Evaluate \(\int_1 ^2(6x^2-2x)dx\)
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Correct Answer: Option D
Explanation:
\(\int_1 ^2(6x^2-2x)dx=[\frac{6x^3}{3}-\frac{2x^2}{2}]_1 ^2\\
= [2x^3 - x^2]_1^2\)
= [2(2)3 - (2)2] – [2(1)3 - (1)2]
= [16-4] – [2-1]
= 12 – 1
= 11
\(\int_1 ^2(6x^2-2x)dx=[\frac{6x^3}{3}-\frac{2x^2}{2}]_1 ^2\\
= [2x^3 - x^2]_1^2\)
= [2(2)3 - (2)2] – [2(1)3 - (1)2]
= [16-4] – [2-1]
= 12 – 1
= 11