The correct option is B 27/2 ∫_1^4(x^2 x)dx = [(x^3/3) (x^2/2)]_1^4 = #10;((64/3) (16/2)) ((1/3) (1/2)) = (63/3)+ #10;((1 16)/2)=21 ...

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Expansions to Remove Indeterminate Form

∫ 14 x2-xdx

Question

$4 \int 1 (x^{2} - x) d x$

\frac{40}{3}

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\frac{1}{6}

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\frac{27}{2}

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\frac{79}{6}

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Solution

The correct option is B $\frac{27}{2}$
$\int_{1}^{4} (x^{2} - x) d x = [(\frac{x^{3}}{3}) - (\frac{x^{2}}{2})]_{1}^{4} = ((\frac{64}{3}) - (\frac{16}{2})) - ((\frac{1}{3}) - (\frac{1}{2})) = (\frac{63}{3}) + (\frac{(1 - 16)}{2}) = 21 - (\frac{15}{2}) = (\frac{27}{2})$

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