Calculus / Unit 7: Antiderivatives and the Integral
7.3 Initial Value Problems
Notes
An initial value problem gives a derivative and one point, so you can find $C$.
- Integrate, then plug in the known point to solve for $C$.
- Example: $f'(x) = 6x$ and $f(1) = 5$. Then $f(x) = 3x^2 + C$, and $3 + C = 5$ gives $C = 2$, so $f(x) = 3x^2 + 2$.
- In motion: integrating velocity gives position, and integrating acceleration gives velocity.
Practice
If $f'(x) = 6x$ and $f(1) = 5$, then $f(x) =$
$f = 3x^2 + C$ and $f(1) = 3 + C = 5$.
A particle has velocity $v(t) = 2t + 1$ and $s(0) = 3$. Find $s(t)$.
$s = t^2 + t + C$ and $s(0) = C = 3$.
A constant acceleration is 6 m/s$^2$ and $v(0) = 2$ m/s. What is $v(3)$?
$v(t) = 6t + 2$, so $v(3) = 20$.
Ready for more? Take the Unit 7 test.