# 4.10 Antiderivatives  (Page 4/10)

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## Evaluating indefinite integrals

Evaluate each of the following indefinite integrals:

1. $\int \left(5{x}^{3}-7{x}^{2}+3x+4\right)dx$
2. $\int \frac{{x}^{2}+4\sqrt[3]{x}}{x}dx$
3. $\int \frac{4}{1+{x}^{2}}dx$
4. $\int \text{tan}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}dx$
1. Using [link] , we can integrate each of the four terms in the integrand separately. We obtain
$\int \left(5{x}^{3}-7{x}^{2}+3x+4\right)dx=\int 5{x}^{3}dx-\int 7{x}^{2}dx+\int 3x\phantom{\rule{0.1em}{0ex}}dx+\int 4dx.$

From the second part of [link] , each coefficient can be written in front of the integral sign, which gives
$\int 5{x}^{3}dx-\int 7{x}^{2}dx+\int 3x\phantom{\rule{0.1em}{0ex}}dx+\int 4dx=5\int {x}^{3}dx-7\int {x}^{2}dx+3\int x\phantom{\rule{0.1em}{0ex}}dx+4\int 1dx.$
Using the power rule for integrals, we conclude that
$\int \left(5{x}^{3}-7{x}^{2}+3x+4\right)dx=\frac{5}{4}{x}^{4}-\frac{7}{3}{x}^{3}+\frac{3}{2}{x}^{2}+4x+C.$
2. Rewrite the integrand as
$\frac{{x}^{2}+4\sqrt[3]{x}}{x}=\frac{{x}^{2}}{x}+\frac{4\sqrt[3]{x}}{x}=0.$

Then, to evaluate the integral, integrate each of these terms separately. Using the power rule, we have
$\begin{array}{cc}\hfill \int \left(x+\frac{4}{{x}^{2\text{/}3}}\right)dx& =\int x\phantom{\rule{0.1em}{0ex}}dx+4\int {x}^{-2\text{/}3}dx\hfill \\ & =\frac{1}{2}{x}^{2}+4\frac{1}{\left(\frac{-2}{3}\right)+1}{x}^{\left(-2\text{/}3\right)+1}+C\hfill \\ & =\frac{1}{2}{x}^{2}+12{x}^{1\text{/}3}+C.\hfill \end{array}$
3. Using [link] , write the integral as
$4\int \frac{1}{1+{x}^{2}}dx.$

Then, use the fact that ${\text{tan}}^{-1}\left(x\right)$ is an antiderivative of $\frac{1}{\left(1+{x}^{2}\right)}$ to conclude that
$\int \frac{4}{1+{x}^{2}}dx=4\phantom{\rule{0.1em}{0ex}}{\text{tan}}^{-1}\left(x\right)+C.$
4. Rewrite the integrand as
$\text{tan}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}x=\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{\text{cos}\phantom{\rule{0.1em}{0ex}}x}\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}x=\text{sin}\phantom{\rule{0.1em}{0ex}}x.$

Therefore,
$\int \text{tan}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}x=\int \text{sin}\phantom{\rule{0.1em}{0ex}}x=\text{−}\text{cos}\phantom{\rule{0.1em}{0ex}}x+C.$

Evaluate $\int \left(4{x}^{3}-5{x}^{2}+x-7\right)dx.$

${x}^{4}-\frac{5}{3}{x}^{3}+\frac{1}{2}{x}^{2}-7x+C$

## Initial-value problems

We look at techniques for integrating a large variety of functions involving products, quotients, and compositions later in the text. Here we turn to one common use for antiderivatives that arises often in many applications: solving differential equations.

A differential equation is an equation that relates an unknown function and one or more of its derivatives. The equation

$\frac{dy}{dx}=f\left(x\right)$

is a simple example of a differential equation. Solving this equation means finding a function $y$ with a derivative $f.$ Therefore, the solutions of [link] are the antiderivatives of $f.$ If $F$ is one antiderivative of $f,$ every function of the form $y=F\left(x\right)+C$ is a solution of that differential equation. For example, the solutions of

$\frac{dy}{dx}=6{x}^{2}$

are given by

$y=\int 6{x}^{2}dx=2{x}^{3}+C.$

Sometimes we are interested in determining whether a particular solution curve passes through a certain point $\left({x}_{0},{y}_{0}\right)$ —that is, $y\left({x}_{0}\right)={y}_{0}.$ The problem of finding a function $y$ that satisfies a differential equation

$\frac{dy}{dx}=f\left(x\right)$

$y\left({x}_{0}\right)={y}_{0}$

is an example of an initial-value problem . The condition $y\left({x}_{0}\right)={y}_{0}$ is known as an initial condition . For example, looking for a function $y$ that satisfies the differential equation

$\frac{dy}{dx}=6{x}^{2}$

and the initial condition

$y\left(1\right)=5$

is an example of an initial-value problem. Since the solutions of the differential equation are $y=2{x}^{3}+C,$ to find a function $y$ that also satisfies the initial condition, we need to find $C$ such that $y\left(1\right)=2{\left(1\right)}^{3}+C=5.$ From this equation, we see that $C=3,$ and we conclude that $y=2{x}^{3}+3$ is the solution of this initial-value problem as shown in the following graph.

## Solving an initial-value problem

Solve the initial-value problem

$\frac{dy}{dx}=\text{sin}\phantom{\rule{0.1em}{0ex}}x,y\left(0\right)=5.$

First we need to solve the differential equation. If $\frac{dy}{dx}=\text{sin}\phantom{\rule{0.1em}{0ex}}x,$ then

$y=\int \text{sin}\left(x\right)dx=\text{−}\text{cos}\phantom{\rule{0.1em}{0ex}}x+C.$

Next we need to look for a solution $y$ that satisfies the initial condition. The initial condition $y\left(0\right)=5$ means we need a constant $C$ such that $\text{−}\text{cos}\phantom{\rule{0.1em}{0ex}}x+C=5.$ Therefore,

$C=5+\text{cos}\left(0\right)=6.$

The solution of the initial-value problem is $y=\text{−}\text{cos}\phantom{\rule{0.1em}{0ex}}x+6.$

I don't understand the formula
who's formula
funny
What is a independent variable
a variable that does not depend on another.
Andrew
solve number one step by step
x-xcosx/sinsq.3x
Hasnain
x-xcosx/sin^23x
Hasnain
how to prove 1-sinx/cos x= cos x/-1+sin x?
1-sin x/cos x= cos x/-1+sin x
Rochel
how to prove 1-sun x/cos x= cos x / -1+sin x?
Rochel
how to prove tan^2 x=csc^2 x tan^2 x-1?
divide by tan^2 x giving 1=csc^2 x -1/tan^2 x, rewrite as: 1=1/sin^2 x -cos^2 x/sin^2 x, multiply by sin^2 x giving: sin^2 x=1-cos^2x. rewrite as the familiar sin^2 x + cos^2x=1 QED
Barnabas
how to prove sin x - sin x cos^2 x=sin^3x?
sin x - sin x cos^2 x sin x (1-cos^2 x) note the identity:sin^2 x + cos^2 x = 1 thus, sin^2 x = 1 - cos^2 x now substitute this into the above: sin x (sin^2 x), now multiply, yielding: sin^3 x Q.E.D.
Andrew
take sin x common. you are left with 1-cos^2x which is sin^2x. multiply back sinx and you get sin^3x.
navin
Left side=sinx-sinx cos^2x =sinx-sinx(1+sin^2x) =sinx-sinx+sin^3x =sin^3x thats proved.
Alif
how to prove tan^2 x/tan^2 x+1= sin^2 x
Rochel
Salim
what is function.
what is polynomial
Nawaz
an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s).
Alif
a term/algebraic expression raised to a non-negative integer power and a multiple of co-efficient,,,,,, T^n where n is a non-negative,,,,, 4x^2
joe
An expression in which power of all the variables are whole number . such as 2x+3 5 is also a polynomial of degree 0 and can be written as 5x^0
Nawaz
what is hyperbolic function
find volume of solid about y axis and y=x^3, x=0,y=1
3 pi/5
vector
what is the power rule
Is a rule used to find a derivative. For example the derivative of y(x)= a(x)^n is y'(x)= a*n*x^n-1.
Timothy
how do i deal with infinity in limits?
f(x)=7x-x g(x)=5-x
Awon
5x-5
Verna
what is domain
difference btwn domain co- domain and range
Cabdalla
x
Verna
The set of inputs of a function. x goes in the function, y comes out.
Verna
where u from verna
Arfan
If you differentiate then answer is not x
Raymond
domain is the set of values of independent variable and the range is the corresponding set of values of dependent variable
Champro
what is functions
give different types of functions.
Paul
how would u find slope of tangent line to its inverse function, if the equation is x^5+3x^3-4x-8 at the point(-8,1)
pls solve it i Want to see the answer
Sodiq
ok
Friendz
differentiate each term
Friendz