Specific antiderivatives
Last updated
Last updated
Normally the antiderivative is in form of f(x) +C
. But actually we could use some additional information to get the C
and get the function only in terms of x
. And we often call the "additional information" as Initial Conditions
, or f₀(x)
.
Solve:
We could Integrate the f'(x)
to get f(x) = 9eˣ + C
.
Since f(8) = 9e⁸ - 8
, we could easily see that C = -8
So the function under this condition would be: f(x) = 9eˣ -8
Then we will get the result f(0) = 9*1 - 8 = 1
.
Solve:
We could integrate f'(x)
to get f(x) = x³ - x² + 7x +C
Since f(6)=200
, so we could substitute 6
into f(x)
:
f(6) = 6³ - 6² + 7*6 +C = 200
which results C = -22
So the function under this condition should be f(x) = x³ - x² + 7x - 22
And f(1) = 1³ - 1² + 7*1 - 22 = -15
It looks confusing, but let's assume f(x)
as y
so dy/dx = 2y
Separate differential equations to get dy/y = 2dx
Take integral of both sides:
Since f(1) = 5
, so:
And f(3) = 5e⁴
, which means m = 5, n = 4
Solve:
Solve: Hint: f(0) = 2
Solve Hint: Don't need to solve y
completely.