Evaluate \( \int_{3}^{2}(x^2-2x)\,dx \)
Question 1 Report
Evaluate \( \int_{3}^{2}(x^2-2x)\,dx \)
Answer Details
To evaluate the integral ?23(x^2 - 2x)dx, we need to find the antiderivative of the integrand and then apply the definite integral between the limits of integration, 2 and 3.
First, we find the antiderivative of x^2 - 2x:
∫(x^2 - 2x)dx = (1/3)x^3 - x^2 + C
where C is the constant of integration.
Next, we apply the limits of integration, 2 and 3:
∫2^3(x^2 - 2x)dx = [(1/3)3^3 - 3^2] - [(1/3)2^3 - 2^2]
= (27/3 - 9) - (8/3 - 4)
= 9 - (8/3)
= (27/3) - (8/3)
= 19/3
Therefore, the value of the integral ?23(x^2 - 2x)dx is 19/3.
So the answer is 4/3.
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