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
$y={\displaystyle \int 6{x}^{2}dx=2{x}^{3}+C}.$
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
$\frac{dy}{dx}=f\left(x\right)$
with the additional condition
$y\left({x}_{0}\right)={y}_{0}$
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.
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
$\text{\u2212}\text{cos}\phantom{\rule{0.1em}{0ex}}x+C=5.$ Therefore,
$C=5+\text{cos}\left(0\right)=6.$
The solution of the initial-value problem is
$y=\text{\u2212}\text{cos}\phantom{\rule{0.1em}{0ex}}x+6.$
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?
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
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
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).
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