What is the integral from 1 to 2 of (e^{3t}i-4/t j) ?

The integral from 1 to 2 of (e^{3t}i-4/t j) is i(-e^3+e^6)/3-4jln(2)

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integral from 1 to 2 of (e^{3t}i-4/t j)

\int_{1}^{2} (e^{3 t} i - \frac{4}{t} j) d t

i \frac{- e ^{3} + e ^{6}}{3} - 4 j ln (2)

Solution steps

\int_{1}^{2} e^{3 t} i - \frac{4}{t} j 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_{1}^{2} e^{3 t} i d t - \int_{1}^{2} \frac{4}{t} j d t

\int_{1}^{2} e^{3 t} i d t = i \frac{- e ^{3} + e ^{6}}{3}

\int_{1}^{2} \frac{4}{t} j d t = 4 j ln (2)

= i \frac{- e ^{3} + e ^{6}}{3} - 4 j ln (2)

What is the integral from 1 to 2 of (e^{3t}i-4/t j) ?
The integral from 1 to 2 of (e^{3t}i-4/t j) is i(-e^3+e^6)/3-4jln(2)