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Evaluating indefinite integrals

Evaluate each of the following indefinite integrals:

  1. ( 5 x 3 7 x 2 + 3 x + 4 ) d x
  2. x 2 + 4 x 3 x d x
  3. 4 1 + x 2 d x
  4. tan x cos x d x
  1. Using [link] , we can integrate each of the four terms in the integrand separately. We obtain
    ( 5 x 3 7 x 2 + 3 x + 4 ) d x = 5 x 3 d x 7 x 2 d x + 3 x d x + 4 d x .

    From the second part of [link] , each coefficient can be written in front of the integral sign, which gives
    5 x 3 d x 7 x 2 d x + 3 x d x + 4 d x = 5 x 3 d x 7 x 2 d x + 3 x d x + 4 1 d x .
    Using the power rule for integrals, we conclude that
    ( 5 x 3 7 x 2 + 3 x + 4 ) d x = 5 4 x 4 7 3 x 3 + 3 2 x 2 + 4 x + C .
  2. Rewrite the integrand as
    x 2 + 4 x 3 x = x 2 x + 4 x 3 x = 0 .

    Then, to evaluate the integral, integrate each of these terms separately. Using the power rule, we have
    ( x + 4 x 2 / 3 ) d x = x d x + 4 x −2 / 3 d x = 1 2 x 2 + 4 1 ( −2 3 ) + 1 x ( −2 / 3 ) + 1 + C = 1 2 x 2 + 12 x 1 / 3 + C .
  3. Using [link] , write the integral as
    4 1 1 + x 2 d x .

    Then, use the fact that tan −1 ( x ) is an antiderivative of 1 ( 1 + x 2 ) to conclude that
    4 1 + x 2 d x = 4 tan −1 ( x ) + C .
  4. Rewrite the integrand as
    tan x cos x = sin x cos x cos x = sin x .

    Therefore,
    tan x cos x = sin x = cos x + C .
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Evaluate ( 4 x 3 5 x 2 + x 7 ) d x .

x 4 5 3 x 3 + 1 2 x 2 7 x + C

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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

d y d x = f ( x )

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 ( x ) + C is a solution of that differential equation. For example, the solutions of

d y d x = 6 x 2

are given by

y = 6 x 2 d x = 2 x 3 + C .

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

d y d x = f ( x )

with the additional condition

y ( x 0 ) = y 0

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

d y d x = 6 x 2

and the initial condition

y ( 1 ) = 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 ( 1 ) = 2 ( 1 ) 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.

The graphs for y = 2x3 + 6, y = 2x3 + 3, y = 2x3, and y = 2x3 − 3 are shown.
Some of the solution curves of the differential equation d y d x = 6 x 2 are displayed. The function y = 2 x 3 + 3 satisfies the differential equation and the initial condition y ( 1 ) = 5 .

Solving an initial-value problem

Solve the initial-value problem

d y d x = sin x , y ( 0 ) = 5 .

First we need to solve the differential equation. If d y d x = sin x , then

y = sin ( x ) d x = cos x + C .

Next we need to look for a solution y that satisfies the initial condition. The initial condition y ( 0 ) = 5 means we need a constant C such that cos x + C = 5 . Therefore,

C = 5 + cos ( 0 ) = 6 .

The solution of the initial-value problem is y = cos x + 6 .

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Questions & Answers

I need help under implicit differentiation
Uchenna Reply
What is derivative of antilog x dx ?
Tanmay Reply
what's the meaning of removable discontinuity
Brian Reply
what's continuous
Brian
an area under a curve is continuous because you are looking at an area that covers a range of numbers, it is over an interval, such as 0 to 4
Lauren
using product rule x^3,x^5
Kabiru
please help me to calculus
World Reply
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Sunny
Luke 17:21 nor will they say, See here or See there For indeed, the kingdom of God is within you. You've never 'touched' anything. The e-energy field created by your body has pushed other electricfields. even our religions tell us we're the gods. We live in energies connecting us all. Doa/higgsfield
Scott
if you have any calculus questions many of us would be happy to help and you can always learn or even invent your own theories and proofs. math is the laws of logic and reality. its rules are permanent and absolute. you can absolutely learn calculus and through it better understand our existence.
Scott
ya doubtless
Bilal
help the integral of x^2/lnxdx
Levis
also find the value of "X" from the equation that follow (x-1/x)^4 +4(x^2-1/x^2) -6=0 please guy help
Levis
Use integration by parts. Let u=lnx and dv=x2dx Then du=1xdx and v=13x3. ∫x2lnxdx=13x3lnx−∫(13x3⋅1x)dx ∫x2lnxdx=13x3lnx−∫13x2dx ∫x2lnxdx=13x3lnx−19x3+C
Bilal
itz 1/3 and 1/9
Bilal
now you can find the value of X from the above equation easily
Bilal
Pls i need more explanation on this calculus
usman
usman from where do you need help?
Levis
thanks Bilal
Levis
integrate e^cosx
Uchenna
-sinx e^x
Leo
Do we ask only math question? or ANY of the question?
Levis Reply
yh
Gbesemete
How do i differentiate between substitution method, partial fraction and algebraic function in integration?
usman
you just have to recognize the problem. there can be multiple ways to solve 1 problem. that's the hardest part about integration
Lauren
test
MOHAMMAD
we asking the question cause only the question will tell us the right answer
Sunny
find integral of sin8xcos12xdx
Levis Reply
don't share these childish questions
Bilal
well find the integral of x^x
Levis
bilal kumhar you are so biased if you are an expert what are you doing here lol😎😎😂😂 we are here to learn and beside there are many questions on this chat which you didn't attempt we are helping each other stop being naive and arrogance so give me the integral of x^x
Levis
Levis I am sorry
Bilal
Bilal it okay buddy honestly i am pleasured to meet you
Levis
x^x ... no anti derivative for this function... but we can find definte integral numerically.
Bilal
thank you Bilal Kumhar then how we may find definite integral let say x^x,3,5?
Levis
evaluate 5-×square divided by x+2 find x as limit approaches infinity
Michagaye Reply
i have not understood
Leo
The answer is 0
Michael
welcome
Sunny
I just dont get it at all...not understanding
Michagaye
0 baby
Sunny
The denominator is the aggressive one
Sunny
wouldn't be any prime number for x instead ?
Harold
or should I say any prime number greater then 11 ?
Harold
just wondering
Harold
I think as limit Approach infinity then X=0
Levis
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No Reply
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Leonito
Fond the value of the six trigonometric function of an angle theta, which terminal side passes through the points(2x½-y)²,4
albert Reply
What's f(x) ^x^x
Emeka Reply
What's F(x) =x^x^x
Emeka
are you asking for the derivative
Leo
that's means more power for all points
rd
if your asking for derivative dy/dz=x^2/2(lnx-1/2)
Levis
iam sorry f(x)=x^x it means the output(range ) depends to input(domain) value of x by the power of x that is to say if x=2 then x^x would be 2^2=4 f(x) is the product of X to the power of X its derivatives is found by using product rule y=x^x introduce ln each side we have lny=lnx^x =lny=xlnx
Levis
the derivatives of f(x)=x^x IS (1+lnx)*x^x
Levis
what is a maximax
Chinye Reply
A maxima in a curve refers to the maximum point said curve. The maxima is a point where the gradient of the curve is equal to 0 (dy/dx = 0) and its second derivative value is a negative (d²y/dx² = -ve).
Viewer
what is the limit of x^2+x when x approaches 0
Dike Reply
it is 0 because 0 squared Is 0
Leo
0+0=0
Leo
simply put the value of 0 in places of x.....
Tonu
the limit is 2x + 1
Nicholas
the limit is 0
Muzamil
limit s x
Bilal
The limit is 3
Levis
Leo we don't just do like that buddy!!! use first principle y+∆y=x+∆x ∆y=x+∆x-y ∆y=(x+∆x)^2+(x+∆x)-x^2+x on solving it become ∆y=3∆x+∆x^2 as ∆x_>0 limit=3 if you do by calculator say plugging any value of x=0.000005 which approach 0 you get 3
Levis
find derivatives 3√x²+√3x²
Care Reply
3 + 3=6
mujahid
How to do basic integrals
dondi Reply
the formula is simple x^n+1/n+1 where n IS NOT EQUAL TO 1 And n stands for power eg integral of x^2 x^2+1/2+1 =X^3/3
Levis
write something lmit
ram Reply
Practice Key Terms 3

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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