Find the area bounded by the curve y = 3x\(^2\) - 2x + 1, the ordinates x = 1 and x = 3 and the x-axis.
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Correct Answer: Option D
Explanation:
\(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 2x + 1\)
\(y = \int_{1} ^{3} (3x^{2} - 2x + 1) \mathrm d x\)
\(y = [x^{3} - x^{2} + x]_{1} ^{3}\)
= \([3^{3} - 3^{2} + 3] - [1^{3} - 1^{2} + 1]\)
= \(21 - 1 = 20\)
\(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 2x + 1\)
\(y = \int_{1} ^{3} (3x^{2} - 2x + 1) \mathrm d x\)
\(y = [x^{3} - x^{2} + x]_{1} ^{3}\)
= \([3^{3} - 3^{2} + 3] - [1^{3} - 1^{2} + 1]\)
= \(21 - 1 = 20\)