What is the integral of [(2t-1)i+4t^3j+3sqrt(t)k] ?

The integral of [(2t-1)i+4t^3j+3sqrt(t)k] is i(t^2-t)+jt^4+2kt^{3/2}+C

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integral of [(2t-1)i+4t^3j+3sqrt(t)k]

\int [(2 t - 1) i + 4 t^{3} j + 3 t k] d t

i (t^{2} - t) + j t^{4} + 2 k t^{\frac{3}{2}} + C

Solution steps

\int (2 t - 1) i + 4 t^{3} j + 3 t 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 (2 t - 1) i d t + \int 4 t^{3} j d t + \int 3 t k d t

\int (2 t - 1) i d t = i (t^{2} - t)

\int 4 t^{3} j d t = j t^{4}

\int 3 t k d t = 2 k t^{\frac{3}{2}}

= i (t^{2} - t) + j t^{4} + 2 k t^{\frac{3}{2}}

Add a constant to the solution

= i (t^{2} - t) + j t^{4} + 2 k t^{\frac{3}{2}} + C

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What is the integral of [(2t-1)i+4t^3j+3sqrt(t)k] ?
The integral of [(2t-1)i+4t^3j+3sqrt(t)k] is i(t^2-t)+jt^4+2kt^{3/2}+C