How to find the integral of a vector function

The form of the integral of a vector function

To find the integral of a vector function $r(t)=r(t)_1\bold i+r(t)_2\bold j+r(t)_3\bold k$, we simply replace each coefficient with its integral.

In other words, the integral of the vector function is

$\int r(t)\ dt=\bold i\int r(t)_1\ dt+\bold j\int r(t)_2\ dt+\bold k\int r(t)_3\ dt$

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If the vector function is given as $r(t)=\langle{r(t)_1,r(t)_2,r(t)_3}\rangle$, then its integral is

$\int{r(t)}=\left\langle{\int{r(t)_1}\ dt,\int{r(t)_2}\ dt,\int{r(t)_3}}\ dt\right\rangle$

How to find the integral of a vector function

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A step-by-step example of how to find the integral of a vector function

Example

Find the integral of the vector function over the interval $[0,\pi]$.

$r(t)=\sin{(2t)}\bold i+2e^{2t}\bold j+4t^3\bold k$

Remember that we’re only taking the integrals of the coefficients, which means $\bold i$, $\bold j$ and $\bold k$ will be left alone.

$\int^{\pi}_0{r(t)}\ dt=\frac{-\cos{(2t)}}{2}\Big|^{\pi}_0\bold i+\frac{2e^{2t}}{2}\Big|^{\pi}_0\bold j+\frac{4t^4}{4}\Big|^{\pi}_0\bold k$

$\int^{\pi}_0{r(t)}\ dt=\frac{-\cos{(2t)}}{2}\Big|^{\pi}_0\bold i+e^{2t}\Big|^{\pi}_0\bold j+t^4\Big|^{\pi}_0\bold k$

To find the integral of a vector function, we simply replace each coefficient with its integral.

Evaluating over the interval $[0,\pi]$, we get

$\int^{\pi}_0{r(t)}\ dt=\left[\frac{-\cos{(2\pi)}}{2}-\frac{-\cos{(2(0))}}{2}\right]\bold i+\left[e^{2\pi}-e^{2(0)}\right]\bold j+\left[\pi^4-0^4\right]\bold k$

$\int^{\pi}_0{r(t)}\ dt=\left[\frac{-\cos{(2\pi)}}{2}+\frac{\cos{0}}{2}\right]\bold i+\left(e^{2\pi}-1\right)\bold j+\left(\pi^4-0\right)\bold k$

$\int^{\pi}_0{r(t)}\ dt=\left(\frac{-1}{2}+\frac{1}{2}\right)\bold i+(e^{2\pi}-1)\bold j+\pi^4\bold k$

$\int^{\pi}_0{r(t)}\ dt=0\bold i+(e^{2\pi}-1)\bold j+\pi^4\bold k$

$\int^{\pi}_0{r(t)}\ dt=(e^{2\pi}-1)\bold j+\pi^4\bold k$

This is the integral of the vector function. We could also write it in the form

$\int^{\pi}_0{r(t)}\ dt=\left\langle0,e^{2\pi}-1,\pi^4\right\rangle$

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