# Find the derivative of given function and hence find its integral

• chwala

#### chwala

Gold Member
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##

Last edited:
$$\int_1^2 LHS\ dx = \int_1^2 RHS\ dx$$

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 !

##\ ##

chwala
For fun and profit and as a nice extension of this exercise (but I understand those who consider it showing off ) :

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## ...

##\ ##

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##
=
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...

##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

##\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##.

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

chwala
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##