 1 = 5.
Find the value of β« from 1 to 2 of (3x^2 - 2) dx.
Practice Questions
Q1
Find the value of β« from 1 to 2 of (3x^2 - 2) dx.
1
2
3
4
Questions & Step-by-Step Solutions
Find the value of β« from 1 to 2 of (3x^2 - 2) dx.
Step 1: Identify the function to integrate, which is 3x^2 - 2.
Step 2: Find the antiderivative of the function. The antiderivative of 3x^2 is x^3, and the antiderivative of -2 is -2x. So, the antiderivative is x^3 - 2x.
Step 3: Evaluate the antiderivative from the lower limit (1) to the upper limit (2).
Step 4: Substitute the upper limit (2) into the antiderivative: (2^3) - 2(2) = 8 - 4 = 4.
Step 5: Substitute the lower limit (1) into the antiderivative: (1^3) - 2(1) = 1 - 2 = -1.
Step 6: Subtract the result of the lower limit from the result of the upper limit: 4 - (-1) = 4 + 1 = 5.
Definite Integral β The question tests the ability to evaluate a definite integral, which involves finding the antiderivative and applying the Fundamental Theorem of Calculus.
Antiderivative Calculation β It requires knowledge of how to compute the antiderivative of a polynomial function.
Evaluation of Limits β The question assesses the skill of correctly substituting the upper and lower limits into the antiderivative.
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