Find the derivative of given function and hence find its integral

[chwala]

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,
$\int [2x ln x +x]\,dx$=$x^2 ln x$
$\int 2x ln x \,dx = x^2ln x$+$\int x\,dx$
$\int_1^2 xln x\,dx =\frac {x^2ln x}{2}$+$\frac{x^2}{4}$=$2ln2+1-0.25$

 

[mitochan]

So from your last formula
\int_1^2 LHS\ dx = \int_1^2 RHS\ dx

 

[BvU]

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 !

$\$

 

[BvU]

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 $>1$ ...

$\$

 

[chwala]

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...

 

[chwala]

$y=x^2ln x-x$
$\frac {dy}{dx}=2x ln x+x-1$
$\int [2xln x+x-1]\,dx$=$x^2ln x-x$, 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 $x=2$.

 

[SammyS]

$y=x^2ln x-x$
$\frac {dy}{dx}=2x ln x+x-1$
$\int [2xln x+x-1]\,dx=x^2ln x-x$ .
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 $\displaystyle \int 2x\ln x\,dx$

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

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

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

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

 

[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 $\displaystyle \int 2x\ln x\,dx$

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

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

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

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

good one Sammy! Brilliant mate!