3.10: Antiderivatives
- Page ID
- 204111
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)3.10 Antiderivatives
- Find the general antiderivative of a given function by applying integration techniques, including basic rules for integrals.
- Apply basic rules for integrals (e.g., power rule, sum rule) to compute antiderivatives of functions.
- Solve initial-value problems by finding the antiderivative of a function and applying the given initial conditions to determine the constant of integration.
A function \( F \) is called an antiderivative of \( f \) on an interval \( I \) if \( F'(x) = f(x) \ \text{for all } x \text{ in } I.\)
For example, if \( f(x) = x^2 \), then \( F(x) = \frac{x^3}{3} \) is an antiderivative of \( f \).
Suppose that \( F \) is an antiderivative of \( f \) on an interval \( I \). Then the most general antiderivative of \( f \) on \( I \) is:
\[
F(x) + C
\]
where \( C \) is a constant. Furthermore, if \( G \) is an antiderivative of \( f \) over \( I \), then there exists a constant \( C \) such that:
\[
G(x) = F(x) + C \quad \text{on } I
\]
Find the general antiderivative of the following functions:
- \( f(x) = e^x \)
- \( f(x) = \sin x \)
- \( f(x) = \cos x \)
- \( f(x) = \dfrac{1}{x} \)
Let \( f \) be a function. The indefinite integral of \( f \), denoted
\[
\int f(x) \, dx
\]
is defined as the most general antiderivative of \( f \). That is, if \( F \) is an antiderivative of \( f \), then
\[
\int f(x) \, dx = F(x) + C
\]
where \( C \) is an arbitrary constant.
| Derivative Formula | Indefinite Integral |
| \( \dfrac{d}{dx}[k] = 0 \) | \( \displaystyle \int k \, dx = kx + C \) |
| \( \dfrac{d}{dx}[x^n] = nx^{n-1} \) | \( \displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), for \( n \ne -1 \) |
| \( \dfrac{d}{dx}[\ln |x|] = \dfrac{1}{x} \) | \( \displaystyle \int \frac{1}{x} \, dx = \ln |x| + C \) |
| \( \dfrac{d}{dx}[e^x] = e^x \) | \( \displaystyle \int e^x \, dx = e^x + C \) |
| \( \dfrac{d}{dx}[\sin x] = \cos x \) | \( \displaystyle \int \cos x \, dx = \sin x + C \) |
| \( \dfrac{d}{dx}[\cos x] = -\sin x \) | \( \displaystyle \int \sin x \, dx = -\cos x + C \) |
| \( \dfrac{d}{dx}[\tan x] = \sec^2 x \) | \( \displaystyle \int \sec^2 x \, dx = \tan x + C \) |
| \( \dfrac{d}{dx}[\sin^{-1} x] = \dfrac{1}{\sqrt{1 - x^2}}\) | \( \displaystyle \int \frac{1}{\sqrt{1 - x^2}} \, dx = \sin^{-1} x + C \) |
| \( \dfrac{d}{dx}[\cos^{-1} x] = -\dfrac{1}{\sqrt{1 - x^2}} \) | \( \displaystyle \int -\frac{1}{\sqrt{1 - x^2}} \, dx = \cos^{-1} x + C \) |
| \( \dfrac{d}{dx}[\tan^{-1} x] = \dfrac{1}{1 + x^2} \) | \( \displaystyle \int \frac{1}{1 + x^2} \, dx = \tan^{-1} x + C \) |
Properties of Indefinite Integrals:
Let \( F \) and \( G \) be antiderivatives of \( f \) and \( g \), respectively. Let \( k \) be a constant.
- \( \displaystyle \int (f(x) \pm g(x)) \, dx = F(x) \pm G(x) + C \)
- \( \displaystyle \int k f(x) \, dx = k F(x) + C \)
Evaluate the following indefinite integrals:
- \( \displaystyle \int (8x^9 + 2x^6 - 4x^2 - 5) \, dx \)
- \( \displaystyle \int (4x^5 + e^x) \, dx \)
- \( \displaystyle \int \frac{3x^5 + 10x^3 + 4x}{x^5} \, dx \)
- \( \displaystyle \int \frac{3x^5 + 7\sqrt[5]{x} + 2}{x} \, dx \)
- \( \displaystyle \int \frac{2}{\sqrt{1 - x^2}} \, dx \)
- \( \displaystyle \int (\cos x - 4\sec^2 x) \, dx \)
Initial-value problem
A differential equation is an equation that relates an unknown function and one or more of its derivatives. The equation
is a simple example of a differential equation. Solving this equation means finding a function \(y\) with a derivative \(f\). Therefore, the solutions of Equation \ref{diffeq1} are the antiderivatives of \(f\). If \(F\) is one antiderivative of \( f\), every function of the form \( y=F(x)+C\) is a solution of that differential equation. For example, the solutions of
are given by
Sometimes we are interested in determining whether a particular solution curve passes through a certain point \( (x_0,y_0)\) —that is, \( y(x_0)=y_0\). The problem of finding a function \(y\) that satisfies a differential equation
with the additional condition
is an example of an initial-value problem. The condition \( y(x_0)=y_0\) is known as an initial condition. For example, looking for a function \( y\) that satisfies the differential equation
and the initial condition
is an example of an initial-value problem. Since the solutions of the differential equation are \( y=x^4+C,\) to find a function \(y\) that also satisfies the initial condition, we need to find \(C\) such that \(y(2)=(2)^4+C=6\). From this equation, we see that \( C=-10\), and we conclude that \( y=x^4-10\) is the solution of this initial-value problem.
Solve the following initial value problems:
- \( \dfrac{dy}{dx} = 4 + \sqrt{x}, \quad y(4) = 20 \)
- \( \dfrac{dy}{dx} = 4x^5 + e^x, \quad y(1) = e \)
- \( \dfrac{dy}{dx} = \cos x - 4\sec^2 x, \quad y\left(\dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2} \)
A particle in motion has acceleration given by:
\[
a(t) = 12t^2 - 6 \ \text{and} \ s(2) = 5.
\]
Find the position function \( s(t) \) of the particle.
A car is traveling at a speed of 66 ft/sec (45 mph) when the brakes are applied. The car begins decelerating at a constant rate of 10 ft/sec².
- How long does it take for the car to stop?
- What distance does the car cover before stopping?


