Integrating Vector-Valued Functions (College Board AP® Calculus BC): Revision Note

When integrating vector-valued functions, it helps to use different letters for the different constants of integration ($+C$, $+D$, ... etc).

The derivative of a vector-valued function is $<12t^3-2t,1+3\sqrt{t}>$.

At $t=1$, the function is $<-3,7>$.

Find the function.

Answer:

You are given the derivative of a vector-valued function, so you need to integrate it to find the original vector-valued function

Integrate both components (write $\sqrt{t}$ as $t^{\frac{1}{2}}$) and add a different constant of integration to each component

$<3t^4-t^2+C,t+2t^{\frac{3}{2}}+D>$

You are given a known vector value, $<-3,7>$, when $t=1$

First substitute in $t=1$

$<3{\left(1\right)}^4-{\left(1\right)}^2+C,\left(1\right)+2{\left(1\right)}^{\frac{3}{2}}+D>=<2+C,3+D>$

Then set this equal to $<-3,7>$

$<2+C,3+D>=<-3,7>$

Solve the two equations to find $C$ and $D$

$\begin{array}{rcl}2+C& =& -3\Rightarrow C=-5\\ 3+D& =& 7\Rightarrow D=4\end{array}$

Substitute the values of $C$ and $D$ back into the function to get the final answer, which you should write in vector notation

$<3t^4-t^2-5,t+2t^{\frac{3}{2}}+4>$