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I am having difficulty parsing this question and visualizing how the picture looks and what tools to use to solve

Question: A flat metal plate is in the shape determined by area by the area under the graph of $f(x) = \frac {1}{1+x}$ between $x = 0$ and $x = 5$. The density of the plate $x$ units from the y-axis is given by $x^2$ grams/cm$^2$. Write down a definite integral which gives the exact value of the total mass of the plate.

From what I can glean, it seems that the plate should rest horizontally on the x axis from zero to five. My first inclination was to find the center of mass, but this question is asking for me to find the total mass. What is the best approach to take with this question?

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In a one variable calculus course, I would say look at the strip of plate from $x$ to $x+\Delta x$. This is almost a rectangle, of area $(\Delta x)\frac{1}{x+1}$. In that region, density is about $x^2$, so mass of strip is about $x^2(\Delta x)\frac{1}{x+1}$. "Add up" (integrate) from $0$ to $5$. Mass is $\int_0^5 x^2\frac{1}{x+1}dx$. – André Nicolas Nov 19 '11 at 5:36
You were asking about the shape. Look at the region below $y=1/(x+1)$, above the $x$-axis, from $x=0$ to $x=5$. That's the shape of the metal plate. The plate is thin, so we are viewing it as being essentially two-dimensional. – André Nicolas Nov 19 '11 at 6:25
up vote 1 down vote accepted

A small square of plate with lower left corner $(x,y)$ and upper right corner $(x+\Delta x,y+\Delta y)$ will have a mass $x^2\Delta x \Delta y$. You need to add these up over the region of interest, giving $\int_0^5\int_0^{\frac{1}{1+x}}x^2\; dy \;dx$. The $y$ integral just gives the difference of the limits because nothing depends on $y$, so you have $\int_0^5\frac{1}{1+x}x^2 \;dx$

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Thanks. And then I must use integration by parts, correct? – Dylan Nov 19 '11 at 5:20
Or wait, substitution rule seems more apt for this... – Dylan Nov 19 '11 at 5:29
Integration by parts is not how I would tackle it, divide the polynomial $x^2$ by $x+1$. We get $x-1+\frac{1}{x+1}$, easy to integrate. – André Nicolas Nov 19 '11 at 5:30
@Andre: I am not sure how you divided the polynomial here... I was trying to use substitution but wasn't sure about what to do with the $x^2$. So far I have $u = 1 + x$ and $du = dx$. But again, I can't figure out what to do with $x^2$ ... – Dylan Nov 19 '11 at 6:15
Get someone to show you polynomial division. It's much like ordinary division. Here is a quick inadequate try at a description: $x+1$ "goes into $x^2$" $x$ times. So the first term of your quotient is $x$. Multiply $x+1$ by $x$. You get $x^2+x$. Subtract from $x^2$. You get $-x$. Now $x+1$ "goes into $-x$" $-1$ times. Multiply $x+1$ by $-1$. You get $-x-1$. Subtract from $-x$. We get $1$. So quotient is $x-1$, remainder is $1$, which gives the result. This stuff, like "long division," is very visual. You need pictures. Undoubtedly they are available on the Internet. Substitution next! – André Nicolas Nov 19 '11 at 6:34

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