What is the integral of (4e^{2t}i+3e^tj) ?

The integral of (4e^{2t}i+3e^tj) is 2ie^{2t}+3je^t+C

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integral of (4e^{2t}i+3e^tj)

\int (4 e^{2 t} i + 3 e^{t} j) d t

2 i e^{2 t} + 3 j e^{t} + C

Solution steps

\int 4 e^{2 t} i + 3 e^{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 4 e^{2 t} i d t + \int 3 e^{t} j d t

\int 4 e^{2 t} i d t = 2 i e^{2 t}

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

= 2 i e^{2 t} + 3 j e^{t}

Add a constant to the solution

= 2 i e^{2 t} + 3 j e^{t} + C

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + 2 y ^{2}} + 3 x)

x \to 1 lim (2 ∣ x - 1 ∣ - 1)

a y^{^{''}} + b y = 0

sin (\frac{π x}{2})

2 sin (3 x)

What is the integral of (4e^{2t}i+3e^tj) ?
The integral of (4e^{2t}i+3e^tj) is 2ie^{2t}+3je^t+C