What is the integral from 0 to 2 of (6t^2i-4tj+3k) ?

The integral from 0 to 2 of (6t^2i-4tj+3k) is 16i-8j+6k

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integral from 0 to 2 of (6t^2i-4tj+3k)

\int_{0}^{2} (6 t^{2} i - 4 t j + 3 k) d t

16 i - 8 j + 6 k

Solution steps

\int_{0}^{2} 6 t^{2} i - 4 t j + 3 k d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{2} 6 t^{2} i d t - \int_{0}^{2} 4 t j d t + \int_{0}^{2} 3 k d t

\int_{0}^{2} 6 t^{2} i d t = 16 i

\int_{0}^{2} 4 t j d t = 8 j

\int_{0}^{2} 3 k d t = 6 k

= 16 i - 8 j + 6 k

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What is the integral from 0 to 2 of (6t^2i-4tj+3k) ?
The integral from 0 to 2 of (6t^2i-4tj+3k) is 16i-8j+6k