ind So Can the Fundamental Theorem of Calculus be used to find 2 2x²-x-3 dx? 2x²- √x O No, f(x) = ² 2x²-x-3 dx is not continuous on the given interval and i T

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Chapter2: Second-order Linear Odes
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### Can the Fundamental Theorem of Calculus be used to find 
\[ \int_{0}^{2} \frac{2x^2 - x - 3}{\sqrt{x}} \, dx? \]

1. ◯ No, \( f(x) = \int_{0}^{2} \frac{2x^2 - x - 3}{\sqrt{x}} \, dx \) is not continuous on the given interval and its antiderivative does not exist.
2. ◯ Yes, \( f(x) = \int_{0}^{2} \frac{2x^2 - x - 3}{\sqrt{x}} \, dx \) is continuous on the given interval and its antiderivative exists.
3. ◯ Yes, \( f(x) = \int_{0}^{2} \frac{2x^2 - x - 3}{\sqrt{x}} \, dx \) is continuous on the given interval but its antiderivative does not exist.
4. ◉ No, \( f(x) = \int_{0}^{2} \frac{2x^2 - x - 3}{\sqrt{x}} \, dx \) is not continuous on the given interval but its antiderivative exists.
Transcribed Image Text:### Can the Fundamental Theorem of Calculus be used to find \[ \int_{0}^{2} \frac{2x^2 - x - 3}{\sqrt{x}} \, dx? \] 1. ◯ No, \( f(x) = \int_{0}^{2} \frac{2x^2 - x - 3}{\sqrt{x}} \, dx \) is not continuous on the given interval and its antiderivative does not exist. 2. ◯ Yes, \( f(x) = \int_{0}^{2} \frac{2x^2 - x - 3}{\sqrt{x}} \, dx \) is continuous on the given interval and its antiderivative exists. 3. ◯ Yes, \( f(x) = \int_{0}^{2} \frac{2x^2 - x - 3}{\sqrt{x}} \, dx \) is continuous on the given interval but its antiderivative does not exist. 4. ◉ No, \( f(x) = \int_{0}^{2} \frac{2x^2 - x - 3}{\sqrt{x}} \, dx \) is not continuous on the given interval but its antiderivative exists.
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