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

The integral from 0 to 3 of (t^2i+tj-2k) is (-12k+9j)/2+9i

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

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

\frac{- 12 k + 9 j}{2} + 9 i

Solution steps

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

\int_{0}^{3} t^{2} i d t = i 9

\int_{0}^{3} t j d t = \frac{9 j}{2}

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

= i 9 + \frac{9 j}{2} - 6 k

Rewrite

i 9 + \frac{9 j}{2} - 6 k

in standard complex form:

\frac{9 j - 12 k}{2} + 9 i

= \frac{- 12 k + 9 j}{2} + 9 i

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What is the integral from 0 to 3 of (t^2i+tj-2k) ?
The integral from 0 to 3 of (t^2i+tj-2k) is (-12k+9j)/2+9i