Prove definite-integrals

Question

So...Try it again. The last time I did task:

Prove equation

$\int_{a}^{b}f(x)dx = \int_{a}^{b}f(a+b-x)dx$

$a,b$ $are$ $ constants$

Solution:

Left part of equation $\int_{a}^{b}f(x)dx = F(b) - F(a)$
Right part of equation $\int_{a}^{b}f(a+b-x)dx = -\int_{a}^{b}f(a+b-x)d(a+b-x)$ then $-(F(a+b-b) - F(a+b-a))$ and result $-(F(a)-F(b)) = F(b)-F(a)$
$F(b) - F(a) = F(b) - F(a)$

We see the same asnwer and proved it.

Now I have new task. Prove equation:

$\int_{0}^{a}x^3f(x^2)dx = \frac12\int_{0}^{a^2}xf(x)dx $

I don't know how to prove it. I hope my question is correct right now.

I'm no expert but I don't think $d(a+b-x)$ makes sense for notation, and especially since $a$ and $b$ are constant (I assume)? — m1cky22, Aug 08 '16 at 16:15

Vishal Gupta · Answer 1 · 2016-08-08T16:28:33.150

1

Use the substitution $\mathrm{u=x^{2}}$. Also, $\mathrm{d(a+b-x)}$ does make sense as notation, the integrator is a+b-x. $\mathrm{d(a+b-x)=d(a)+d(b)-d(x)=-d(x)}$ as a and b are constants.

edited Aug 08 '16 at 16:28

answered Aug 08 '16 at 16:26

Vishal Gupta

83
6

$\mathrm{du=2xdx}$ so $\mathrm{udu=2x^3dx}$ – Vishal Gupta Aug 08 '16 at 16:31
The limits are $\mathrm{0^2=0}$ and $\mathrm{a^2}$. After the substitution, change the dummy variable of integration back to x to get the result you posted above. – Vishal Gupta Aug 08 '16 at 16:35

John Wayland Bales · Accepted Answer · 2016-08-08T16:42:06.100

0

Let $u=a+b-x$. then $dx=-du$ and \begin{eqnarray} \int_a^bf(x)\,dx&=\int_{(a+b)-a}^{(a+b)-b}f(a+b-u)(-du)\\&=-\int_b^af(a+b-u)\,du\\&=\int_a^bf(a+b-u)\,du \end{eqnarray}

Then let $u=x$ and you're done with the first.

For the second, let $u=x^2$. Then $du=2xdx$, so

\begin{eqnarray} \int_0^ax^3f(x^2)\,dx&=\frac{1}{2}\int_0^ax^2f(x^2)(2xdx)\\&=\frac{1}{2}\int_0^{a^2}uf(u)\,du \end{eqnarray}

Then substitute $x=u$ and you are finished.

edited Aug 08 '16 at 16:42

answered Aug 08 '16 at 16:30

John Wayland Bales

21,814

The last expression: Why $a^2 = a$ ? – Boris Molostvov Aug 08 '16 at 16:48
The last integral is in terms of $u$ which equals $x^2$ so $x=a$ becomes $u=a^2$. – John Wayland Bales Aug 08 '16 at 17:11
The limits on integrals are actually equations. $\displaystyle\int_a^bf(x),dx$ actually means $\displaystyle\int_{x=a}^{x=b}f(x),dx$. – John Wayland Bales Aug 08 '16 at 17:12
Yes, you're right. But, can we use newton leibnitz theorem in the last integral? – Boris Molostvov Aug 08 '16 at 17:31
Yes, for some function $y=G(x)$ satisfying $\frac{d}{dx}G(x)=xf(x)$ it is true that $\displaystyle\int_0^{a^2}xf(x),dx=G(a^2)-G(0)$. – John Wayland Bales Aug 08 '16 at 17:40
Okay, I see. But $\int_{0}^{a^2}xf(x)dx $ I think its not simple integral – Boris Molostvov Aug 08 '16 at 17:53

score 0 · Answer 3 · answered Aug 08 '16 at 16:35

Hint:

Simply use substitution of variables, and find the new limits of integration.

For the first use the substitution: $$ a+b-x=t $$ so that: $$ x=a \iff t=b \qquad;\qquad x=b \iff t=a $$ and $$ -dx=dt $$ so the integral becomes: $$ \int_a^bf(a+b-x)dx=-\int_b^af(t)dt=\int_a^b f(t)dt $$

for the second integral use the substitution: $$ x^2=t $$ that gives $$ x=0 \iff t=0 \qquad x=a\iff t=a^2 $$ and $$ 2xdx=dt$$

Can you do from this?

score 0 · Answer 4 · answered Aug 08 '16 at 17:06

0

Here is a solution that relies on the Leibniz rule instead of substitution:

Let $\phi(t) = \int_a^t f(a+t-x) dx -\int_a^t f(x)dx $. Then $\phi(a)=0$, and $\phi'(t) = f(a) + \int_a^t f'(a+t-x) dx - f(t) = f(a)+f(t)-f(a) -f(t)= 0$, and so $\phi = 0$.

Let $\eta(t) = \int_{0}^{t}x^3f(x^2)dx - \frac12\int_{0}^{t^2}xf(x)dx$, we have $\eta(0) = 0$ and $\eta'(t) = t^3f(t^2)-{1 \over 2} (t^2 f(t^2) 2t) = 0$ and so $\eta = 0$.

answered Aug 08 '16 at 17:06

copper.hat

172,524

Well..its correct. But i don't think that it's true way. But thank you! – Boris Molostvov Aug 08 '16 at 17:10
@BorisMolostvov: It is good to have alternative approaches :-) – copper.hat Aug 08 '16 at 17:28

Prove definite-integrals

4 Answers4