Find the derivative of given function and hence find its integral

In summary, we discussed the differentiation of the function y=x^2ln x-x, which resulted in the equation dy/dx=2x ln x+x-1. We then proceeded to integrate the function by breaking it down into simpler terms and using the substitution method, ultimately arriving at the solution of x^2ln x-x. This was further verified by a different approach, which resulted in the same solution.
  • #1
chwala
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Homework Statement
find the derivative of ##y=x^2ln x-x## hence evaluate ##\int_1^2 xln x \ dx##
Relevant Equations
differentiation
##y=x^2ln x-x##
##\frac {dy}{dx}=2x ln x+x-1##
##\int [2xln x+x-1]\,dx##=##x^2ln x-x##, since ##\int -1 dx= -x##
it follows that,
##\int [2x ln x +x]\,dx##=##x^2 ln x##
##\int 2x ln x \,dx = x^2ln x##+##\int x\,dx##
##\int_1^2 xln x\,dx =\frac {x^2ln x}{2}##+##\frac{x^2}{4}##=##2ln2+1-0.25##
 
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  • #2
So from your last formula
[tex]\int_1^2 LHS\ dx = \int_1^2 RHS\ dx[/tex]
 
  • #3
Hello Chwala,

$$\frac {dy}{dx}=2x \ln x+x-1$$was correct. I think you had an equation following that looking like $$x \ln x = {1\over 2} \frac {dy}{dx} - {x\over 2} + {1\over 2} $$which was correct as well. And prompted @mitochan to 'hint' $$\int_1^2 LHS\ dx = \int_1^2 RHS\ dx\ .$$ As the exercise text suggests.

Your $$\int [2x\ln x+x]\,dx =x^2\ln x$$ is still fine.

But then you fall into a trap. You want to subtract ##\int x## on both sides, so you should get $$\int [2x\ln x]\,dx =x^2\ln x- \int x$$

Other than that, bravo !

##\ ##
 
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  • #4
For fun and profit and as a nice extension of this exercise (but I understand those who consider it showing off :wink: ) :

Note that you could have done a coarse check on the answer with the help of mr. Taylor:

The Taylor series of ##\ln(1+\epsilon)## is pretty common knowledge:$$
\ln(1+\epsilon) = \epsilon - {\epsilon^2\over 2} + {\epsilon^3\over 3} - \ ... $$ so in the range ##\ [1,2]\ ## we have for ##\ \epsilon = x-1 \ ## that ##\ x\ln x < x(x-1) \ ## and therefore $$
\int_1^2 x\ln x \;dx < \int_1^2 x(x-1) \;dx =
\tfrac{1}{3} x^3- \tfrac {1}{2} x^2 \; \Bigg |_1^2 = {7\over 3 } - {3\over 2} = {5\over 6} $$ whereas your answer is clearly ##>1## ...

1614874673850.png

##\ ##
 
  • #5
BvU said:
Hello Chwala,

$$\frac {dy}{dx}=2x \ln x+x-1$$was correct. I think you had an equation following that looking like $$x \ln x = {1\over 2} \frac {dy}{dx} - {x\over 2} + {1\over 2} $$which was correct as well. And prompted @mitochan to 'hint' $$\int_1^2 LHS\ dx = \int_1^2 RHS\ dx\ .$$ As the exercise text suggests.

Your $$\int [2x\ln x+x]\,dx =x^2\ln x$$ is still fine.

But then you fall into a trap. You want to subtract ##\int x## on both sides, so you should get $$\int [2x\ln x]\,dx =x^2\ln x- \int x$$

Other than that, bravo !

##\ ##

aaargh! i see that, $$x \ln x = {1\over 2} \frac {dy}{dx} - {x\over 2} + {1\over 2} $$
$$\int_1^2 LHS\ dx = \int_1^2 RHS\ dx\ .$$
##\int_1^2 xln x \,dx=\int_1^2 [\frac {dy}{dx} - {x\over 2} + {1\over 2}] dx##
=
BvU said:
Hello Chwala,

$$\frac {dy}{dx}=2x \ln x+x-1$$was correct. I think you had an equation following that looking like $$x \ln x = {1\over 2} \frac {dy}{dx} - {x\over 2} + {1\over 2} $$which was correct as well. And prompted @mitochan to 'hint' $$\int_1^2 LHS\ dx = \int_1^2 RHS\ dx\ .$$ As the exercise text suggests.

Your $$\int [2x\ln x+x]\,dx =x^2\ln x$$ is still fine.

But then you fall into a trap. You want to subtract ##\int x## on both sides, so you should get $$\int [2x\ln x]\,dx =x^2\ln x- \int x$$

Other than that, bravo !

##\ ##
let me correct that...
 
  • #6
##y=x^2ln x-x##
##\frac {dy}{dx}=2x ln x+x-1##
##\int [2xln x+x-1]\,dx##=##x^2ln x-x##, since ##\int -1 dx= -x##
it follows that,
##\int [2x ln x +x]\,dx##=##x^2 ln x##
##\int 2x ln x \,dx = x^2ln x##-##\int x\,dx##
##\int_1^2 xln x\,dx =\frac {x^2ln x}{2}##-##\frac{x^2}{4}##=##2ln2-1+0.25##=##2ln2-0.75##...1

alternatively as you had suggested,

##\frac {dy}{dx}-x +1=2x ln x##
##\frac {1}{2}\frac {dy}{dx}-\frac {1}{2}x +\frac {1}{2}=xln x##
##\int x ln x\,dx##=## \int [ \frac {1}{2}\frac {dy}{dx}-\frac {1}{2}x +\frac {1}{2}\,]dx##
##\int_1^2 x ln x\,dx##=## \int_1^2 [ \frac {1}{2}\frac {dy}{dx}-\frac {1}{2}x +\frac {1}{2}\,]dx##
##\int_1^2 x ln x\,dx##=## \frac {1}{2}y##-##\frac {1}{4}x^2## +##\frac {1}{2}x## with limits ##x=1## and ##x=2## in mind, it follows that,
##\int_1^2 x ln x\,dx##=## \frac {1}{2}[x^2ln x-x]##-##\frac {1}{4}x^2## +##\frac {1}{2}x##
##\int_1^2 x ln x\,dx##=##\frac {1}{2}x^2ln x##-##\frac {1}{4}x^2##...which will realize the same result as equation 1 above by substitution of limits ##x=1## and ##x=2##.
 
  • #7
chwala said:
##y=x^2ln x-x##
##\frac {dy}{dx}=2x ln x+x-1##
##\int [2xln x+x-1]\,dx=x^2ln x-x## .
Since ##\int -1 dx= -x## ,
it follows that, ##\int [2x ln x +x]\,dx=x^2 ln x##
##\int 2x ln x \,dx = x^2ln x##-##\int x\,dx##
Good. (Of course there is also a Constant of Integration. But that loses importance for the definite integral.)

A slight tweak to the above method.

By doing the suggested differentiation you have established that ##\displaystyle \int (2x\ln x+x-1)\,dx=x^2\ln x-x ## .

So ##\displaystyle \int 2x\ln x\,dx##

##\quad \quad = \displaystyle \int (2x\ln x+(x-1)-(x-1))\,dx##

##\quad \quad = \displaystyle \int (2x\ln x+x-1)\, dx + \int -(x-1)\,dx##

##\quad \quad = \displaystyle x^2 \ln (x)-x - \frac{x^2}{2} +x + C##

##\quad \quad = \displaystyle x^2 \ln (x) - \frac{x^2}{2} + C##
 
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  • #8
SammyS said:
Good. (Of course there is also a Constant of Integration. But that loses importance for the definite integral.)

A slight tweak to the above method.

By doing the suggested differentiation you have established that ##\displaystyle \int (2x\ln x+x-1)\,dx=x^2\ln x-x ## .

So ##\displaystyle \int 2x\ln x\,dx##

##\quad \quad = \displaystyle \int (2x\ln x+(x-1)-(x-1))\,dx##

##\quad \quad = \displaystyle \int (2x\ln x+x-1)\, dx + \int -(x-1)\,dx##

##\quad \quad = \displaystyle x^2 \ln (x)-x - \frac{x^2}{2} +x + C##

##\quad \quad = \displaystyle x^2 \ln (x) - \frac{x^2}{2} + C##

good one Sammy! Brilliant mate!:cool:
 

FAQ: Find the derivative of given function and hence find its integral

What is the definition of a derivative?

The derivative of a function is the rate of change of the function at a specific point. It represents the slope of the tangent line to the function at that point.

How do you find the derivative of a given function?

To find the derivative of a function, you use the rules of differentiation such as the power rule, product rule, quotient rule, and chain rule. These rules allow you to find the derivative of any function, as long as it is continuous and differentiable.

Why is finding the derivative important?

Finding the derivative is important because it helps us understand the behavior of a function and its rate of change. It is also used in many real-life applications, such as calculating velocity and acceleration, optimizing functions, and solving differential equations.

What is the relationship between a derivative and an integral?

The derivative and integral are inverse operations of each other. This means that the integral of a function is the reverse process of finding its derivative. The integral represents the area under the curve of a function, while the derivative represents the slope of the curve at a specific point.

How do you find the integral of a given function?

To find the integral of a function, you use the rules of integration such as the power rule, substitution rule, and integration by parts. These rules allow you to find the antiderivative of a function, which is the function that, when differentiated, gives the original function.

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