Question help please (1 Viewer)

fake43 · Oct 29, 2021

Answer is C but I'm getting pi..please help

I looked at the solution and im stuck up to here:

I don't get the next step after

5uckerberg · Oct 29, 2021

Okay, fake43, now this is what we call a dummy variable, a variable that I would call a "useful idiot" until you are done with it. Here y is the "useful idiot" or the dummy variable because the integral of ysiny is just like the integral of xsinx. So therefore you bring that to the other side. At this point you will have 2 times the original integral given= $\pi\sin{x}dx$ with the boundaries $[0,\pi]$ . The rest should be history. Often these questions you have to envision flipping the graph for the boundaries given.

Answer is C

CM_Tutor · Oct 29, 2021

$\begin{align*} \int_0^{\pi} x\sin{x}\ dx &= \int_{x=0}^{x=\pi} (\pi - y)\sin{(\pi - y)} \times -1\ dy \qquad \text{putting $x = \pi - y \implies dx = -dy$} \\ &= -\int_{\pi}^0 (\pi - y)\sin{(\pi - y)}\ dy \qquad \text{as $x = \pi - y$, when $x = 0,\ y = \pi$ and when $x = \pi,\ y = 0$} \\ &= \int_0^{\pi} (\pi - y)\sin{y}\ dy \qquad \text{as $\sin(\pi - \theta) = \sin\theta$} \\ &= \pi \int_0^{\pi} \sin{y}\ dy - \int_0^{\pi} y\sin{y}\ dy \\ \int_0^{\pi} x\sin{x}\ dx &= \pi \int_0^{\pi} \sin{x}\ dx - \int_0^{\pi} x\sin{x}\ dx \qquad \text{after recognising the dummy variable} \\ 2\int_0^{\pi} x\sin{x}\ dx &= \pi \int_0^{\pi} \sin{x}\ dx \\ \int_0^{\pi} x\sin{x}\ dx &= \cfrac{\pi}{2} \int_0^{\pi} \sin{x}\ dx \end{align*}$

which, if you keep integrating, will simplify to

$\int_0^{\pi} x\sin{x}\ dx = \pi$

Life'sHard · Oct 29, 2021

This is basically what the dummy variable is. So using this you can solve.

Screen Shot 2021-10-29 at 12.28.21 pm.png

fake43 · Oct 29, 2021

So, they're equal because the "~~equations~~ integrand are the same" and has the same limits, making them a dummy variable?

I get everything else but am not too sure if I'm correctly understanding why the two are equal/what is a dummy variable.

5uckerberg · Oct 29, 2021

fake43 said:
So, they're equal because the "equations are the same" and has the same limits, making them a dummy variable?

I get everything else but am not too sure if I'm correctly understanding why the two are equal/what is a dummy variable.

To put it simply a dummy variable is what we call a variable that serves as a placeholder.
The two are equal because simply put the letters are basically conveying the same message eg xsindx is basically integrating by a minuscule amount of x and then ysinydy is integrating by a minuscule amount of y and that x and y are like two football players who are subbing on and off for each other as y is the dummy variable.

Using that the two are equal and as such ysiny which is just xsinx in disguise can be moved to the LHS in your work and then integrate the leftover.

fake43 · Oct 29, 2021

5uckerberg said:
To put it simply a dummy variable is what we call a variable that serves as a placeholder.
The two are equal because simply put the letters are basically conveying the same message eg xsindx is basically integrating by a minuscule amount of x and then ysinydy is integrating by a minuscule amount of y and that x and y are like two football players who are subbing on and off for each other as y is the dummy variable.

Using that the two are equal and as such ysiny which is just xsinx in disguise can be moved to the LHS in your work and then integrate the leftover.

Thank you

CM_Tutor · Oct 29, 2021

Put another way... If:

$I = \int_0^{\pi} x\sin{x}\ dx$

then $I$ is the signed area enclosed by the curve $y = x\sin{x}$ , the $x$ -axis, and the boundaries $x = 0$ and $x = \pi$ .

Now, consider

$J = \int_0^{\pi} t\sin{t}\ dt$

$J$ is the signed area enclosed by the curve $y = t\sin{t}$ , the $t$ -axis, and the boundaries $t = 0$ and $t = \pi$

...which, if you think about it, has to be $I$ , because both integrals are representing exactly the same region, thus $I = J$ .

fake43 · Oct 29, 2021

CM_Tutor said:
Put another way... If:

$I = \int_0^{\pi} x\sin{x}\ dx$

then $I$ is the signed area enclosed by the curve $y = x\sin{x}$ , the $x$ -axis, and the boundaries $x = 0$ and $x = \pi$ .

Now, consider

$J = \int_0^{\pi} t\sin{t}\ dt$

$J$ is the signed area enclosed by the curve $y = t\sin{t}$ , the $t$ -axis, and the boundaries $t = 0$ and $t = \pi$

...which, if you think about it, has to be $I$ , because both integrals are representing exactly the same region, thus $I = J$ .

Ooh thanks for this

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