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Wize University Calculus 1 Textbook > Integrals

Initial Conditions

Antiderivatives of 1x\frac{1}{x}Previous Section
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Initial Conditions

We don't always get unique answers for ∫f(x)dx\displaystyle \int f(x)dx. Instead we get F(x)+CF(x) +C. If we are given some initial conditions for the function, we can find this constant by substituting the coordinates and solving for CC .

Finding f(x) from f'(x)

To recover f(x)f(x) from f′(x)f'(x), givenf(a)=bf(a)=b:
  1. Antiderive f′(x)f'(x)to obtain f(x)f(x). (Don't forget +C+C)
  2. Substitute f(a)=bf(a)=b to solve for CC.
  3. Rewrite f(x)f(x) with the value ofCC.

Finding f(x) from f''(x)

To recover f(x)f(x) from f′′(x)f''(x), given two initial conditions, we have to integrate twice:
  1. Antiderive f′′(x)f''(x) to obtain f′(x)f'(x). (Don't forget +C+C)
  2. Antiderive f′(x)f'(x) to obtain f(x)f(x).
  3. The second time we antiderive we pick up another constant of integration DD , and we get Cx+DCx+D.
  4. Substitute the given conditions to find CC and DD.
  5. Rewrite f(x)f(x)with the values of CC and DD.
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Example: Initial Conditions

Find f(x)f(x) if f(0)=1f(0)=1 and f′(x)=x2+xf'(x)=x^2+x.

f(x)=∫(x2+x)dxf(x)=\displaystyle \int (x^2+x) dx

f(x)=x33+x22+Cf(x)\displaystyle =\frac{x^3}{3}+\frac{x^2}{2}+C

Now substitutte the intial condition f(0)=1f(0)=1

  ⟹  033+022+C=1\implies \displaystyle \frac{0^3}{3}+\frac{0^2}{2}+C=1

  ⟹  C=1\implies C=1

f(x)=x33+x22+1\displaystyle \boxed{f(x)=\frac{x^3}{3}+\frac{x^2}{2}+1}
Given f′′(x)=5ex+2sin⁡xf''(x)=5e^x+2\sin x, f(0)=0f(0)=0, and f(π)=0f(\pi)=0, find f(x)f(x).
Extra Practice
Indefinite Integral
Find f(x)f\left(x\right), given that f′(x)=x2−2+sin⁡xf'\left(x\right)=x^2-2+\sin x and f(0)=3f\left(0\right)=3.
Applications of Integration: Position, Velocity and Acceleration
If the accelertion of a particle is given by a(x)=2x+xa\left(x\right)=2^x+x, and v(0)=s(0)=0v\left(0\right)=s\left(0\right)=0, then the displacement of the particle s(x)s\left(x\right) is
Practice: Indefinite Integral
Find f(x)f\left(x\right), given that f′(x)=x2−2+sin⁡xf'\left(x\right)=x^2-2+\sin x and f(0)=3f\left(0\right)=3.
Practice: Indefinite Integral
Practice: Indefinite Integral
Find f(x)f\left(x\right), given that f′(x)=x2−2+sin⁡xf'\left(x\right)=x^2-2+\sin x and f(0)=3f\left(0\right)=3.
Indefinite Integral
Find f(x)f\left(x\right), given that f′(x)=x2−2+sin⁡xf'\left(x\right)=x^2-2+\sin x and f(0)=3f\left(0\right)=3.
Practice: Indefinite Integral (~F2018 Final Q26)
Practice: Indefinite Integral
Find f(x)f\left(x\right), given that f′(x)=x2−2+sin⁡xf'\left(x\right)=x^2-2+\sin x and f(0)=3f\left(0\right)=3.
Exponential Functions: $f$ from $f''$
Given f′′(x)=5ex+2sin⁡xf''(x)=5e^x+2\sin x and given f(0)=0f(0)=0, f(π)=0f(\pi)=0, find f(x)f(x). Enter "f(x)=..."
Practice: Indefinite Integral
Find f(x)f\left(x\right), given that f′′(x)=x2−2+sin⁡xf''\left(x\right)=x^2-2+\sin x, f(0)=3f\left(0\right)=3 and f′(3)=1f'(3)=1
Practice: Indefinite Integral
Find f(x)f\left(x\right), given that f′(x)=x2−2+sin⁡xf'\left(x\right)=x^2-2+\sin x and f(0)=3f\left(0\right)=3.
Applications of Integration: Position, Velocity and Acceleration
If the acceleartion of a particle is given by a(x)=2x+xa\left(x\right)=2^x+x, and v(0)=s(0)=0v\left(0\right)=s\left(0\right)=0, then the displacement of the particle s(x)s\left(x\right) is
Initial Conditions
Find the function y=f(x)y=f(x) if
y′′=3x−5/2y′(1)=1y(4)=6y''=3x^{-5/2} \quad y'(1)=1 \quad y(4)=6
Initial Conditions
Solve the initial value problem
y′=x2+1xy'=\frac{x^2+1}{\sqrt{x}}
y(1)=3y(1)=3
Initial Conditions
Solve the initial value problem
y′′=ex/2y′(0)=3y(0)=5y''=e^{x/2} \quad y'(0)=3 \quad y(0)=5
Initial Conditions
Find the function f(x)f(x) whose derivative is 4x−1/x24x-1/x^2 forx>0 x>0 and for which f(1)=−1f(1)=-1.
Solve the initial value problem
f(x)=4x+x22+5,F(1)=1/3\displaystyle f(x)=4x+\frac{x^2}{2}+5, F(1)=1/3
Solve the initial value problem
f(x)=x+x4x2,F(0)=2\displaystyle f(x)=\frac{\sqrt{x}+x^4}{x^2}, F(0)=2
Solve the initial value problem y′=sec⁡2x−e−x,    y(0)=2y'=\sec^2x-e^{-x},\ \ \ \ y\left(0\right)=2
Consider Newton's law of cooling dTdt=2−T10\displaystyle\frac{\text{d}T}{\text{d}t}=2-\frac{T}{10} . Determine T(t)T(t) if T(0)=100T(0)=100
Indefinite Integral
Find f(x)f\left(x\right), given that f′(x)=x2−2+sin⁡xf'\left(x\right)=x^2-2+\sin x and f(0)=3f\left(0\right)=3.
Practice: Rearrange First
Q:\bf{Q:} Solve the initial value problem t2dydt+1=t3t^2\dfrac{dy}{dt} + 1 = t^3 with y(1)=1y(1) = 1
Q:\bf{Q:} Solve the initial value problem y′=x3+sin⁡2xy'=x^3+\sin2x given y(π2)=0y\left(\dfrac{\pi}{2}\right)=0
Q:\bf{Q:}Solve the initial value problem y′=sec⁡2x−e−xy'=\sec^2x-e^{-x} with y(0)=2y\left(0\right)=2
Solve the initial-value problem
y′=x2+1xy(1)=3\begin{aligned} y' &=\dfrac{x^2 + 1}{\sqrt{x}}\\ y(1) & = 3 \end{aligned}
Practice: From Second Derivative to Function
Q:\bf{Q:} Solve the initial-value problem y′′=sin⁡2xy'' = \sin 2x, y′(0)=3y'(0) = 3 , y(0)=5y(0) = 5.
Exponential Functions: $f$ from $f''$
Given f′′(x)=5ex+2sin⁡xf''(x)=5e^x+2\sin x and given f(0)=0f(0)=0, f(π)=0f(\pi)=0, find f(x)f(x). Enter "f(x)=..."
Solve the initial value problem
y′′=ex/2y′(0)=3y(0)=5\begin{aligned} y''& = e^{x/2}\\ y'(0)& = 3\\ y(0) &= 5 \end{aligned}
Solve the initial value problem y′=sec⁡2x−e−x,  y(0)=2y'=\sec^2x-e^{-x},\ \ y\left(0\right)=2.
Practice: Indefinite Integral
Practice: Indefinite Integral
Find f(x)f\left(x\right), given that f′(x)=x2−2+sin⁡xf'\left(x\right)=x^2-2+\sin x and f(0)=3f\left(0\right)=3.
Find the antiderivative FF
f(x)=x+x4x2,F(1)=2\displaystyle f(x)=\frac{\sqrt{x}+x^4}{x^2}, F(1)=2
Find the antiderivative FF
f(x)=4x+x22+5,F(1)=1/3\displaystyle f(x)=4x+\frac{x^2}{2}+5, F(1)=1/3