Evaluate (int_1 ^2(6x^2-2x)dx)
-
A.
16 -
B.
13 -
C.
12 -
D.
11
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