What is the integral from 0 to 1 of (8ti+tj-k) ?

The integral from 0 to 1 of (8ti+tj-k) is (j-2k)/2+4i

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integral from 0 to 1 of (8ti+tj-k)

\int_{0}^{1} (8 t i + t j - k) d t

\frac{j - 2 k}{2} + 4 i

Solution steps

\int_{0}^{1} 8 t i + t j - 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}^{1} 8 t i d t + \int_{0}^{1} t j d t - \int_{0}^{1} k d t

\int_{0}^{1} 8 t i d t = 4 i

\int_{0}^{1} t j d t = \frac{j}{2}

\int_{0}^{1} k d t = k

= 4 i + \frac{j}{2} - k

Rewrite

4 i + \frac{j}{2} - k

in standard complex form:

\frac{j - 2 k}{2} + 4 i

= \frac{j - 2 k}{2} + 4 i

\int_{- 2}^{1} \frac{1}{( 11 + 5 x )} d x

\int_{0}^{2} e^{2 x} d x

\int_{2}^{3} e^{x - 2} - e d x

\int_{0}^{3} t 1 + t^{2} d t

\int_{\infty}^{0} e^{x} d x

What is the integral from 0 to 1 of (8ti+tj-k) ?
The integral from 0 to 1 of (8ti+tj-k) is (j-2k)/2+4i